Solutions for Chapter 2 End-of-Chapter Problems Problem 2.1. Student answers will vary. This is an example. Use a pipette to collect samples from various places throughout the mixture. Analyze each sample for properties like density, boiling point, freezing point, or specific gravity. Compare the results to each other. If the solution is homogeneous all the results should be the same (within the experimental uncertainty). Problem 2.2. Some examples of homogeneous solutions you might find at the supermarket or a pharmacy might be: soft drinks, saline solutions for contact lenses, rubbing alcohol, liquid bleach, and vinegar. Some examples of heterogeneous solutions (suspensions) might be: milk, Milk of Magnesia, Peptobismol, calamine lotion, and paint. Problem 2.3. (a) An energy diagram for the change of a liquid (water) going to a gas is shown in Figure 1.35 and this problem is here to help recall the earlier introduction. The energy diagram for this general case with the energy change labeled is:

(b) For gaseous solvent molecules solvating a solute to form a liquid solution, we can think about the reaction occurring two steps, first condensing the gas to a liquid and then dissolving the solute in the liquid. The first step is exothermic, because condensing a gas to a liquid always releases energy. The second step, dissolution of the solute, may be either exothermic or endothermic, although not endothermic enough to make the overall two-step process endothermic, because the energy released in condensation of solvents is quite substantial. Thus, the overall process will be exothermic, but the solvation step may by endothermic or exothermic. If solvation by the liquid solvent is endothermic, we have:

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ACS Chemistry Chapter 2 suggested solutions

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If solvation by the liquid solvent is exothermic, we have:

Problem 2.4. (a) The solution feels cool to the touch because the process is drawing needed thermal energy from your hand, providing the energy needed for the overall solution process. This is a sketch of the energy diagram showing this endothermic solution process:

(b) The solution feels warm to the touch because the process is adding thermal energy to your hand, because the overall solution process releases energy. This is a sketch of the energy diagram showing this exothermic solution process:

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ACS Chemistry Chapter 2 suggested solutions

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Problem 2.5. The energy diagram for the overall solution process for the exothermic dissolution of calcium chloride, CaCl2, in water is:

Problem 2.6. Whether you can predict that a substance will be soluble in water by looking at its line formula depends to some extent on how the formula is actually written. For example, the line formula C2H6O, does not tell you how the atoms are connected, so the best you can do is predict that the molecule will probably be polar (because of the oxygen bonded to other less electronegative atoms) and, since it is a small molecule is likely to be reasonably soluble in water. However, if the line formula is written as C2H5OH, you can predict that this low molecular mass alcohol should be very soluble (actually it is miscible – mixes in any ratio) in water. If the line formula is written as CH3OCH3, you can predict that this low molecular mass ether should be somewhat soluble in water, but probably not as soluble as the alcohol (which can act as donor of an H to form H bonds as well as an acceptor of H from water to form other H bonds. A structural formula will always show the connectivity of the atoms and will, therefore, always provide the kind of information we just discussed for simple line formulas written to help us understand the connectivity. The regions of more positive and more negative charge in the molecule will be relatively easy to locate and an estimate of the polar and non-polar contributions to the solubility (or insolubility) will help you predict whether the substance will dissolve in water or not. Problem 2.7. The like-dissolves-like expression reflects the fact that attractions between solute molecules and some attractions between solvent molecules must be replaced by solute-solvent attractions when a solution forms. If the new attractions are similar to those replaced, we expect that a solution will be more easily formed. A polar liquid, such as water, is generally the best solvent for polar compounds, especially those with H bonding sites. Non-polar liquids, such as hexane are better solvents for non-polar compounds like, for example, wax, whose molecules are mostly attracted by dispersion forces.

ACS Chemistry Chapter 2 suggested solutions

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Problem 2.8. (a) Two representations of the testosterone structure are:

The dipoles are easier to show on the structure on the left (with fewer atoms shown explicitly):

(b) The acetylsalicylic acid (aspirin) structure, with polar bonds circled and direction of dipoles shown, is:

(c) The methyl salicylate (oil of wintergreen) structure, with polar bonds circled and direction of dipoles shown, is:

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ACS Chemistry Chapter 2 suggested solutions

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Problem 2.9. (a) The dashed lines in this diagram represent the network of hydrogen bonds that can form between ethanol and water.

Note that there are two H bonds between H atoms in water molecules and nonbonding electron pairs on the oxygen of ethanol, and another H bond between the –OH group’s hydrogen and a nonbonding electron pair on the oxygen in water. The water molecules have other nonbonding pairs on oxygen and covalently bonded hydrogen, all of which are capable of extending the network of hydrogen bonds. In Chapter 1, we found that there are, on the average, fewer than four H bonds per water molecule in liquid water. The three H bonds formed by ethanol fits right into this structure and the rather small ethyl group, –CH2CH3, probably does not perturb the structure very much, so ethanol molecules fit well into the liquid water structure and thus account for their miscibility with water. (b) The alcohol group in pentanol, CH3CH2CH2CH2CH2OH, can also H bond with water, as shown for ethanol in part (a). However, the solubility of pentanol is expected to be a good deal lower than the solubility of ethanol because the nonpolar hydrocarbon part of the molecule is larger and has a much larger effect on the structure of the liquid water. The solubility of pentanol in water is about 0.3 M (about 27 g·L–1). Problem 2.10. (a) Cyclohexane, C6H12, is a nonpolar liquid while methanol, CH3OH, is a polar liquid. The interactions among cyclohexane molecules in the liquid are induced dipole attractions (dispersion forces). For methanol, the largest interactions among the molecules in the liquid are H bonds; methanol acts much like water as a solvent. In a mixture of methanol and cyclohexane, the molecules of the two liquids lose some of their freedom of movement to make way for each other. This type of reorganization of “unlike” molecules is unfavorable and limits their mutual solubility, so they are not miscible. They are, however, rather soluble in one another: 100 mL of methanol dissolves 57 g of cyclohexane. Interestingly, ethanol (with its slightly larger alkyl group) and cyclohexane are miscible with one another, so we have to be careful not to push the “like dissolves like” (or “unlikes do not dissolve”) simplistic idea to extremes. (b) Naphthalene, C10H8, is a nonpolar solid and water is a polar, hydrogen-bonding liquid. The reorganization of the H-bonding structure of liquid water in order to accommodate naphthalene molecules is unfavorable. In this case water molecules would lose some of their freedom of movement while making way for naphthalene molecules. (c) Both naphthalene, C10H8(s), and benzene, C6H6(l), are nonpolar molecules. The reorganization involved in mixing two nonpolar compounds, both of which interact mainly by

ACS Chemistry Chapter 2 suggested solutions

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induced dipole attractions (dispersion forces), favors the mixed state, so the solid naphthalene dissolves in the liquid benzene. (d) Molecules of water can form hydrogen bonds with 1-propanol, CH3CH2CH2OH, like those shown for methanol in the solution to Problem 2.9(a). The nonpolar part of the molecule, like those of methanol and ethanol, apparently does not disturb the liquid water structure enough to make the interactions unfavorable. Thus mixing of the two liquids, water and 1-propanol, is not impaired and they mix in all proportions (are miscible). Problem 2.11. Since gasoline, C8H18, is a nonpolar molecule, we predict that it will be pretty insoluble in water. Its interactions with water molecules will be unfavorable as the freedom of movement of the water molecules is impaired. Problem 2.12. (a) The Lewis structures for 1-hexanol, CH3(CH2)5OH, and 1,6-hexanediol, HO(CH2)6OH, with their regions for hydrogen bonding identified, are:

1-hexanol

1,6–hexanediol, HO(CH2)6OH (b) 1,6-hexanediol is predicted to be more soluble in water than 1-hexanol, because compounds with multiple polar groups present more opportunity for hydrogen bonding with water molecules. Problem 2.13. Recall (from Section 2.2) that, for alcohols, the solubilizing and hydrogen bonding effects of the polar hydroxy (–OH) group becomes less and less important as the hydrocarbon portion of the alcohol increases in size. To a first approximation, the solubility is related directly to the ratio of carbon atoms to polar groups. In part (a), the polar group is an alcohol and in part (b), the polar group is an amine, –NH2, which can also hydrogen bond with water. The ratios and rank orders of solubilities are shown (#1 is the most soluble) here: carbon/polar grp ratio solubility rank (a) CH3-CH2-CH2-CH2-OH 4/1 3 CH3-CH2-CH2-CH2-CH2-OH 5/1 4 4/2 = 2/1 2 HO-CH2-CH2-CH2-CH2-OH HO- CH2-CH2-CH2-OH 3/2 = 1.5/1 1

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ACS Chemistry Chapter 2 suggested solutions

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Aqueous Solutions and Solubility carbon/polar grp ratio

solubility rank

4/1 5/1 6/1 8/1

1 2 3 4

(b) CH3CH2CH2CH2NH2 (CH3)2CHCH2CH2NH2 (CH3)3CCH2CH2NH2 (CH3 CH2)3CCH2NH2

Problem 2.14. [There are many books (especially beginning biochemistry texts) and web resources that have the structures of these sugars. We are not interested here in the exact stereochemistry of these molecules (for which most students are unprepared at this stage), but in the connectivity that shows the several alcohol groups in each one. The structures here are simply modeled after that for glucose in Figure 2.6 without showing the nonbonding electrons.] The structure of fructose might be found in either the pyranose (six-membered ring) or furanose (five-membered ring) form: CH2OH HO HO C O H C C H H C C HO OH H H

HOJ2C O CH2OH C H HO C C C HO OH H H

In both structures, there are five alcohol (–OH) groups that can H bond with water, as well as a ring oxygen atom with nonbonding electrons that can also donate electron pairs to H bonds with water. Looking, as in the solution to Problem 2.13, at the ratio of number of carbons to number of alcohol groups, we have 6/5 = 1.2/1. This ratio is comparable to that for methanol (1/1) and ethanol (2/1), both of which are miscible with water. The high solubility of fructose is expected. The structures of lactose and sucrose are a bit more complicated, because they are disaccharides combinations of two simpler sugars, two glucose molecules in lactose and glucose and fructose in sucrose: CH2 OH CH2 OH H H C O OH HO C O O C C C C H C C HH C C H HO H OHH HO H OHH

lactose

CH2 OH H CH 2OH HO C O O C O CH2 OH C C C H C C H C H C HO H OHH H OH OH H

sucrose

Lactose and sucrose both have eight alcohol (–OH) groups that can H bond with water, as well as the two ring oxygen atoms and the oxygen atom bonding the rings together which have nonbonding electrons that can also donate electron pairs to H bonds with water. (One alcohol group from each of the simple sugars has been lost in forming the disaccharides via a reaction we can write as R1–OH + R2–OH R1–O–R2 + H2O.) Looking again at the ratio of number of carbons to number of alcohol groups, we have 12/8 = 1.5/1. Again, this ratio is comparable to that for the simplest alcohols and helps to explain the high solubility of these sugars.

ACS Chemistry Chapter 2 suggested solutions

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Problem 2.15. The more polar groups present in a molecule the more soluble it will be in water. Conversely, the fewer the polar groups and the larger the nonpolar portion of a molecule the less soluble it will be in water and the more soluble it will be in nonpolar solvents. H H H H H H H H H H H H C H C H H H C C H H C C C C C C C C C C C C O H H

H

C

C

C

H H

H

H

H

vitamin A

H

H

O

H

O

H H H C C C O C H C O O C H H O H

vitamin C

Vitamin A has a single polar alcohol group and a substantial nonpolar part, so it will be insoluble in water, but relatively soluble in nonpolar solvents like the fats in our body. Vitamin A is a fat-soluble vitamin. Vitamin C has three polar alcohol groups that can H bond with water as well as two other oxygen atoms with nonbonding electrons that can also donate electron pairs to H bonds with water. Using our ratio of number of carbon atoms to number of alcohol groups (as in the solutions to Problems 2.13 and 2.14), we have 5/4 = 1.25/1, which indicates that vitamin C should be quite soluble in water. Vitamin C is a water-soluble vitamin. Problem 2.16. (a) Vitamins that are soluble in the fatty tissues in our bodies will tend to stay in the body for a substantial period of time. Vitamins that are soluble in water will dissolve in fluids like our blood plasma and can be transported to the kidneys, removed from the blood stream, and excreted. The fat-soluble vitamins D, E, and K can be stored in your body, although they are slowly being used and lost and need to be part of your diet (which is part of the definition of what constitutes a vitamin), but not in large amounts. Water soluble, vitamin B is not stored and needs to be included in your daily diet at a larger dosage, in order to maintain an appropriate level in your body. (b) Vitamin A can be stored in your body, while vitamin C should be included in your daily diet in relatively large amounts. (c) Since only fat-soluble vitamins (vitamins A, D, E, and K) can be stored in your body, true hypervitaminosis has been observed only for these vitamins. This does not mean that you are entirely safe ingesting excessive doses of water-soluble vitamins. Vitamin C is a case in point. One fad is to take massive doses of vitamin C, in order to prevent colds. The problems with this regimen occur when the dosage is lowered. Your body becomes used to the high level and, if it is reduced to the normal level, you begin to show the signs of scurvy, a vitamin C deficiency disease. Problem 2.17. In order for the light bulb to glow when testing a solution for electrical conductivity, as in Investigate This 2.10, ions must be present in the solution.

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ACS Chemistry Chapter 2 suggested solutions

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Problem 2.18. In the representations of the Fe3+(aq) and NO3–(aq) ions in the Web Companion, Chapter 2, Section 2.3, page 1, movies, both ions have six water molecules around them in their hydration layer. The positively charged Fe3+(aq) ions attract the negative (oxygen) ends of the water molecules, so the water molecules are oriented with their oxygen atoms directed toward the ion. The negatively charged NO3–(aq) ions attract the positive (hydrogen) ends of the water molecules, so the water molecules are oriented with their hydrogen atoms directed toward the ion. Problem 2.19. Solution A, ethanoyl chloride (acetyl chloride), CH3C(O)Cl, dissolved in water, must contain ions, because it conducts and electric current. Solution B, 2-chloroethanol, ClCH2CH2OH, dissolved in water, does not conduct an electric current, so does not contain ions. The identity of the ions in solution A cannot be determined from these data. Problem 2.20. (a) Barium chloride, BaCl2(s), dissolves in water to give barium cations, Ba2+(aq), and chloride anions, Cl–(aq). (b) Potassium chloride, KCl(s), dissolves in water to give potassium cations, K+(aq), and chloride anions, Cl–(aq). (c) Sodium triphosphate, Na3PO4(s), dissolves in water to give sodium cations, Na+(aq), and phosphate anions, PO43–(aq). (e) Ammonium chloride, NH4Cl(s), dissolves in water to give ammonium cations, NH4+(aq), and chloride anions, Cl–(aq). (f) Sodium sulfide, Na2S(s), dissolves in water to give sodium cations, Na+(aq), and sulfide anions, S2–(aq). (g) Magnesium sulfate, MgSO4(s), dissolves in water to give magnesium cations, Mg2+(aq), and sulfate anions, SO42–(aq). Problem 2.21. (a) If current flow through a conducting solution were simply caused by a flow of electrons through the solution, we would not expect any chemical changes to occur in the solution. However, in Investigate This 2.10, you probably observed bubbles of gas formed at the electrodes when electric current passed through the conducting solutions. Formation of the gases is a sign that chemistry occurs as a result of current passing through the solution. Some mechanism for chemistry to occur, coupled with current flow, is necessary to explain the observations. Movement of ions in the solution provides a mechanism that explains both current flow and the observation that chemical reactions occur at the electrodes, presumably when the ions pick up electrons from or deliver them to the electrodes. See Chapter 10, Section 10.1 for a discussion of the chemistry that occurs at electrodes. (b) The Web Companion, Chapter 2, Section 2.3, page 2, represents the motion of cations and anions in a solution, so you can see how they are always in motion. You have to indicate which ion moves which way when responding to this page, so the assumption is made that the ions move and are the source of the electric current in the solution. If your friend is not convinced by the argument in part (a), s/he may not be convinced by an illustration of the mechanism that is

ACS Chemistry Chapter 2 suggested solutions

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in the textbook. The movement of colored ions in a conducting medium, as in electrophoresis (introduced in Chapter 8, Section 9.5), might be more convincing Problem 2.22. Neither solid NaCl nor solid HgCl2 will conduct electricity because any ions present are not free to move and transport the charge. An aqueous solution of sodium chloride is a good conductor because sodium cations, Na+(aq), and chloride anions, Cl–(aq), surrounded by polar water molecules, are free to move about in the solution. Although HgCl2 is soluble in water, the aqueous solution does not conduct electricity. Evidently no ions are present when this compound goes into solution. The molecule itself stays together in solution as HgCl2(aq). This is not the usual circumstance for what appears to be a salt, but mercury(II) chloride molecules stay bonded together even in aqueous solution. Problem 2.23. This is a representation of a positively charged ion surrounded by polar molecules like our simple ellipsoids with positive and negative ends in Chapter 1, Figures 1.15 and 1.16.

If you are the positive ion, you will feel a good deal of attraction from the partial negative charges on the surrounding solvation sphere of polar molecules. Other positive or negative ions elsewhere in the solution are farther away and have little influence on your behavior. They, too, are surrounded by solvating polar solvent molecules and are little influenced by your charge. Problem 2.24. (a) MgBr2(s) dissolves in water to produce Mg2+(aq) and Br–(aq) ions, so the solution will conduct an electric current (by movement of the ions). (b) CH3OH(l) dissolves in water and forms H bonds with the water molecules, but produces no ions, so the solution will not conduct an electric current (there are no ions to carry the current). (c) NaOH(s) dissolves in water to produce Na+(aq) and OH–(aq) ions, so the solution will conduct an electric current (by movement of the ions). (d) CH3OCH3(g) dissolves in water and forms some H bonds with the water molecules, but produces no ions, so the solution will not conduct an electric current (there are no ions to carry the current). (e) KNO3(s) dissolves in water to produce K+(aq) and NO3–(aq) ions, so the solution will conduct an electric current (by movement of the ions). (h) CH3CH2CH2CH3(g) dissolves to a very limited extent in water and produces no ions, so the solution will not conduct an electric current (there are no ions to carry the current). 10

ACS Chemistry Chapter 2 suggested solutions

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Aqueous Solutions and Solubility

Problem 2.25. (a) Alkali metals (column I of the periodic table) lose their single valence electron to form positive ions (cations) with a 1+ charge. Familiar examples are Na+ and K+ and members of a family ( column of the table) have similar properties. (b) Oxygen family elements (column VI of the periodic table) gain two electrons in their valence shell to form negative ions (anions) with a 2– charge. The oxide, O2–, and sulfide, S2–, anions are familiar examples and we expect other members of the family to have similar properties. (c)Alkaline earth metals (column II of the periodic table) lose their two valence electrons to form positive ions (cations) with a 2+ charge. Familiar examples are Mg2+ and Ca2+ and we expect members of the family to have similar properties. (d) Halogens (elements in column VII of the periodic table) gain an electron in their valence shell to form negative ions (anions) with a 1– charge. The chloride, Cl–, and bromide, Br–, anions are familiar examples and we expect other members of the family to have similar properties. Problem 2.26. (a) The magnesium cation, Mg2+, and bromide anion, Br–, combine to form the electrically neutral ionic compound magnesium bromide, MgBr2. (b) The calcium cation, Ca2+, and nitrate anion, NO3–, combine to form the electrically neutral ionic compound calcium nitrate, Ca(NO3)2. (c) The magnesium cation, Mg2+, and sulfate anion, SO42–, combine to form the electrically neutral ionic compound magnesium sulfate, MgSO4. (d) The potassium cation, K+, and oxide anion, O2–, combine to form the electrically neutral ionic compound potassium oxide K2O. Problem 2.27. (a) Na2SO4 is sodium sulfate, an ionic compound of the Na+ cation and SO42– anion. (b) MgCl2 is magnesium chloride, an ionic compound of the Mg2+ cation and Cl– anion. (c) (NH4)2CO3 is ammonium carbonate, an ionic compound of the NH4+ cation and CO32– anion. (d) Al2S3 is aluminum sulfide, an ionic compound of the Al3+ cation and S2– anion. Problem 2.28. (a) The electrically neutral formula for barium nitrate, a combination of the Ba2+ cation and NO– anion, is Ba(NO3)2. (b) The electrically neutral formula for ammonium phosphate, a combination of the NH4+ cation and PO43– anion, is (NH4)3PO4. (c) The electrically neutral formula for calcium oxide, a combination of the Ca2+ cation and O2– anion, is CaO. (d) The electrically neutral formula for potassium sulfate, a combination of the K+ cation and SO42– anion, is K2SO4.

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Problem 2.29. (a) MgS is magnesium sulfide, an ionic compound of the Mg2+ cation and S2– anion. (b) Na3PO4 is sodium phosphate (or sometimes trisodium phosphate), an ionic compound of the Na+ cation and PO43– anion. (c) NH4NO3 is ammonium nitrate, an ionic compound of the NH4+ cation and NO3– anion. (d) LiOH is lithium hydroxide, an ionic compound of the Li+ cation and OH– anion. Problem 2.30. (a) The electrically neutral formula for calcium iodide, a combination of the Ca2+ cation and I– anion, is CaI2. (b) The electrically neutral formula for sodium fluoride, a combination of the Na+ cation and F– anion, is NaF. (c) The electrically neutral formula for potassium carbonate, a combination of the K+ cation and CO32– anion, is K2CO3. (d) The electrically neutral formula for barium hydroxide, a combination of the Ba2+ cation and OH– anion, is Ba(OH)2. Problem 2.31. The completed grid for ionic compound formation by the cation/anion pairs is: Mg2+ NH4+ Al3+

Na+

CO32–

PO43–

F–

MgCO3 magnesium carbonate (NH4)2CO3 ammonium carbonate Al2(CO3)3 aluminum carbonate Na2CO3 sodium carbonate

Mg3(PO4)2 magnesium phosphate (NH4)3PO4 ammonium phosphate AlPO4 aluminum phosphate

MgF2 magnesium fluoride NH4F ammonium fluoride AlF3 aluminum fluoride

Na3PO4 sodium phosphate

NaF sodium fluoride

Problem 2.32. (a) Milk of Magnesia® is a suspension of the sparingly soluble ionic compound Mg(OH)2, magnesium hydroxide. (b) Epsom salt is MgSO4, magnesium sulfate. (c) Plaster of Paris is CaSO4, calcium sulfate. (d) Caustic soda (often called lye) is NaOH, sodium hydroxide. (e) Soda ash is Na2CO3, sodium carbonate.

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Problem 2.33. The reaction equations for formation of the common cations or anions of these elements are: (a) potassium K(g) K+(g) + e– (b) calcium Ca(g) Ca2+(g) + 2e– (d) sulfur S(g) + 2e– S2–(g) (e) bromine Br(g) + e– Br-(g) Problem 2.34. By definition, ionization always involves the separation of a negative charge (an electron) and a positive charge (the cation that remains after the electron has departed). If energy is released when opposite charges come together (energy has a negative value for the process), then the reverse process requires the input of energy (energy has a positive value for the process). Mathematically, the energy of coulombic attraction, E (Q1·Q2)/d, equation (2.2), expresses the energy of attraction of opposite charges when Q1 and Q2 have opposite signs. This attraction must be overcome (a reversal of the mathematical sign from negative to positive) in order to separate the opposite charges. Problem 2.35. We are asked whether lattice energies, such as those in Table 2.3, are consistent with electrical attraction energies characterized by equation (2.2), E (Q1·Q2)/d. If they are consistent, we would expect the lattice energies to be roughly proportional to the product, |Q1·Q2| for a series of ionic compounds. The distance separating the ions, d in equation (2.2), might also affect the lattice energies, but simple cations are about the same size, as are simple anions, so the distances of separation are probably not very different and we will focus on the effects of charge. We are interested only in relative values, so let us assign values of Q as the charges we write on the ions. For our comparisons, two of the ionic compounds from Table 2.3 are chosen here (but any other of the 1:1 cation-to-anion compounds could be chosen): Compound NaBr MgS

Cation Na+ Mg+2

Q1

Anion

1+ 2+

Br– S2–

Q2

Q1·Q2

12-

1 4

Elattice, kJ·mol– 1

751 3406

We see here that the lattice energies are roughly proportional to Q1·Q2 . An increase by a factor of four in the product of the charges is accompanied by an increase of about 4.5 in the lattice energy. Thus, the lattice energy data are consistent with coulombic attraction energy. Other factors, such as the distance of separation of the ions and the geometric arrangement of the ions with respect to one another, also affect lattice energies, but charge is most important. Problem 2.36. (a) Based on coulombic electrical attraction, CaBr2(s) would have greater forces of attraction and repulsion than KBr(s), if the distance separating the charges is the same in both crystals. The double charge on Ca2+ will result in larger coulombic forces, because they are directly related to the size of the charges interacting.

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(b) The data in Table 2.3 supports our answer in part (a). The lattice energy for CaBr2, 2176 kJ·mol–1, is about three times the lattice energy of KBr, 689 kJ·mol–1. Problem 2.37. The problem statement gives us Elattice = 2176 kJ·mol–1 and Eion form = 966 kJ·mol–1 for calcium bromide, CaBr2(s). Combining these values in an energy diagram, we have:

We equate the energies for the two pathways from separated atoms to the ionic crystal to find Extal form: Extal form = Eion form + (– Elattice) = (966 kJ·mol–1) + (–2176 kJ·mol–1) Extal form = –1210 kJ·mol–1 (as shown on the energy diagram) Problem 2.38. This is the table of lattice energies (in kJ·mol–1) we are to use for this problem. F–

Cl–

Br–

Li+

1046

861

818

Na+

929

787

751

K+

826

717

689

(a) As we go across any row of this table, the size of the anion increases, because the size of the ions in a group (column) of the periodic table increases as we go down the group and that is what we are doing going from F– to Br–. We note that the lattice energy decreases across each row, indicating that the lattice energy decreases as the size of the anion increases (and the cation remains constant). This makes sense, because the distance between the cations and anions in the crystal increases and we see from equation 2.2 that the energy of attraction between unlike charges decreases as the distance between them increases. (b) Analysis of the data for any column of this table is exactly like that in part (a) with the roles of the cation and anion reversed. As we go down a column the size of the cation increases while the size of the anion remains constant. The lattice energy decreases as the size of the cation increases, because the distance between the cations and anions in the crystal increases and the energy of attraction between unlike charges decreases as the distance between them increases. 14

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(c) The lattice energy of a salt decreases as the size of its ions increases. The lattice energies are largest when the ions are the smallest. (d) For CsI, the cation, Cs+, is larger than any of the alkali metal cations in the table and the anion, I–, is larger than any of the anions in the table. The lattice energy for CsI should be even lower than the lowest value in the table (that for KBr) and certainly considerably smaller than that of NaCl. The lattice energy for CsI is 604 kJ·mol–1. Problem 2.39. This is the energy diagram for the formation of one mole of ionic crystals of MgCl2 that we are to use for this problem.

(a) The lattice energy, Elattice, for MgBr2 is the energy required to convert one mole of the compound from its solid crystal to separated ions in the gas phase. On the diagram, we see that 2524 kJ·mol–1 is released when the gaseous ions come together to form the ionic solid. Therefore, Elattice = 2524 kJ·mol–1, the amount of energy to get the ions apart. (b) From the energy diagram, the energy required to change gaseous atoms to gaseous ions, Eion form = 1490 kJ·mol–1. (c) The energy of formation of ionic crystals from gaseous atoms, Extal form, is the difference between the two energies in parts (a) and (b): Extal form = –1034 kJ·mol–1 as shown on the diagram. (d) This table shows the comparison between these energies for MgBr2 and those for NaCl from Figure 2.14. Ionic compound MgCl2 NaCl

Elattice kJ·mol–1 2524 787

Eion form kJ·mol–1 1490 145

Extal form kJ·mol–1 1034 642

Note that the large difference between the lattice energies is largely a result of the much higher energy required to remove two electrons from magnesium atoms to form the doubly charged Mg2+(g) cations compared to removal of only one electron to form the Na+(g) cation. This large amount of energy is about 60% of the energy released when the coulombic attraction of the doubly-charged cations and singly-charged anions brings them together to form the ionic solid.

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Chapter 2

Problem 2.40. (a) We can use the information that the energy required to remove electrons from gaseous silver atoms to form gaseous silver cations is 731 kJ·mol–1 and that 296 kJ·mol–1 of energy is released when gaseous iodine atoms gain electrons to form gaseous iodide anions, to construct an energy diagram to find the energy change for the net reaction: Ag(g) + I(g) Ag+(g) + I–(g). The loss of an electron by each Ag(s) atom and gain of an electron by each I(g) atom is represented on the diagram by the slanting arrow showing that the electron lost by a silver atom is gained by an iodine atom, so there is no net loss or gain of electrons in the overall reaction The diagram shows that the net reaction energy is 435 kJ·mol–1.

(b) Use the result from part (a) and the lattice energy for AgI(s) crystals, Elattice = 887 kJ·mol–1, to draw an energy diagram analogous to Figure 2.14 and use it to find Extal form for the formation of ionic crystals of AgI(s) from the gaseous atoms.

Problem 2.41. In the energy diagrams we have seen in the textbook or constructed for the formation of solid ionic compounds, we have seen this relationship among the energies represented on the diagrams: 16

ACS Chemistry Chapter 2 suggested solutions

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Aqueous Solutions and Solubility Extal form = Eion form + (– Elattice) = Eion form – Elattice

For KBr(s) crystals, the problem statement says that Elattice = 689 kJ·mol–1. The formation of separate gaseous atoms of potassium, K(g), and bromine, Br(g), from the ionic crystal requires 594 kJ·mol–1, so Extal form (energy change for the reverse of process, forming the ionic crystal from the gaseous atoms) = –594 kJ·mol–1. Substituting these values in the preceding equation, gives: –594 kJ·mol–1 = Eion form – (689 kJ·mol–1) Eion form = 95 kJ·mol–1 An energy diagram incorporating these data is shown here (not quite to scale, so the numeric values can be included with the arrows representing them):

Problem 2.42. As in the solution for Problem 2.41, to find the lattice energy, Elattice, for magnesium fluoride, MgF2(s), we use the energy relationship: Extal form = Eion form + (– Elattice) = Eion form – Elattice Substituting the values given in the problem statement, Extal form = –1424 kJ·mol–1 and Eion form = 1533 kJ·mol–1, we get: –1424 kJ·mol–1 = (1533 kJ·mol–1) – Elattice Elattice = 2957 kJ·mol–1 An energy diagram incorporating these data is shown here:

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Problem 2.43. (a) When ammonium acetate, NH4C2H3O2 [= (NH4+)(C2H3O2–)], is dissolved in water, the mixture becomes quite cold. This observation means that the dissolution process is taking thermal energy from its surroundings, the molecules in the solution, the container, and your hand, if you are holding the container. A reaction that requires an input of thermal energy, Ereaction > 0, is endothermic. (b) The energy change for the process of dissolving ionic solutes in water can be broken into two parts. Lattice energy is required to break the coulombic electrical attractions between cations and anions in the lattice while hydration energy is released as water molecules surround these ions and are attracted to them by coulombic attractions. The difference between these two energies determines whether thermal energy will be absorbed (endothermic) or released (exothermic) by the dissolving process. For ammonium acetate, the dissolving process absorbs energy, so breaking the lattice attractions must take more energy than is gained back by hydration of the ions. An energy diagram that represents this case is:

[Note that water is shown separately with the solid and gaseous ions as a reminder that the system as a whole contains the water into which the solid dissolves and which hydrates the ions. Sometimes, as in Figures 2.15 and 2.16 in the text and the solution to Problem 2.44 below, the water is omitted for simplicity, but this is probably not a good idea.] (c) The ions present in a solution of ammonium acetate are the same as the ions in the solid crystal, ammonium cation and acetate anion, which we write as NH4+(aq) and C2H3O2–(aq) to 18

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signify that the ions are hydrated, that is, surrounded by polar water molecules that are attracted by the ionic charges. (d) The molecular level interactions of hydrated ions are represented in Figure 2.9 by showing the negative oxygen end of water molecules oriented toward the positive cations and the positive hydrogen end of water molecules oriented toward the negative cations. We expect similar orientations and interactions for the NH4+ and C2H3O2– ions, but there is an added factor for these ions, because they can hydrogen bond with the water molecules and form even more directed interactions. Some of these H bonds are illustrated here: H O H

H

H H O H

H N H H

H

O H H

O H

O

H H

H

O

O CH3 C O

H O H

H H O

Problem 2.44. (a) Use the data in Table 2.3 for LiCl, Elattice = 861 kJ·mol–1 and Ehydration = –898 kJ·mol–1, to sketch an energy diagram for the dissolution process (broken into two steps). Use the diagram to find that Edissolve = –37 kJ·mol–1. Energy is released, so the dissolution reaction is exothermic.

(b) Use the data in Table 2.3 for KBr, Elattice = 689 kJ·mol–1 and Ehydration = –670 kJ·mol–1, to sketch an energy diagram for the dissolution process (broken into two steps). Use the diagram to find that Edissolve = 19 kJ·mol–1. Energy is required, so the dissolution reaction is endothermic.

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Problem 2.45. We are asked to try to explain why lithium sulfate, Li2SO4, is quite soluble in water (261 g·L–1) while calcium sulfate, CaSO4, is essentially insoluble (4.9 mg·L–1). To see what factors might be responsible for such a difference, let’s compare the data for a pair calcium ionic compounds in Table 2.3: Compound

Elattice kJ·mol–1

Ehydration kJ·mol–1

CaCl2(s)

2260

–2337

CaCO3(s)

2804

–2817

In CaCl2(s) and CaCO3(s), we have the cation with a 2+ charge and anions with a 1– charge and a 2– charge, respectively. CaCl2(s) is quite soluble in water, as you found in Investigate This 2.22, and CaCO3(s), marble or chalk, is quite insoluble. As a first approximation, these compounds are rather like Li2SO4(s) and CaSO4(s), where we have the anion with a 2– charge and cations with a 1+ charge and 2+ charge, respectively. The ion-ion interactions in the solids and ion-polar solvent interactions in aqueous solution should be roughly the same for an ionic compound with 1+ cations and a 2– anion as for a compound with a 2+ cation and 1– anions. If this supposition is correct, we see from the comparison in this table that the hydration energy should substantially outweigh the lattice energy, for a 2:1 ionic compound like Li2SO4(s) compared to a 2:2 ionic compound like CaSO4(s). As a first suggestion about the factor responsible for the high solubility of Li2SO4(s) compared to CaSO4(s), we would probably say that the hydration energy (due to solvation of the ions) favors dissolution of Li2SO4(s), Edissolve = –67 kJ·mol–1, more than CaSO4(s), Edissolve = –13 kJ·mol–1. This is a large factor in this case, but we have to be careful making to much of this argument, since we know that there are soluble ionic compounds whose dissolution is endothermic, so energetics cannot be the whole picture.

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Problem 2.46. We asked to describe or represent in three different ways what happens when sodium sulfate, Na2SO4(s), dissolves in water. First we describe the dissolution in words. Water molecules are attracted to the Na+ and SO42– ions in the solid crystal lattice, especially those at the edges and corners of the crystals. The negative oxygen ends of the water molecules are attracted to the Na+ and positive hydrogen ends in water are attracted to SO42-. These ion-dipole attractions compete with the ion-ion attractions, lattice energy, that holds Na+ and SO42- in the crystal. In the case of a soluble ionic compound like Na2SO4(s), the ion-dipole attractions are finally successful in the tug of war and the ions are broken away from the crystal. Once the ions are broken away from the crystal, more water molecules surround each ion creating a hydration layer. These hydration layers create a shield, making it difficult for hydrated ions that are oppositely charged to get too close to each other. In our example, hydrated Na+ and hydrated SO42- do not interact much with one another (until their concentrations become so high that they are forced close to one another because there are so many of them). An ionic equation that succinctly represents this process is usually written as: Na2SO4(s) 2Na+(aq) + SO42-(aq) Although water is obviously involved in the process, it is usually left out of such equations, because its stoichiometry is not easily represented. If we just add H2O(l) to the reactant side of the equation, it might be interpreted as one molecule of water for each formula unit of Na2SO4 dissolved, and that would be misleading. We just have to remember that water is an active reactant in the dissolution process, not an inert by-stander. A molecular level representation of the dissolution process is shown in this very rough sketch showing a two-dimensional crystal interacting with solvent (water): S

+

Na+ cation

2–

SO4 2– anion

S

solvent, water molecules

+

+

2–

2– +

+ 2– +

S + S S S + S

S

2– S + S +

+

S

S S S

S

2–

S

2– +

S

+

S

S

S

S

For simplicity, the many water molecules that are not interacting with the ions are omitted and the polar ends of the water molecules are not represented. See Figure 2.9(b) for a representation of the orientation of the water molecules with respect to the cations and anions.

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Problem 2.47. (a) The mixing process described in this problem is represented in this table, modeled after Table 2.4 in the text: Before Mixing

Positive ion(s) Negative ion(s) Conductivity? Precipitate?

Na3PO4 solution Na+ (aq) PO4-(aq) yes

CaBr2 solution Ca2+(aq) Br–(aq) yes

After Mixing Na3PO4 and CaBr2 Na+(aq)+ Ca2+(aq) PO4- (aq)+ Br– (aq) yes yes

(b) After mixing, two new combinations of cations and anions are possible: NaBr and Ca3(PO4)2. Our solubility rules say that ionic compounds of alkali metal cations and halide anions are soluble, so sodium bromide, NaBr, is likely to be a soluble compound. Calcium phosphate, Ca3(PO4)2, must be the precipitate and it fits our solubility rule that says that ionic compounds of multiply-charged cations and anions are likely to be insoluble. (c) This diagram is a simple molecular level representation of this mixing and reaction (precipitate formation) with only enough of each ion shown to represent the stoichiometry of the reaction:

Note that each of the individual solutions that are mixed contains ions and there are also ions in the solution remaining after the solid has precipitated. The presence of the ions in all three solutions explains why they are all conduct an electric current, as shown in the table above. (d) The complete ionic equation that represents the precipitation reaction involves all four ions: 6Na+(aq) + 2PO4–(aq) + 3Ca2+(aq) + 6 Br–(aq) 6Na+(aq) + 6Br--(aq) + Ca3(PO4)2(s) (e) The net ionic equation that represents the precipitation reaction involves only the two ions that react to form the precipitate (the spectator ions are omitted): 3Ca2+(aq) + 2PO4–(aq) Ca3(PO4)2(s) Problem 2.48. (a) When aqueous solutions of potassium chloride, KCl, and sodium bromide, NaBr, are mixed, the solution contains all four ions, Na+(aq), K+(aq), Cl–(aq), and Br–(aq), and no precipitate is formed. We can conclude that both NaCl(s) and KBr(s) are water soluble, since the solids could

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have formed from this mixture, but did not. We also note that all of these ionic compounds are combinations of an alkali metal cation and a halide anion, which our solubility rules predict will be soluble. (b) This molecular level representation, similar to Figure 2.17, illustrates the result when the two solutions are mixed.

(c) The complete ionic equation representing what happens when the two solutions are mixed involves all four ions: K+(aq) + Cl-(aq) + Na+(aq) + Br-(aq) K+(aq) + Cl-(aq) + Na+(aq) + Br-(aq) The ions are present in two separate solutions on the left-hand (reactant) side of this equation and together in a single solution on the right-hand (product) side. The separation on the left is not represented in this standard ionic equation, but we can amend it slightly to suggest the separation by bracketing the separate solution components on the left: [K+(aq) + Cl-(aq)] + [Na+(aq) + Br-(aq)] K+(aq) + Cl-(aq) + Na+(aq) + Br-(aq) Since there is no net reaction (no apparent reaction of any kind), there is no net ionic equation for this mixing. In a sense, all the ions are spectator ions. Problem 2.49. The solubility rules indicate that multiple charged cations and anions tend to form insoluble ionic compounds. Barium sulfate, BaSO4, falls into this category. The mixture that doctors use to x-ray the gastrointestinal (GI) tract is a suspension of insoluble solid barium sulfate in water. The solubility of the solid ionic compound is so low that the suspension contains only a tiny amount of Ba2+ cation. Its concentration is below the toxic level of Ba2+, so the patient is not harmed (except by having to swallow a substantial amount of chalky suspension).

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Problem 2.50. We are asked to suggest a sequence of selective precipitation reactions to separate Ag+, Ba2+, and Fe3+ from solution in which all three cations are present. The information we have to work with is in this table: Cation Test Solution

Ag+(aq)

Ba2+(aq)

Fe3+(aq)

NaCl

ppt

no ppt

no ppt

NaOH

ppt

no ppt

ppt

Na2SO4

no ppt

ppt

no ppt

The objective of our sequence of additions of the test solutions (which we might also call the separation reagents) is to precipitate the cations one at a time without forming any mixtures of precipitates that contain more than one cation. For example, if we were to add the NaOH test solution to the original mixture of cations, both the Ag+(aq) and Fe3+(aq) cations would react to form precipitates, so the solid product would be a mixture of solids and we would not have separated the cations from one another. However, if the Ag+(aq) cation had already been removed from the solution, addition of the NaOH test solution would result in a precipitate containing the Fe3+(aq) cations and we would have the Fe3+(aq) cations as a separate precipitate. Our strategy is to look for a test solution that will precipitate only one of the cations, use it to precipitate that cation and then use a second test solution that will precipitate one or the other of the remaining cations. There are two ways to start our sequence. (1) We can add the NaCl test solution to precipitate the Ag+(aq) cation as AgCl(s), leaving behind a solution containing the Ba2+(aq) and Fe3+(aq) cations [as well as the Na+(aq) cations added with the test solution]. Or, (2) we can begin by adding the Na2SO4 test solution to precipitate the Ba2+(aq) cations as BaSO4(s), leaving behind a solution containing the Ag+(aq) and Fe3+(aq) cations [as well as the Na+(aq) cations added with the test solution]. To pick the next step in sequence (1), consider this table that tells us how our test solutions react with the remaining cations: Cation 2+

Test Solution

Ba (aq)

Fe3+(aq)

NaOH

no ppt

ppt

Na2SO4

ppt

no ppt

We have two choices, both of which will separate these two cations from one another. In the pathway that we will label (1a), we add NaOH test solution to the solution we get after removing the AgCl(s) in order to precipitate the Fe3+(aq) as Fe(OH)3(s). The solution now contains the Ba2+(aq) [and the Na+(aq) cations added with the test solutions]. The cations are now separated as two precipitates, AgCl(s) and Fe(OH)3(s), with the remaining cation, Ba2+(aq), in solution. If we need to have the Ba2+ separated from all the Na+(aq) also in the solution, we can add the Na2SO4 test solution to precipitate it as BaSO4(s). The net ionic reactions for the three steps of pathway/sequence (1a) [and the alternative (1b)] are:

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Pathway (1a) Ag+(aq) + Cl–(aq) 3+

–

Fe (aq) + 3OH (aq)

AgCl(s) Fe(OH)3(s)

Pathway (1b) Ag+(aq) + Cl–(aq) 2+

2–

Ba (aq) + SO4 (aq)

AgCl(s) BaSO4(s)

Ba2+(aq) + SO42–(aq) BaSO4(s) Fe3+(aq) + 3OH–(aq) Fe(OH)3(s) The net ionic reactions for the three steps of pathway (2) are given here. You can reason through the logic that leads to this sequence and the reason why there is no alternative possibility for this sequence. Pathway (2) Ba2+(aq) + SO42–(aq) BaSO4(s) Ag+(aq) + Cl–(aq) 3+

–

Fe (aq) + 3OH (aq)

AgCl(s) Fe(OH)3(s)

Problem 2.51. (a) When an aluminum nitrate, Al(NO3)3, solution is mixed with a sodium oxalate, Na2C2O4, solution the solution formed contains aluminum, Al3+(aq), and sodium, Na+(aq), cations, and nitrate, NO3–(aq), and oxalate, C2O42–(aq), anions. A precipitate forms from this mixed solution. The possibilities for this solid ionic compound are Al2(C2O4)3(s) and NaNO3(s). Our solubility rules indicate that ionic compounds with an alkali metal cation and nitrate anion are likely to be soluble, so the precipitate is not NaNO3(s). The precipitate, Al2(C2O4)3(s), is an ionic compound formed from a multiply-charged cation and a multiply-charged anion, which our rules suggest is likely to be insoluble. (b) The net ionic equation for the reaction that forms Al2(C2O4)3(s) has the aqueous ions with appropriate stoichiometric coefficients as reactants and the solid as product: 2Al3+(aq) + 3C2O42-(aq) Al2(C2O4)3(s) Problem 2.52. When a solution of lithium nitrate, LiNO3, is mixed with a solution of sodium phosphate, Na3PO4, the solution contains lithium, Li+(aq), and sodium, Na+(aq), cations and nitrate, NO3– (aq), and phosphate, PO43–(aq), anions. A white precipitate is observed to form from this mixture. The possibilities for this solid ionic compound are Li3PO4(s) and NaNO3(s). Our solubility rules indicate that ionic compounds with an alkali metal cation and nitrate anion are likely to be soluble, so the precipitate is not NaNO3(s). The precipitate must be Li3PO4(s). Although ionic compounds with alkali metal cations are generally expected to be soluble, note that, in Worked Example 2.34, lithium is discussed as an exception to the general rule. (b) The net ionic equation for the precipitation reaction is: 3Li+(aq) + PO43–(aq) Li3PO4(s) Problem 2.53. [Note that there seems to be no symbol for the equilibrium double arrow that is common to both PC and Macintosh computer platforms, so the symbol “ ” is used in these notes to make them cross-platform compatible.] When the reactants and products are separated by a forward arrow over a backward arrow, “ ” in a reaction equation, it means that the reaction can or is going in both directions. That is, the

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reactant species are combining to yield products and the product species are combining to give reactants and both processes are occurring simultaneously. Problem 2.54. (a) When a solution of cadmium chloride, CdCl2, is mixed with a solution of ammonium sulfide, (NH4)2S, the solution contains cadmium, Cd2+(aq), and ammonium, NH4+(aq), cations and chloride, Cl–(aq), and sulfide, S2–(aq), anions. A yellow-orange precipitate is observed to form from this mixture. The possibilities for this solid ionic compound are CdS(s) and NH4Cl(s). Our solubility rules indicate that ionic compounds with a singly-charged cation and halide anion are likely to be soluble, so the precipitate is not NH4Cl(s). The precipitate must be CdS(s). (Cadmium sulfide -- cadmium yellow -- is used as a pigment in paints.) (b) The net ionic equation for the precipitation reaction is: Cd2+(aq) + S2-(aq) CdS(s) Problem 2.55. The objective of this problem is to use our solubility rules, including the few exceptions we have noted, to predict the products of mixing four pairs of aqueous solutions and to write complete and net ionic equations for those reactions that produce a precipitate. (a) barium chloride(aq) + sodium sulfate(aq) The mixed solution contains barium, Ba2+(aq), and sodium, Na+(aq), cations and chloride, Cl– (aq), and sulfate, SO42–(aq), anions. Our solubility rules indicate that an ionic compound with a multiply-charged cation and anion is likely to be insoluble. Thus, we can predict that BaSO4(s) is insoluble and write this complete and net ionic reaction equation: Ba+2(aq) + 2Cl-(aq) + 2Na+(aq) + SO42-(aq) BaSO4(s) + 2Na+(aq) + 2Cl-(aq) Ba+2(aq) + SO42-(aq)

BaSO4(s)

(b) silver nitrate(aq) + magnesium chloride(aq) The mixed solution contains silver, Ag+(aq), and magnesium, Mg2+(aq), cations and nitrate, NO3–(aq), and chloride, Cl–(aq), anions. Our solubility rules indicate that ionic compounds with a halide or nitrate anion are likely to be soluble, so the initial instinct is to predict that no precipitate will form. However, when we recall that the silver ion is a notable exception to the rules, we predict that silver chloride, AgCl(s), is insoluble and write this complete and net ionic reaction equation: 2Ag+(aq) + 2NO3–(aq) + Mg2+(aq) + 2Cl–(aq) 2AgCl(s) + Mg2+(aq) + 2NO3–(aq) 2Ag+(aq) + 2Cl–(aq)

2AgCl(s)

(c) strontium nitrate(aq) + potassium nitrate(aq) The mixed solution contains strontium, Sr2+(aq), and potassium, K+(aq), cations and nitrate, NO3–(aq), anions. No new products are possible, because there is only one anion that is common to both soluble starting compounds. Thus we can write: Sr2+(aq) + 3NO3-(aq) + K+(aq) + NO3-(aq) NO APPARENT REACTION (d) ammonium phosphate(aq) + calcium bromide(aq) The mixed solution contains barium, NH4+(aq), and calcium, Ca2+(aq), cations and phosphate, PO43–(aq), and bromide, Br–(aq), anions. Our solubility rules indicate that an ionic compound

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with a multiply-charged cation and anion is likely to be insoluble. Thus, we can predict that Ca3(PO4)2(s) is insoluble and write this complete and net ionic reaction equation: 6NH4+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Br-(aq) Ca3(PO4)2(s) + 6NH4+(aq) + 6Br-(aq) 3Ca2+(aq) + 2PO43-(aq)

Ca3(PO4)2(s)

Problem 2.56. The objective of this problem is to use your knowledge of the solubility rules to propose a way to prepare (synthesize) several solid ionic compounds by precipitation reactions from mixtures of soluble ionic compounds. (a) The net ionic reaction for preparing BaSO4(s) is: Ba2+(aq) + SO42–(aq) BaSO4(s) Mixing aqueous solutions of the soluble ionic compounds Ba(NO3)2 (nitrates are soluble) and Na2SO4 (alkali metal salts are soluble) will provide the reactant ions necessary for this reaction. (b) The net ionic reaction for preparing AgCl(s) is: Ag+(aq) + Cl–(aq) AgCl(s) Mixing aqueous solutions of the soluble ionic compounds AgNO3 (nitrates are soluble) and KCl (alkali metal halide salts are soluble) will provide the reactant ions necessary for this reaction. (c) The net ionic reaction for preparing Ca3(PO4)2(s) is: 3Ca2+(aq) + 2PO43- (aq) Ca3(PO4)2 (s) Mixing aqueous solutions of the soluble ionic compounds CaCl2 (halides, except silver, are soluble) and K3PO4 (alkali metal salts are soluble) will provide the reactant ions necessary for this reaction. (d) The net ionic reaction for preparing CaC2O4(s) is: Ca2+ (aq) + C2O42-(aq) CaC2O4 (s) Mixing aqueous solutions of the soluble ionic compounds CaCl2 (halides, except silver, are soluble) and K2C2O4 (alkali metal salts are soluble) will provide the reactant ions necessary for this reaction. Problem 2.57. Since the concentration of a solution (assuming it is well mixed after preparation) is uniform throughout the entire volume, spilling some of it will not change the concentration of the remaining solution. Concentration is an intensive variable (like density or temperature) that does not depend upon the amount of solution you have. Problem 2.58. (a) If it is properly labeled, the solution in a bottle labeled, "0.5 M CaCl2" contains 0.5 moles of CaCl2(aq) per liter of solution. There are 0.5 moles of Ca2+(aq) per liter of solution and 1 mole of Cl–(aq) per liter of solution. (b) If a 0.5-L bottle is about half full, then it contains approximately 0.25 L of 0.5 M CaCl2 solution. Use the definition of molarity, moles of solute per liter of solution, to determine the number of moles of the solute in about one-quarter liter of solution:

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mol CaCl2 in 0.25 L = (0.25 L)

0.5 mol 1L

= 0.13 mol CaCl2

(c) Use the molar formula mass of CaCl2, 111 g·mol–1, to convert the number of moles of CaCl2 from part (b) to mass:

0.13 mol CaCl2 = (0.13 mol CaCl2)

111 g 1 mol CaCl 2

= 14 g CaCl2

Problem 2.59. (a) Counting the number of atoms of each element in the molecular structure for vitamin C in Problem 2.15 gives the molecular formula C6H8O6. (b) Find the molar mass of vitamin C by summing the masses of each element in a mole of the compound: 12.01 g (6 mol C) = 72.06 g 1 mol C 1.008 g = 8.06 g (8 mol H) 1 mol H 16.00 g = 96.00 g (6 mol O) 1 mol O

molar mass vitamin C = 176.12 g = 176 g (accurate enough for the rest of the data) (c) To find the number of moles of vitamin C in a tablet that contains 500-mg of the vitamin, we use the molar mass to convert this mass to moles. The actual mass of vitamin in a vitamin tablet is only accurate to a few percent, so we will assume that the mass of vitamin C in the tablet is about 0.50 g [= (500 mg) 1 g 1000 mg ], with an implied accuracy of 1 part in 50 or about 2%.

0.50 g vit C = (0.50 g vit C)

1 mol vit C 176 g

= 0.0028 mol vit C = 2.8

10–3 mol vit C

(d) To find the number of molecules of vitamin C in a tablet that contains 500-mg of the vitamin, we use Avogadro’s number to convert number of moles from part (c) to number of molecules:

2.8

–3

10 mol vit C = (2.8 = 1.7

6.02 10 23 molec 10 mol vit C) 1 mol –3

1021 molec vit C

Problem 2.60. (a) To convert moles of aspartame to mass, we need the molar mass of aspartame, C14H18N2O5: 12.01 g (14 mol C) = 168.14 g 1 mol C 1.008 g (18 mol H) = 18.14 g 1 mol H 28

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14.01 g 1 mol N 16.00 g (5 mol O) 1 mol O (2 mol N)

= 28.02 g = 80.00 g

molar mass aspartame = 294.30 g = 294 g (accurate enough for the rest of the data) The mass of aspartame in 2.5 mol of aspartame is: 294 g 2.5 mol = (2.5 mol) = 735 g = 7.4 1 mol aspartame

102 g = 0.74 kg

(b) To convert moles of aspirin to mass, we need the molar mass of aspirin, C9H8O4: 12.01 g (9 mol C) = 108.09 g 1 mol C 1.008 g (8 mol H) = 8.06 g 1 mol H 16.00 g (4 mol O) = 64.00 g 1 mol O

molar mass aspirin = 180.15 g = 180 g (accurate enough for the rest of the data) The mass of aspirin in 0.040 mol of aspirin is: 0.040 mol = (0.040 mol)

180 g = 7.2 g 1 mol aspirin

(c) To convert number of molecules of cholesterol, C27H46O, to mass, we first need to convert number of molecules to number of moles (using Avogadro’s number) and then number of moles to mass [using the molar mass, as in parts (a) and (b) of the problem]. The number of moles in 2.5 1023 molecules of cholesterol is:

2.5

1023 molec = (2.5

1023 molec)

1 mol 23 6.02 10 molec

= 0.42 mol

The molar mass of cholesterol is: 12.01 g (27 mol C) = 324.27 g 1 mol C 1.008 g (46 mol H) = 46.37 g 1 mol H 16.00 g (1 mol O) = 16.00 g 1 mol O molar mass cholesterol = 386.64 g = 387 g (accurate enough for the rest of the data) The mass of cholesterol in 0.42 mol cholesterol is: 387 g = 163 g = 1.6 0.42 mol = (0.42 mol) 1 mol cholesterol

102 g = 0.16 kg

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(d) To convert number of molecules of caffeine, C8H10N4O2, to mass, we first need to convert number of molecules to number of moles (using Avogadro’s number) and then number of moles to mass [using the molar mass, as in parts (a) and (b) of the problem]. The number of moles in 1.2 1022 molecules of caffeine is:

1.2

1022 molec = (1.2

1022 molec)

1 mol 6.02 10 23 molec

= 0.020 mol

The molar mass of caffeine is: 12.01 g (8 mol C) = 96.08 g 1 mol C 1.008 g (10 mol H) = 10.08 g 1 mol H 14.01 g (4 mol N) = 56.04 g 1 mol N 16.00 g (2 mol O) = 32.00 g 1 mol O molar mass caffeine = 194.20 g = 194 g (accurate enough for the rest of the data) The mass of caffeine in 0.020 mol of caffeine is: 194 g = 3.9 g 0.020 mol = (0.020 mol) 1 mol caffeine Problem 2.61. To find the number of atoms of carbon in 5 mg of niacin, we need a strategy to get from the mass of a substance to the number of atoms of an element in that mass of the substance. This problem involves understanding and applying the concept of the mole. One way to plan your work is to reason backward from the desired answer, using the information given in the problem, the mass of niacin, 5 mg, and its molecular structure: H H H

C C

C N

O C C

C H

N

H

H

The molecular formula for niacin, from its structure, is C6H6N2O, so each molecule of niacin contains six atoms of carbon. (Note that you might write the formula a different way, ON2C6H6 for example, and that is OK. The format we used, C6H6N2O, is the conventional way chemists write formulas for carbon-containing compounds: carbon first, hydrogen second, and then all other elements in alphabetical order. Any format that shows the correct number of each atom in the molecule is fine for stoichiometric problems like the one here.) If we know the number of molecules of niacin in 5 mg of niacin, there are six times as many atoms of carbon in the sample. If we know the number of moles of niacin in the sample, we use Avogadro’s number to get the number of molecules in the sample. Finally, use the mass (in grams) of the sample of

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niacin and its molar mass (from the formula) to get the number of moles of niacin. Going forward through the solution, our strategy is summarized in this sequence: mass moles molecules atoms To find the number of moles of niacin in a 0.005 g [= (5 mg) 1 g 1000 mg ] sample, we need the molar mass of niacin: 12.01 g (6 mol C) = 72.06 g 1 mol C 1.008 g (6 mol H) = 6.05 g 1 mol H 14.01 g (2 mol N) = 28.02 g 1 mol N 16.00 g (1 mol O) = 16.00 g 1 mol O molar mass niacin = 122.13 g = 122 g (accurate enough for the rest of the data) The number of moles of niacin is: 0.005 g niacin = (0.005 g niacin)

1 mol niacin =4 122 g

10–5 mol niacin

The number of molecules of niacin is: 4

–5

10 mol niacin = (4 = 2.4

6.02 10 23 molec 10 mol niacin) 1 mol –5

1019 molec niacin

The number on atoms of carbon in the sample is: 2.4

1019 molec niacin = (2.4 = 1.4

1019 molec niacin)

6 atoms carbon 1 molec niacin

1020 atoms carbon

Problem 2.62. In order to determine the molar concentration, mol·L–1, of DNA in the given bacterium, we need to know the number of moles of DNA and the volume of the bacterium in which it is found. We are told that there is one molecule of DNA in the bacterium, so we can use Avogadro’s number to find the number of moles of DNA: 1 mol = 1.7 10–24 mol 1 molec DNA = (1 molec DNA) 23 6.02 10 molec

We are told that the bacterium is spherical and has a diameter, d, of 1 10–6 m. The formula for the volume of a sphere is 4 3 r3 = 1 6 d3 (where radius, r, = d/2). We want the volume in liters and one way to get it is to recall that 1 L = 1 dm3. Therefore, if we express the diameter of the bacterium in decimeters, 1 dm = 10–1 m, the volume we get will be in dm3 (= liters).

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(Another approach is to calculate the volume in cubic centimeters, which are equal to milliliters, and thence to liters.) The required diameter of the bacterium is: 1 dm d = 1 10–6 m = (1 10–6 m) = 1 10–5 dm 10 –1 m The volume, V, of the bacterium is: V = 1 6 d3 = 1 6 (1 10–5 dm)3 = 5 10–16 dm3 = 5 10–16 L The DNA concentration is: mol DNA 1.7 10 –24 mol [DNA] = = = 3 10–9 M V 5 10 –16 L Problem 2.63. To find the number of sodium ions in 50 mL of blood serum, we need a strategy to get from the volume of a solution to the number of ions of an element in that volume of the solution. One way to plan your work is to reason backward from the desired answer, using the information given in the problem, the volume of solution (serum), 50 mL, and its molarity, 0.14 M in NaCl. If we know the number of moles of sodium ion in the sample, we use Avogadro’s number to get the number of sodium ions in the sample. We know that every mole of NaCl dissolved in the serum produces a mole of sodium ions, Na+(aq). Finally, use the molarity of the serum sample and its volume (in liters) to get the number of moles of NaCl dissolved. Going forward through the solution, our strategy is summarized in this sequence: volume solution moles NaCl moles ions number ions

The volume of the serum sample is: volume = 50 mL = (50 mL)

1L 1000 mL

= 0.050 L

The number of moles of NaCl in the serum sample is: 0.14 mol 0.050 L = ( 0.050 L) = 7.0 10–3 mol 1L The number of moles of sodium ions in the serum sample is: 1 mol Na + –3 –3 7.0 10 mol NaCl = (7.0 10 mol NaCl) = 7.0 1 mol NaCl The number of molecules of sodium ions in the serum sample is: 6.02 10 23 molec –3 + –3 + 7.0 10 mol Na = (7.0 10 mol Na ) 1 mol

10–3 mol Na+

= 4.2

1021 Na+

Problem 2.64. For each solution in this problem we are asked to calculate the mass of solute present in a known volume of solution of a known molar concentration. The route map shown in Worked Example 2.50 (read from bottom to top) shows the sequence of steps required to get these masses: volume solution moles solute mass solute

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These steps can be combined: mol solute g solute 1 L solution 1 mol solute In each case, we will need the molar mass of the solute and will calculate this first. mass solute, g = (volume solution, L)

(a) The molar mass of K2Cr2O7 is: 39.10 g (2 mol K) = 78.20 g 1 mol K 52.00 g = 104.00 g (2 mol Cr) 1 mol Cr 16.00 g = 112.00 g (7 mol O) 1 mol O molar mass K2Cr2O7 = 294.20 g = 294 g (accurate enough for the rest of the data)

The mass of K2Cr2O7 in 350 mL of 0.105 M K2Cr2O7 is: mass K2Cr2O7 = (0.350 L)

0.105 mol K2 Cr2O 7 1 L solution

294 g 1 mol K2 Cr2O 7

= 10.8 g

(b) The molar mass of FeCl3·6H2O is: 55.85 g (1 mol Fe) = 55.85 g 1 mol Fe 35.45 g = 106.35 g (3 mol Cl) 1 mol Cl 1.008 g (12 mol H) = 12.10 g 1 mol H 16.00 g = 112.00 g (6 mol O) 1 mol O molar mass FeCl3·6H2O = 286.30 g = 286 g (accurate enough for the rest of the data)

The mass of FeCl3·6H2O in 50 mL of 1.0 M FeCl3·6H2O is: mass FeCl3·6H2O = (0.050 L)

1.0 mol FeCl3 ·6H2 O 1 L solution

286 g 1 mol FeCl3 ·6H2 O

= 14.3 g

(c) The molar mass of KCl is: 39.10 g (1 mol K) = 39.10 g 1 mol K 35.45 g = 35.45 g (1 mol Cl) 1 mol Cl molar mass KCl = 74.55 g = 74.6 g (accurate enough for the rest of the data)

The mass of KCl in 0.3 L of 1.70 M KCl is: 1.70 mol KCl mass KCl = (0.3 L) 1 L solution

74.6 g 1 mol KCl

= 38.0 g

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Or we get about 4

Chapter 2

102 g, if the volume is really only known to about 1 part in 3 ( 33%).

Problem 2.65. For each solution in this problem we are asked to calculate the molar concentration, molarity = moles per liter of solution, of a known mass of solute present in a known volume of solution. To get the molarity, we need to convert the given mass to moles and divide by the solution volume (in liters) to find the number of moles per liter: mass solute moles solute molarity These steps can be combined: 1 mol solute 1 molarity solute, mol·L–1 = (mass solute, g) volume solution, L g solute In each case, we will need the molar mass of the solute and will calculate this first. (a) The molar mass of NaCl is: 22.99 g (1 mol Na) = 22.99 g 1 mol Na 35.45 g (1 mol Cl) = 35.45 g 1 mol Cl molar mass NaCl = 58.44 g = 58.4 g (accurate enough for the rest of the data)

The molarity of NaCl in 0.120 L of solution containing 4.5 g NaCl is: molarity NaCl = (4.5 g NaCl)

1 mol NaCl 58.4 g

1 0.120 L

= 0.64 M

(b) The molar mass of NH4Cl is: 14.01 g (1 mol N) = 14.01 g 1 mol N 1.008 g = 4.03 g (4 mol H) 1 mol H 35.45 g = 35.45 g (1 mol Cl) 1 mol Cl molar mass NH4Cl = 53.49 g = 53.5 g (accurate enough for the rest of the data)

The molarity of NH4Cl in 0.25 L of solution containing 1.3 g NH4Cl is: 1 mol NH 4 Cl 1 = 0.097 M 53.5 g 0.25 L Or we get about 0.10 M, if the mass is only known to a precision of about 1 part in 13. molarity NH4Cl = (1.3 g NH4Cl)

(c) The molar mass of AgNO3 is: 107.9 g = 107.9 g (1 mol Ag) 1 mol Ag 14.01 g = 14.01 g (1 mol N) 1 mol N

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16.00 g = 48.00 g 1 mol O molar mass AgNO3 = 169.9 g = 170 g (accurate enough for the rest of the data)

(3 mol O)

The molarity of AgNO3 in 1.3 L of solution containing 1.85 g AgNO3 is: 1 1 mol AgNO3 = 0.0084 M 1.3 L 170 g Or we get about 0.008 M, if the volume is only known to a precision of about 1 part in 13. molarity AgNO3 = (1.85 g AgNO3)

Problem 2.66. (a) We are asked to calculate the molar concentration, molarity = moles per liter of solution, of a known mass of glucose present in a known volume of solution. To get the molarity, we need to convert the given mass to moles and divide by the solution volume (in liters) to find the number of moles per liter: mass glucose moles glucose molarity These steps can be combined: 1 mol glucose 1 molarity glucose, mol·L–1 = (mass glucose, g) g glucose volume solution, L

We will need the molar mass of glucose, C6H12O6, and will calculate this first. 12.01 g (6 mol C) = 72.06 g 1 mol C 1.008 g = 12.10 g (12 mol H) 1 mol H 16.00 g = 96.00 g (6 mol O) 1 mol O molar mass glucose = 180.16 g molarity glucose = (5.405 g glucose)

1 mol glucose 180.16 g

1 1.000 L

= 0.03000 M

(b) There are many ways to solve this part of the problem and any one that you can explain and justify to get the correct answer is OK. The way presented here may be useful in other situations involving moles/millimoles (mmol) and liters/milliliters (mL); these are especially common in biochemistry. Since 1 L = 1000 mL, a solution that contains x moles of a solute in one liter of solution will contain x millimoles of a solute in one milliliter of solution (one-thousandth of the solute in one-thousandth of the solution). For example, the solution in part (a) contains 0.03000 mmol of solute in each 1.000 mL of solution. The “rule” is this: the numeric value of a solution concentration in mmol·mL–1 equals the concentration in mol·L–1. In the problem we have here, we wish to convert a desired number of millimoles to the equivalent volume (in mL) of solution of a known concentration: 1.000 mL 0.950 mmol glucose = (0.950 mmol glucose) = 31.67 mL 0.03000 mmol

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You may have chosen a different route to calculate this volume, and, as we said at the beginning, that’s OK. Whatever way you choose is perfectly acceptable — as long as you understand the concepts and arrive at the correct answer. Problem 2.67. In order to prepare a 1.00 M solution of any solute, you have to dissolve one molar mass of the solute in one liter of solution (or an equivalent fraction of the molar mass and the volume). For this problem, we need to know the molar mass of the solutes each student used. The molar mass of CuSO4 is: 63.55 g (1 mol Cu) = 63.55 g 1 mol Cu 32.07 g (1 mol S) = 32.07 g 1 mol S 16.00 g = 64.00 g (4 mol O) 1 mol O

molar mass CuSO4 = 159.62 g The molar mass of CuSO4·5H2O adds the mass of the five moles of water, 90.08 g, to the mass of the CuSO4 for a total of 249.70 g·mol–1. Both students used the correct procedure for making a solution of known concentration, but the male student neglected to account for the water of hydration and weighed too little of the solute: so prepared a solution of too low a concentration: 1 mol 159.60 g CuSO4·5H2O = (159.60 g CuSO4·5H2O) 249.70 g CuSO4 5H2 O = 0.639 mol He prepared a solution of too low a concentration: 0.639 mol·L–1 = 0.639 M. The female student used a solid of the correct composition, CuSO4, and prepared a solution of the correct composition: 1.000 mol·L–1 = 1.000 M. Problem 2.68. To make a solution with a fairly exact concentration of a solid solute, we need to weigh the solute carefully, add the solid to a volumetric flask, dissolve the solid in some water in the flask, and then make up the solution to the exact volume of the flask. Volumetric flasks are commonly available in a variety of sizes: 10 , 25 , 50 , 100 , 250 , 500 , and 1000 mL. We need about 170 mL of 0.10 M NaOH. In order to waste as little as possible of the solid reagent, we should use the smallest volumetric flask, 250-mL, that will give us enough of the solution. This is a problem for which we know the concentration of the desired solution, 0.10 M, and its volume, 0.250 L, so we need to convert the volume to the equivalent number of moles of solute and then moles to mass using the molar mass of NaOH (40.0 g·mol–1) to get the mass of NaOH that must be dissolved. The calculation may be done in a single step by combining the two conversions: 0.10 mol NaOH 40.0 g 0.250 L soln = (0.250 L solution) = 1.00 g NaOH 1 L solution 1 mol NaOH You may not have known the standard sizes of volumetric flasks and have chosen a different volume of solution to prepare. That’s OK. If you substitute your volume into this equation and get the result you calculated, you solved the problem correctly.

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Problem 2.69. We are first asked to determine the mass of sodium chloride, NaCl, required to make 250. mL of a 0.90% (mass to volume %) sodium chloride (normal saline) solution and then to find the molarity of this solution. The mass to volume % concentration is not discussed in the textbook, but its meaning is clear: the mass of solute is a given percentage (parts in 100) of the volume of the solution. The only ambiguities are the units to be used for mass and volume. In context, it makes sense to use grams and milliliters, because they will give a reasonable result (where “reasonable” means physically significant or possible). A 0.90% solution will have 0.90 g of solute per 100. mL of solution. In this case, volume of solution is given, 250. mL, and we wish to find the mass of NaCl solute: 0.90 g NaCl 250. mL solution = (250. mL solution) = 2.25 g NaCl 100. mL solution

To find the molarity of this solution, we need to convert the mass of NaCl in a given volume (in liters) to moles of NaCl, using the molar mass of NaCl (58.4 g·mol–1), and then divide by the volume to get molarity (mol·L–1). We can either work with the mass of NaCl in 100. mL of solution (0.90 g) or the mass in 250. mL (2.25 g). Let’s use the former: 0.90 g NaCl = (0.90 g NaCl)

1 mol 58.4 g NaCl

1 0.100 L

= 0.154 M

Check to see that using 2.25 grams of NaCl in 250. mL of aqueous solution to calculate the molarity gives the same answer. Problem 2.70. We have a mass of solute, 25 g of urea, (NH2)2CO, in 2.5 L of solution (urine) and are asked to calculate the molarity of the solution. We need to convert the mass of urea to moles, using the molar mass of urea (60 g·mol–1), and then divide by the volume to get molarity (mol·L–1):

25 g urea = (25 g urea)

1 mol 60 g urea

1 2.5 L

= 0.17 M

Problem 2.71. The label on a sports drink tells us that 240 mL of the solution contains 30 mg of potassium. KH2PO4 is the only solution ingredient listed on the label that can provide this potassium. We are asked to determine the mass of KH2PO4 required to provide 30 mg of potassium [as K+(aq)] in the solution and to find the molarity of KH2PO4 in the solution. Note that each mole of KH2PO4 that dissolves in the solution provides a mole of K+(aq) in the solution. There are several ways to approach this problem, but they all require knowing the molar masses of K+ (39.1 g·mol–1) and of KH2PO4 (136 g·mol–1), because we need to convert masses to moles and moles to masses in this problem. In one approach, we follow these pathways: mass K+ moles K+ = moles KH2PO4 moles KH2PO4 mass KH2PO4 moles KH2PO4 molarity KH2PO4

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Another approach takes these pathways: mass K+ mass KH2PO4 (using relative masses in one mole KH2PO4) mass K+ moles K+ molarity K+ = molarity KH2PO4 Note that the actual number of conversions in each approach is the same, so neither is more efficient or preferred, although the second approach can be condensed into two steps. We will go through both to show that the results are the same. In the first approach, we have (after converting milligrams to grams and milliliters to liters): 0.030 g K+ = (0.030 g K+)

1 mol 39.1 g K+

= 7.7

10–4 mol K+ = 7.7

10–4 mol KH2PO4

7.7

10–4 mol KH2PO4 = (7.7

10–4 mol KH2PO4)

136 g KH2 PO 4 1 mol

= 0.10 g

7.7

10–4 mol KH2PO4 = (7.7

10–4 mol KH2PO4)

1 0.240 L

10–3 M KH2PO4

KH2PO4 = 3.2

In the second approach we have: 0.030 g K+ = (0.030 g K+)

136 (g KH 2 PO4 ) (mol KH2 PO4 ) + –1 39.1 (g K ) (mol KH2 PO 4 )

0.030 g K+ = (0.030 g K+)

1 mol 39.1 g K+

1 0.240 L

= 3.2

–1

= 0.10 g KH2PO4

10–3 M K+

= 3.2 10–3 M KH2PO4 Thus, the results are the same, no matter which approach we use. If you used yet a different approach, got the same results, and can explain the concepts behind what you did, that’s fine. Problem 2.72. We need to remember that the relationship of numbers of atoms of the elements to a given number of molecules of a compound is the same as the relationship of numbers of moles of atoms of the elements to the given number of moles of the compound. For example, in two molecules of ammonium acetate, NH4C2H3O2, there are two atoms of nitrogen, N, and in two moles of NH4C2H3O2 there are two moles of nitrogen atoms, N. Also, in two moles of NH4C2H3O2, we have:

2 mol NH4C2H3O2 = (2 mol NH4C2H3O2)

2 mol C 1 mol NH 4 C2 H 3O2

= 4 mol C

2 mol NH4C2H3O2 = (2 mol NH4C2H3O2)

7 mol H 1 mol NH 4 C2 H 3O2

= 14 mol H

2 mol NH4C2H3O2 = (2 mol NH4C2H3O2)

2 mol O 1 mol NH 4 C2 H 3O2

= 4 mol O

Summing the number of moles of all atoms (N, C, H, and O) in 2 mol NH4C2H3O2, we have 24 (= 2 + 4 + 14 + 4) mol of atoms. 38

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We are also asked for the total number of moles of ions in two moles of NH4C2H3O2. Although no discrete molecules of NH4C2H3O2 are present in a crystal of the ionic compound, the chemical formula indicates that the ratio of cations to anions is 1:1, so one formula unit (equivalent to a molecule) contains one ammonium cation, NH4+, and one acetate anion, C2H3O2–. Two formula units (molecules) of NH4C2H3O2 contain four ions (two cations and two anions), so two moles of NH4C2H3O2 contain four moles of ions: 2 mol NH4C2H3O2 = (2 mol NH4C2H3O2)

2 mol ions 1 mol NH 4 C2 H 3O2

= 4 mol ions

There is more than one way to solve a problem like this and, if you solved it differently, but got the correct answers and can explain the concepts you used, that’s OK. Problem 2.73. We are asked to determine the mass of calcium phosphate that can be made by mixing 125. mL of 0.100 M calcium chloride with 125. mL of 0.100 M sodium phosphate. To solve this problem, we will first need the formulas for calcium chloride (CaCl2), sodium phosphate (Na3PO4), and calcium phosphate (Ca3(PO4)2). The formulas are based on balancing out the known charges on the cations and anions, which we can get from Table 2.2, if we don’t remember them. Combined with the solution concentration data, the first two formulas provide us enough information to determine the molarity of the reactant ions, Ca2+(aq) and PO43–(aq), in the solutions and, hence, the number of moles of each of these ions present in the mixed solution. We see that there is one mole of Ca2+(aq) in solution for each mole of CaCl2 that is dissolved. Thus, the concentration and number of moles of calcium cation is:

0.100 M CaCl2 =

0.100 mol CaCl2 1 L solution

2+

1 mol Ca 1 mol CaCl 2

= 0.100 M Ca2+(aq)

0.100 mol Ca 2+ 0.100 M Ca (aq) = (0.125 L solution) = 1.25 1 L solution 2+

10–2 mol Ca2+(aq)

The same chain of reasoning gives the concentration and number of moles of phosphate anion: 0.100 mol Na 2 PO4 0.100 M Na3PO4 = 1 L solution

3–

1 mol PO4 1 mol Na 2 PO4

= 0.100 M PO43–(aq)

0.100 mol PO3– 4 0.100 M PO4 (aq) = (0.125 L solution) = 1.25 1 L solution 3–

10–2 mol PO43–

(aq) Use the formula for the product of reaction, a precipitate of calcium phosphate (Ca3(PO4)2(s)), to write a balanced reaction for its formation from the ions: 3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s) The equation tells us that the mole ratio of Ca2+(aq) to PO43–(aq) ions that react is 3:2. Thus, 1.5 (= 3 2 ) mol of Ca2+(aq) is required to react completely with 1 mol of PO43–(aq). Since our

mixed solution contains an equal number of moles of each ion, there is not enough Ca2+(aq) to react with all the PO43–(aq), so Ca2+(aq) is the limiting reactant in the mixture. We use the ratio

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of moles of calcium ions in a mole of calcium phosphate to convert number of moles of calcium ion that react to moles of solid calcium phosphate precipitated: 1 mol Ca 3 (PO 4 )2 1.25 10–2 mol Ca2+ = (1.25 10–2 mol Ca2+) 3 mol Ca 2+ = 4.17 10–3 mol Ca3(PO4)2(s) The molar mass of Ca3(PO4)2 is 310.2 g, so the mass of Ca3(PO4)2(s) formed is: 4.17

10–3 mol Ca3(PO4)2 = (4.17

10–3 mol Ca3(PO4)2)

310.2 g Ca 3 (PO4 )2 1 mol Ca 3 (PO4 )2

= 1.29 g Ca3(PO4)2(s) Problem 2.74. The coefficients in the balanced ionic equation give the relative number of moles of each reactant and product; 2Fe3+(aq) + SO32–(aq) + 3H2O(l) 2Fe2+(aq) + SO42–(aq) + 2H3O+(aq) These ratios can also be expressed in millimoles (1 mol = 1000 mmol). See the solution for Problem 2.66. The volume of 0.100 M SO32–(aq) needed to react exactly and completely with 24.0 mL of 0.200 M Fe3+(aq) is: 24.0 mL Fe3+ = (24.0 mL 2– 0.200 mmol Fe 3+ 1 mmol SO2– 1.000 mL SO3 3 Fe3+) 2– 1.000 mL Fe 3+ 2 mmol Fe3+ 0.100 mmol SO 3

= 24.0 mL SO32– solution Problem 2.75. (a) We are asked to determine the number of moles of each of the four ions, Na+, SO42–, Ba2+, and Cl–, in a mixture prepared by mixing 50.0 mL of a 0.45 M Na2SO4 solution with 50.0 mL of a 0.36 M BaCl2 solution. The number of moles of an ion in the mixture is the same as the number moles of that ion in its original solution. We need to convert the volumes (in liters) to moles of ionic compound solute, using the solution molarity, and then to moles of individual ions, using the ratio of moles of ions to moles of solute:

0.45 mol Na 2 SO 4 0.050 L Na2SO4 = (0.050 L Na2SO4) 1L

+

2 mol Na 1 mol Na 2 SO4

= 0.045 mol Na+ 0.050 L Na2SO4 = (0.050 L Na2SO4)

0.45 mol Na 2 SO 4 1L

2–

1 mol SO4 1 mol Na 2 SO4

= 0.023 mol SO42– 0.050 L BaCl2 = (0.050 L BaCl2)

0.36 mol BaCl2 1L

= 0.018 mol Ba2+

40

2+

1 mol Ba 1 mol BaCl 2

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0.36 mol BaCl2 0.050 L BaCl2 = (0.050 L BaCl2) 1L

2 mol Cl 1 mol BaCl 2

= 0.036 mol Cl– (b) We are asked to determine how many moles of Ba2+(aq) are required to react with all the SO42–(aq) in the mixture, if the SO42–(aq) reacts with Ba2+(aq) to give BaSO4(s). We use the 1:1 reactant ratio to find the number of moles of Ba2+(aq) required: 2–

0.023 mol SO4 (aq) = (0.023 mol SO4

2–

2+

1 mol Ba ) 2– 1 mol SO 4

= 0.023 mol Ba2+(aq)

(c) We are asked to determine how many moles of SO42–(aq) are required to react with all the Ba2+(aq) in the mixture, if the SO42–(aq) reacts with Ba2+(aq) to give BaSO4(s). We use the 1:1 reactant ratio to find the number of moles of SO42–(aq) required:

1 mol SO2– 4 0.018 mol Ba (aq) = (0.018 mol Ba ) 1 mol Ba 2+ 2+

2+

= 0.018 mol SO42–(aq)

(d) To determine whether Ba2+(aq) or SO42–(aq) is the limiting reactant in this mixture, we compare the number of moles of each ion in the solution, from part (a), to the number of moles of that ion, from parts (b) and (c), required to react completely with the other ion. If the solution does not contain the requisite number of moles of one of the ions, that ion is the limiting reactant, since it will be used up before all the other ion has reacted. In this case, we see that 0.023 mol Ba2+(aq) are required to react with all the SO42–(aq), but the solution contains only 0.018 mol Ba2+(aq). Therefore, Ba2+(aq) is the limiting reactant in this mixture and it will react with 0.018 mol SO42–(aq), leaving about 0.005 mol SO42–(aq) behind in solution. Problem 2.76. The objective of this problem is to use our solubility rules, including the few exceptions we have noted, to predict the products of mixing four pairs of aqueous solutions and to determine the limiting reagent for each reaction and the mass of the precipitate (assuming that all precipitation reactions go to completion). To find the limiting reagent in each case we will also need to write a net ionic equation for the reaction that occurs (a) 125 mL of 0.15 M BaBr2 are mixed with 125 mL of 0.15 M Na3PO4. The mixed solution contains a multiply-charged cation, Ba2+(aq) and a multiply-charged anion, PO43–(aq), which react to form an insoluble precipitate: 3Ba2+(aq) + 2PO43–(aq) Ba3(PO4)2(s) The number of moles of each of these ions in the mixture is: 0.15 mol BaBr2 0.125 L BaBr2 = (0.125 L BaBr2(aq)) 1 L BaBr2 (aq)

2+

1 mol Ba 1 mol BaBr2

= 0.019 mol Ba2+(aq) 0.125 L Na3PO4 = (0.125 L Na3PO4(aq))

0.15 mol Na 3 PO 4 1 L Na 3 PO 4 ( aq)

3–

1 mol PO4 1 mol Na 3PO4

= 0.019 mol PO43–(aq)

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The equation tells us that the mole ratio of Ba2+(aq) to PO43–(aq) ions that react is 3:2. Thus, 1.5 (= 3 2 ) mol of Ba2+(aq) is required to react completely with 1 mol of PO43–(aq). Since our

mixed solution contains an equal number of moles of each ion, there is not enough Ba2+(aq) to react with all the PO43–(aq), so Ba2+(aq) is the limiting reactant in the mixture. We use the ratio of moles of barium ions in a mole of barium phosphate to convert number of moles of barium ion that react to moles of solid barium phosphate precipitated: 1 mol Ba 3 (PO 4 )2 0.019 mol Ba2+ = (0.019 mol Ba2+) = 6.3 10–3 mol Ba3(PO4)2(s) 2+ 3 mol Ba The molar mass of Ba3(PO4)2 is 602 g, so the mass of Ba3(PO4)2(s) formed is: 6.3

10–3 mol Ba3(PO4)2 = (6.3

10–3 mol Ba3(PO4)2)

602 g Ba 3 (PO4 )2 1 mol Ba 3 (PO4 )2

= 3.8 g Ba3(PO4)2(s) (b) 85 mL of 0.40 M NH4Cl are mixed with 65 mL of 0.50 M KNO3. The mixed solution contains only singly-charged cations and anions which form soluble ionic compounds, so there is NO APPARENT REACTION in this case. (c) 85 mL of 0.40 M (NH4 )2S are mixed with 65 mL of 0.50 M ZnCl2. The mixed solution contains a multiply-charged cation, Zn2+(aq) and a multiply-charged anion, S2–(aq), which react to form an insoluble precipitate: Zn2+(aq) + S2–(aq) ZnS(s) The number of moles of each of these ions in the mixture is:

0.065 L ZnCl2 = (0.065 L ZnCl2(aq))

0.50 mol ZnCl2 1 L ZnCl2 (aq )

2+

1 mol Zn 1 mol ZnCl 2

= 0.033 mol Zn2+(aq) 0.085 L (NH4 )2S = (0.085 L (NH4 )2S(aq))

0.40 mol (NH4 )2 S 1 L (NH4 )2 S(aq)

2–

1 mol S 1 mol (NH 4 )2 S

= 0.034 mol S2–(aq) These ions react in a 1:1 ratio, so the ion whose concentration is lower, the Zn2+(aq), will be completely used up in the reaction. Thus, Zn2+(aq) is the limiting reactant and the mass of ZnS(s), molar mass = 97.4 g, that will be formed is: 1 mol ZnS 97.4 g ZnS 0.033 mol Zn2+ = (0.033 mol Zn2+) = 3.2 g ZnS(s) 2+ 1 mol Zn 1 mol ZnS (d) 15.0 mL of 0.20 M AgNO3 are mixed with 15.0 mL of 0.40 M NaBr. The mixed solution contains only singly-charged cations and anions, but these include Ag+(aq) and Br–(aq), which form a precipitate, AgBr(s), that is an exception to our rules: Ag+(aq) + Br–(aq) AgBr(s) The number of moles of each of these ions in the mixture is:

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0.20 mol AgNO3 0.0150 L AgNO3 = (0.0150 L AgNO3(aq)) 1 L AgNO 3 (aq )

+

1 mol Ag 1 mol AgNO3

= 0.0030 mol Ag+(aq) 0.40 mol NaBr 1 mol Br – 0.0150 L NaBr = (0.0150 L NaBr(aq)) 1 mol NaBr 1 L NaBr(aq) – = 0.0060 mol Br (aq) These ions react in a 1:1 ratio, so the ion whose concentration is lower, the Ag+(aq), will be completely used up in the reaction. Thus, Ag+(aq) is the limiting reactant and the mass of AgBr(s), molar mass = 188 g, that will be formed is: 0.0030 mol Ag+ = (0.030 mol Ag+)

1 mol AgBr 1 mol Ag+

188 g AgBr 1 mol AgBr

= 5.6 g AgBr(s)

Problem 2.77. (a) What is the white precipitate formed when 50. mL of an aqueous 0.1 M SrCl2 solution are mixed with 50. mL of an aqueous 0.1 M Na3PO4 solution? The mixed solution contains a multiply-charged cation, Sr2+(aq) and a multiply-charged anion, PO43–(aq), which react to form an insoluble precipitate, Sr3(PO4)2(s). The complete ionic equation representing the reaction that occurs in the mixed solution is: 3Sr2+(aq) + 6Cl–(aq) + 6Na+(aq) + 2PO43–(aq) Sr3(PO4)2(s) + 6Cl–(aq) + 6Na+(aq) (b) The complete ionic reaction shows that all the chloride ion from its original solution is still present in the mixture when the precipitation is complete. The number of moles of chloride and its mass may be calculated from the number of moles in its original solution volume:

0.050 L SrCl2(aq) = (0.050 L SrCl2(aq))

0.1 mol SrCl 2 1 L SrCl2 (aq )

–

2 mol Cl 1 mol SrCl 2

= 0.01 mol Cl–(aq) 0.01 mol Cl– = (0.01 mol Cl–)

35.45 g 1 mol Cl –

= 0.35 g Cl–

The concentrations of the solutions are only given with one significant digit, so we really should not report this result with two figures, but we will need the greater precision for the later calculation to show the mixed solution’s electrical neutrality. (c) The number of moles of each of the other ions added to the mixed solution is: 0.1 mol SrCl 2 0.050 L SrCl2(aq) = (0.050 L SrCl2(aq)) 1 L SrCl2 (aq )

2+

1 mol Sr 1 mol SrCl 2

= 0.005 mol Sr2+(aq) 0.1 mol Na 3 PO4 0.050 L Na3PO4(aq) = (0.050 L Na3PO4(aq)) 1 L Na 3 PO 4 (aq)

+

3 mol Na 1 mol Na 3PO4

= 0.015 mol Na+(aq)

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Aqueous Solutions and Solubility 0.1 mol Na 3 PO4 0.050 L Na3PO4(aq) = (0.050 L Na3PO4(aq)) 1 L Na 3 PO 4 (aq)

Chapter 2 3–

1 mol PO4 1 mol Na 3PO4

= 0.005 mol PO43–(aq) The equation in part (a) tells us that all of the Na+(aq) added to the mixture is still present after the precipitation reaction is complete. Now we need to find out whether any Sr2+(aq) or PO43– (aq) ions remain unreacted. Without doing any calculations, we can predict which ion will still be present after the precipitation is complete. Note that we found in part (b) that 0.01 mol of Cl– (aq) is present in the mixture and now we have 0.015 mol of Na+(aq) also left in the mixture. But the solution must be electrically neutral, so there must be more negative ions present. The only possibility is that some PO43–(aq) ions remain unreacted. Thus, it must be the case that Sr2+(aq) is the limiting reactant. Let’s see if that is the case and find out how much PO43–(aq) remains unreacted. The equation in part (a) tells us that the mole ratio of Sr2+(aq) to PO43–(aq) ions that react is 3:2. Thus, 1.5 (= 3 2 ) mol of Sr2+(aq) is required to react completely with 1 mol of PO43–(aq). Since

our mixed solution contains an equal number of moles of each ion, there is not enough Sr2+(aq) to react with all the PO43–(aq), so Sr2+(aq) is the limiting reactant in the mixture, as we reasoned in the previous paragraph. We use the ratio of moles of strontium ions in a mole of strontium phosphate to convert number of moles of strontium ion that react to moles of solid strontium phosphate precipitated and then to find the number of moles of PO43–(aq) left unreacted: 1 mol Sr3 (PO4 )2 = 0.0017 mol Sr3(PO4)2(s) 0.005 mol Sr2+ = (0.005 mol Sr2+) 3 mol Sr 2+ Since each mole of Sr3(PO4)2 contains two moles of PO43–, we have 0.0034 mol of PO43– removed from solution as Sr3(PO4)2(s). The amount of PO43–(aq) remaining in solution is 0.0016 mol [= (0.005 mol) – (0.0034 mol)]. (d) The total number of moles of negative charge on the remaining PO43–(aq) anions, from part (c), is 0.005 mol [= (3 charges·mol–1)(0.0016 mol)]. This amount of negative charge added to that from Cl–(aq), 0.01 mol, is equivalent to the positive charge from the Na+(aq), 0.015 mol, so the solution is electrically neutral, as we said in the reasoning in part (c). Problem 2.78. We are asked whether gases are very soluble in water. There is no simple answer to this question, because the answer depends upon the interactions of the gas with water. Non-polar gas molecules like nitrogen and oxygen have only limited solubility because they are so volatile and interact only weakly by dispersion forces with the water. At the other end of the solubility scale are gases like ammonia and hydrogen chloride that hydrogen bond strongly with and/or react with water to form quite concentrated solutions. Intermediate solubilities are shown by gases like carbon dioxide that have polar bonds (dipole-dipole attractions with water) and react somewhat with water (as CO2(aq) does to form small amounts of H2CO3(aq)). In all cases, the solubility of gases is temperature dependent and decreases as temperature increases. Problem 2.79. We are asked to predict whether the noble gases (He, Ne, Ar, Kr, and Xe) have a low solubility in water (less than 1 g·L–1) or a high solubility in water (greater than 10 g·L–1). Recall from Section 2.11 that only when water reacts with a gas will the gas have a very high solubility.

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Since the noble gases are all rather unreactive (that is why they are called noble -- aloof and not mingling well with the more “common” elements), they show no tendency to react with or bond with water. The electron distribution around the noble gas nuclei is spherically symmetric, which makes them non-polar and further suggests that they are insoluble in water. Problem 2.80. (a) We are asked to determine how many moles of nitrogen gas, N2(g), dissolve in 10.0 L of water at a temperature of 25 C and gas pressure of 101 kPa (one atmosphere). The data in Table 2.6 give the solubility of N2(g) under these conditions as 0.018 g·kg–1 (grams of gas per kilogram of water). The density of water at 25 C and one atmosphere pressure is 1.00 kg·L–1 [Chapter 1, Figure 1.29(a)], so 10.0 L of water is 10.0 kg of water and the mass of N2 (g) that dissolves in this much water is 0.18 g [= (0.018 g·kg–1)(10.0 kg)]. Thus, the number of moles of nitrogen dissolved in 10.0 L of water is:

0.18 g N2 = (0.18 g N2)

1 mol N2 28 g N2

= 6.4

10–3 mol N2

(a) We are asked to determine how many moles of oxygen gas, O2(g), dissolve in 0.100 L of water at a temperature of 25 C and gas pressure of 101 kPa (one atmosphere). The data in Table 2.6 give the solubility of O2(g) under these conditions as 0.039 g·kg–1. As in part (a), 0.100 L of water is 0.100 kg of water and the mass of O2 (g) that dissolves in this much water is 0.0039 g [= (0.039 g·kg–1)(0.100 kg)]. Thus, the number of moles of oxygen dissolved in 0.100 L of water is:

0.0039 g O2 = (0.0039 g O2)

1 mol O2 32 g O2

= 1.2

10–4 mol O2

Problem 2.81. The solubility of H2(g) in water is temperature dependent. The solubility at 25 C is 7.68 10–4 mol L–1 and, at 0 C, the solubility is 9.61 10–4 mol L–1. Note that the solubility of this nonpolar molecule is quite low at both temperatures. There is evidently very little attraction of water molecules for hydrogen molecules. There are only dispersion interactions and with only two electrons, it is difficult to induce dipoles in hydrogen molecules. Since there is very little attraction between H2 and H2O, perhaps the explanation for the temperature dependence of the solubility lies in the motion of H2 at different temperatures. At 25 C, H2 are moving faster than 0 C. Thus, more molecules will have enough energy to escape into the gas phase at 25 C than at 0 C. This means that the solubility of H2 should be greater at 0 C than at 25 C, as observed. Problem 2.82. The solution that is formed when hydrogen bromide gas, HBr(g), dissolves in water to form an acidic solution is called hydrobromic acid, just as the analogous solution of HCl(g) is called hydrochloric acid, Problem 2.83. A saturated solution of HBr(g) in water (hydrobromic acid) is approximately 8.9 M in HBr(aq) and the density of the solution is about 1.5 kg·L–1. We are asked to express the solubility of

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HBr(g) in units of g·kg–1. The density of the solution tells us that a liter of the solution has a mass of 1.5 kg and we know, from the molarity, that this liter contains 8.9 mol HBr. The molar mass of HBr is 80.9 g, so the mass of HBr dissolved in the liter of solution is: 80.9 g HBr 8.9 mol HBr = (8.9 mol HBr) = 7.2 102 g HBr = 0.7 kg HBr 1 mol HBr The mass of water in the solution is 0.8 kg [= (1.5 kg solution) – (0.7 kg solute)], so the concentration of HBr(aq) (in units of g·kg–1)is: 7.2 10 2 g HBr = 9 102 g·kg–1 0.8 kg water Note that the mass of HBr(g) that dissolves in a kilogram of water is greater than the mass of HCl(g) that dissolves in water, 695 g·kg–1, but the number of moles of HBr(g) that dissolve, about 11, is only about half as much as the moles of HCl(g) that dissolve, about 19. On a molar basis, HCl(g) is much more soluble. Problem 2.84. (a) We are asked to illustrate each of the potential interactions that might explain the high solubility of HCl(g) in diethyl ether. For simplicity, diethyl ether, C2H5O C2H5, is written as R– O–R in these sketches showing (i) dipole-dipole, (ii) hydrogen bonding, and (iii) ionization of HCl in ether solution:

(b) An experiment that might be done to eliminate or confirm one or more of these possibilities is testing the electrical conductivity of the solution, If it conducts, then we know that at least some HCl ionization, interaction (iii), must occur. This does not rule out contributions to the solubility from the other interactions. If the solution does not conduct, ionization is ruled out as a contributor to the solubility. The other interactions are hard to distinguish experimentally. NOTE: Stephen Branz, San Jose State University, has provided an interesting discussion of the possible reasons for the high solubility of HCl(g) in diethyl ether. If ionization reactions are primarily responsible for the solubility of HCl(g), we would expect the dielectric strength of the solvent to greatly affect the solubility. Solubility data for HCl(g) in water, MeOH, EtOH, and Et2O are from the Merck Index, 9th ed., and all other data are from standard reference tables.

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Solvent

Solubility g·(100 mL)–1 at 30 °C

Dielectric Constant

water 67.3 78.4 methanol 43.0 33.6 ethanol 38.1 25.0 diethyl ether 19.5 4.3 benzene “soluble” * 2.3 pentane n/a # 1.8 chloroform n/a # 4.8 * no quantitative data available; several sources report “soluble.” # presumably very low Although there is indeed a reduction in the solubility of HCl going from water to the alcohols and ether, the differences are (qualitatively) not reflective of the differences in dielectric constants. Another way to look at the data is to calculate the solubility relative to the number of oxygen atoms in a 100 mL sample of the solvent. Solvent

g HCl / mol “O” (in 100 mL solvent)

water methanol ethanol diethyl ether

12.1 17.4 22.2 20.3

Assuming that there is a specific interaction between HCl and the oxygen of the solvent (and this might be anywhere from formation of the oxonium ion, R2OH+, to a strong hydrogen bond), it is justifiable to consider the solubility relative to the number of oxygen atoms in the solvent. By this measure, of the four oxygenated solvents, water is the worst solvent for HCl! A (perhaps naive) qualitative explanation is that of the four solvents in the table, only water has two H’s and two lone pairs for hydrogen bonding (or oxonium ion formation). Water must sacrifice H bonds (among water molecules) in order to interact with/ionize HCl. The other three solvents each have an “excess of lone pairs” and can more effectively solvate the HCl. The foregoing discussion is based on correlations with no physical data (such as conductivity, for which we have found no data) to back it up, but the correlation is quite good. Other support for a possible specific interaction (or interactions) between HCl and ether molecules should be noted: (a) Although we don’t have quantitative data for benzene, strong hydrogen bonding to the pi system is quite well documented and undoubtedly accounts for the high solubility of HCl in benzene. (b) The boiling points of ether and pentane (35-36 °C) are nearly identical, reflective of their very similar sizes and shapes. If the dipole moment of ether were of great importance in interactions with itself or other molecules, we would predict that its boiling point would be higher than that of pentane.

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(c) There are many well-known organic reactions that support the formation of low concentrations of ionic species in non-polar solvents (for example, electrophilic aromatic substitution). (d) Some non-polar solvents can be surprisingly good solvents for ionic species (for example, chloroform dissolves very high concentrations of simple organic ammonium chlorides even though those same salts are quite insoluble in ether -- these solvents have very similar dielectric constants). Problem 2.85. We would expect hydrogen chloride gas, HCl(g), to be less soluble in hexane than in water. Hydrogen chloride can react with water to form hydronium cations and chloride anions, equation (2.16). This reaction with the solvent increases the solubility of HCl(g) in water. There is no parallel reaction of HCl with hexane, CH3CH2CH2CH2CH2CH3, a non-polar solvent, so its hexane solubility is much lower. Problem 2.86. (a) The data in Table 2.6 are for the solubility of the gases at 101 kPa (one atmosphere) pressure. Assume that the amount of a gas that dissolves in water is directly proportional to its pressure over the solution; the lower the pressure, the less gas dissolved. (This is Henry’s Law for the solubility of non-polar gases in liquids.) In air, the nitrogen pressure is 80% of one atmosphere, so only 80% as much nitrogen will dissolve: (0.80)[0.018 g·(kg water)–1] = 0.014 g·(kg water)–1. Oxygen pressure is 20% of one atmosphere: (0.20)[0.039 g·(kg water)–1] = 0.008 g·(kg water)–1. (b) In one kilogram of water saturated with gases from the air, there are 0.014 g of nitrogen and 0.008 g of oxygen, or 0.022 g total of these gases. The mass percent of oxygen is:

0.008 g O2 100% = 36% 0.22 g air (c) In a mole of air (Avogadro’s number of nitrogen and oxygen molecules), 80% of the molecules are nitrogen and 20% are oxygen. The mass of nitrogen in a mole of air is (0.80 mol)(28 g·mol–1) = 22.4 g. The mass of oxygen is (0.20 mol)(32 g·mol–1) = 6.4 g. The total mass of a mole of air is 28.4 g and the mass percent of oxygen in air is:

6.4 g O2 100% = 22% 28.8 g air The mass percent of oxygen in the gases dissolved in water is greater than the mass percent of oxygen in the (major) gases in the atmosphere. Problem 2.87. An acid is a compound whose aqueous solution has a pH below 7. Problem 2.88. A base is a compound whose aqueous solution has a pH above 7. Problem 2.89. An acid solution that has a pH of 1 contains a higher concentration (100 times higher) of hydronium ions than an acid solution that has a pH of 3. From the data given, this is the only

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conclusion we can draw. We have no way to characterize acid A relative to acid B, because we do not know the concentrations of acid A and acid B in the solutions. Problem 2.90. Aqueous solutions with pH < 7, that is [H3O+(aq)] > 1.0 10-7 M and [OH–(aq)] < 1.0 10-7 M, are acidic. Aqueous solutions with pH > 7, that is [H3O+(aq)] < 1.0 10-7 M and [OH–(aq)] > 1.0 10-7 M, are basic. Aqueous solutions with pH = 7, that is [H3O+(aq)] = 1.0 10-7 M and [OH–(aq)] = 1.0 10-7 M, are neither acidic nor basic. Each of these solutions is characterized as acidic, basic, or neither on the basis of the definitions just stated. (a) pH < 7 acidic + -7 (b) [H3O (aq)] = 1.0 10 M neither – -7 (c) [OH (aq)] > 1.0 10 M basic (d) pH > 7 basic (e) [H3O+(aq)] > 1.0 10-7 M acidic (f) [OH–(aq)] < 1.0 10-7 M acidic + -7 (g) [H3O (aq)] < 1.0 10 M basic – -7 (h) [OH (aq)] = 1.0 10 M neither Problem 2.91. We calculate the pH of each of these solutions using the definition of pH [equation (2.18)]: pH = –log10[H3O+(aq)]. This is very easy to do on any scientific calculator, because they all have a “log” key that returns the logarithm to the base 10, log10, of whatever value is displayed. That is how these results were calculated. The number of significant digits following the decimal point in a pH value is determined by the number in the pre-exponential factor in the concentration value. In all the cases here, the pre-exponential has two digits and the pH has two digits after the decimal point. (a) [H3O+(aq)] = 1.0 10-2 M pH = 2.00 + -10 (b) [H3O (aq)] = 1.0 10 M pH = 10.00 + -4 (c) [H3O (aq)] = 5.0 10 M pH = 3.30 + -8 (d) [H3O (aq)] = 5.0 10 M pH = 7.30 Problem 2.92. If the reaction of HCl(g) and water goes to completion to form H3O+(aq) and Cl–(aq), the molar concentration of HCl(aq) in the solution is the same as the molar concentration of H3O+(aq). In this problem, we are asked to find the [HCl(aq)], which is equal to [H3O+(aq)], that will result in a solution of a given pH. Use Figure 2.24 to find the [H3O+(aq)] that corresponds to a given pH and that will also be the appropriate [HCl(aq)]. (a) For a pH = 4, Figure 2.24 shows that [H3O+(aq)] = 10–4 M, so we need [HCl(aq)] = 10–4 M. (b) For a pH = 2, Figure 2.24 shows that [H3O+(aq)] = 10–2 M, so we need [HCl(aq)] = 10–2 M. Problem 2.93. By analogy with HCl(g) being named hydrogen chloride and HCl(aq) named hydrochloric acid, we can name other gaseous hydrides that dissolve to form acidic solutions: HI(g) is hydrogen iodide. HI(aq) is hydroiodic acid.

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H2S(g) is hydrogen sulfide. H2S(aq) is hydrosulfuric acid. Problem 2.94. The two movies in the Web Companion, Chapter 2, Section 2.12, page 3, are an attempt to show in animation what the figure at the bottom of that page and Figure 2.26 in the textbook try to illustrate. What the figures and movies try to show is how hydronium cations and hydroxide anions “migrate” in aqueous solution by transfer of a hydrogen ion, H+ (a proton), from one molecule or ion to another ion or molecule. Four frames from the hydronium ion movie are shown here:

In the first frame (on the left), one of the hydrogens from the hydronium at the far left of the frame is hydrogen bonded to the adjacent water molecule and by a small movement of this hydrogen atomic core, the hydronium ion has transferred a proton to form a new hydronium ion (center of second frame). This new hydronium ion goes on in frames three and four to transfer a proton (not the one it got in the previous transfer) to a third water molecule. Although the water molecules and hydronium are moving about, they don’t move far during these transfers, but a hydronium ion has moved almost all the way across the frame and, in the movie, is shown finally bonded to the oxygen atom on the far right of the frame. A similar sequence of proton transfers is shown in the hydroxide ion movie where transfers of a proton from water to hydroxide result in the movement of a hydroxide ion. [Theoretical calculations and experimental reaction rates are consistent with this mechanism for the migration of hydronium and hydroxide ions through aqueous solutions. It is only a matter of time until very fast spectroscopic methods will be able to observe events like these “directly” to see if they actually occur as we imagine.] Problem 2.95. (a) The balanced chemical reaction equation for the reaction of phosphorus pentoxide, P2O5(s), with water to form a solution of phosphoric acid, (HO)3PO(aq) (or H3PO4(aq)) is: P2O5(s) + 3H2O(l) 2H3PO4(aq) (b) To find the molarity of the phosphoric acid solution formed when 1.42 g of phosphorus pentoxide is mixed with 250. mL of water, we need to convert moles of phosphorus pentoxide (molar mass = 142 g) to moles of phosphoric acid, using the reaction equation in part (a), and then divide by the volume of solution, in liters, to get the molarity. We assume that the volume of liquid does not change significantly when this small amount of phosphorus pentoxide is added and reacted.

1.42 g P2O5 = (1.42 g P2O5) = 8.0

50

1 mol P2 O5 142 g P2 O5

2 mol H3 PO 4 (aq) 1 mol P2 O5

10-2 M H3PO4(aq)

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1 0.250 L

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Aqueous Solutions and Solubility

Problem 2.96. The name is given for each of these ionic compounds with oxyanions. The arsenate oxyanion, AsO43–, is named by analogy with its phosphorus analog, the phosphate oxyanion, PO43–. (a) Ca(HSO4)2 calcium hydrogen sulfate (b) Na2CO3 sodium carbonate (c) Al2(HPO4)3 aluminum hydrogen phosphate (d) Ce2SO4 cesium sulfate (e) KHCO3 potassium hydrogen carbonate (f) Na3AsO4 sodium arsenate Problem 2.97. The Lewis structures (showing all nonbonding electron pairs as a pair of dots and all covalent bonds as lines) for the nitrate, ethanoate (acetate), and hydrogen sulfate oxyanions are: O

O O

H O

C

C

N

H

O O

H

O

S

O

H

O

nitrate ethanoate hydrogen sulfate The hydrogen sulfate oxyanion is shown with six bonding electron pairs around the central sulfur atomic core, as in Table 2.7. The marginal note on page 126 of the textbook indicates that, in Chapters 5 and 6, we will find that third (and higher) period elements often accommodate more than four bonding electron pairs and that the structure with six bonding electron pairs and several double bonds seems to fit the properties of the oxyanion better than a structure with four bonding pairs (an octet). Problem 2.98. Each Brønsted-Lowry acid and base is identified with an A or B, respectively, below the species in these reactions. When necessary, the reaction in the problem is written out in is complete ionic form to make identification of the acid and base possible. (a)

H2S(g) + H2O(l) A B

HS–(aq) + H3O+(aq) B A

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) + – Na (aq) + OH (aq) + H3O+(aq) + Cl–(aq) Na+(aq) + Cl–(aq) + H2O(l) + H2O(l) B A B A The Na+(aq) and Cl–(aq) ions are spectator ions in this reaction, which, in aqueous solution, is just a reaction between hydronium cation and hydroxide anion to produce two molecules of water. Water can act as either a Brønsted-Lowry acid or base, depending upon whether it is a proton donor or proton acceptor in the reaction. (b)

NH3(g) + HCl(g) NH4+Cl–(s) B A A B In this case, the molecule, HCl(g) is the acid (proton donor). This is unlike part (b), where HCl(aq) does not exist as a molecule in aqueous solution, but as the hydronium cation and chloride anion. (c)

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Problem 2.99. Each Brønsted-Lowry acid and base is identified with an A or B, respectively, below the species in these reactions. When necessary, the reaction in the problem is written out in is complete ionic form to make identification of the acid and base possible. (a)

NO2-(aq) + H3O+(aq) B A

HNO2(aq) + H2O(l) A B

(b) 2H3O+(aq) + 2ClO4–(aq) + Mg2+(OH–)2(s) Mg2+(aq) + 2ClO4–(aq) + 4H2O(l) 2H3O+(aq) + 2ClO4–(aq) + Mg2+(OH–)2(s) Mg2+(aq) + 2ClO4–(aq) + 2H2O(l) + 2H2O(l) A B B A In this reaction, the hydronium ion from a perchloric acid, HOClO3(aq), solution reacts with the hydroxide ion in solid magnesium hydroxide, Mg2+(OH–)2(s). The magnesium hydroxide formula is written out this way to make explicit the presence of the ions in the solid. Like the perchlorate anion, the magnesium cation is essentially a spectator ion in this reaction. See the solution for Problem 2.98(b) for a discussion of the role of water molecules.

HNO3(aq) + Al3+(OH–)3(s) Al3+(aq) + 3NO–4(aq) + 3H2O(l) + 3H2O(l) 3H3O+(aq) + 3NO3–(aq) + Al3+(OH–)3(s) A B B A This reaction is modeled after the one in part (b), which was written out in full as a model. (c)

(d) HCN(aq) + NaOH(aq) Assuming that hydrocyanic acid, HCN(aq), like hydrochloric acid, transfers a proton to a water molecule to form H3O+(aq) and CN–(aq), we can write the reaction equation as: H3O+(aq) + CN–(aq) + Na+(aq) + OH–(aq) Na+(aq) + CN–(aq) + H2O(l) + H2O(l) A B B A

[At this point in this chapter, weak acids have not been introduced, so we are not forced to consider whether the reaction could also be written with the aquated hydrogen cyanide molecule, HCN(aq), as the acid: HCN(aq) + Na+(aq) + OH–(aq) Na+(aq) + CN–(aq) + H2O(l) A B B A This, of course, changes the interpretation of the reaction somewhat and makes it clearer, as will be discussed in Chapter 6, that there is a competition between OH–(aq) a CN–(aq) for protons in this solution.] Problem 2.100. The Lewis structures (showing all nonbonding electron pairs as a pair of dots and all covalent bonds as lines) for HOCO2–, (HO)2PO2–, and HOPO32– (modeled on the structures for the corresponding oxyacids in Table 2.7) are: O O

C

H

O

H

hydrogen carbonate

52

O

P

2

O

O O

H

O

dihydrogen phosphate

O

P

O

H

O

monohydrogen phosphate

ACS Chemistry Chapter 2 suggested solutions

Chapter 2

Aqueous Solutions and Solubility

Problem 2.101. (a) The Brønsted-Lowry acids and bases in two possible acid-base reactions of ammonia and water are identified by placing an A below each acid and a B below each base. H2O(l) + NH3(g) H3O+(aq) + NH2–(aq) B A A B

OH–(aq) + NH4+(aq) H2O(l) + NH3(g) A B B A The ammonia and water molecules switch roles in these two possible reactions. (b) We predict that the second equation represents the correct acid-base reaction. Because oxygen is more electronegative than nitrogen, the O–H bonding electrons in water are held more closely by the oxygen than the N–H bonding electrons in ammonia. Thus, in comparison with the H+ in the N–H bond, the H+ in the O–H bond can interact more effectively with non-bonding electron pairs on other molecules. The result is that H+ from a water molecule is transferred to a Brønsted-Lowry base more readily than the H+ from an ammonia molecule. If the second reaction is the correct one, then solutions of NH3 in water will have extra OH–(aq) and will be basic. The fact that the pH > 7 in aqueous ammonia solutions indicates that there is hydroxide present in concentrations greater than found in neutral water (pH = 7) and confirms our prediction. Problem 2.102. To explain the electrical conductivity observed in a solution of methylamine, CH3NH2(g), dissolved in water, we need to consider how ions are formed in the interaction of the methylamine and water molecules. By analogy with the reaction of ammonia with water, we can write: CH3NH2(aq) + H2O(l) CH3NH3+(aq) + OH–(aq) Since the electrical conduction by the solution is weak, we conclude that only a few ions are produced, that is, the reaction does not proceed far toward products. We indicate this with an equilibrium arrow in the balanced equation to show that the system comes to equilibrium before all the methylamine has reacted. Problem 2.103. To predict/decide whether an aqueous solution of ethylene glycol, HOCH2CH2OH will be basic, acidic, or neutral and whether it will conduct an electric current, we need to consider how the molecules will interact with water molecules. Other compounds we know about with –OH groups bonded to carbon are alcohols, like methanol (CH3OH) and ethanol (CH3CH2OH), and compounds with multiple –OH groups, like glucose (se Figure 2.6 in the textbook). These compounds are miscible with water (the alcohols) or very soluble (glucose), so it is not surprising that ethylene glycol, a small molecule with two –OH groups, should be miscible with water. The reason all these compounds are so soluble is that they can form several hydrogen bonds with water and the non-polar part of the molecule does not disrupt the water structure enough to be a major factor in their solubility. For all these molecules, including ethylene glycol, the interactions with water do not involve any proton transfers that would form ionic products, so the solutions do not conduct an electric current and are not acidic or basic.

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Problem 2.104. Our task in this problem is to use Le Chatelier’s principle to predict and clearly explain the outcome of changes in the reaction conditions for this equilibrium reaction system: CH3C(O)OH + HOCH2CH3 CH3C(O)OCH2CH3 + H2O acetic acid ethanol ethyl acetate (a) Starting with 0.1 mole of acetic acid and 0.1 mole of ethanol, would more, less, or the same amount of ethyl acetate be formed, if water is added to the reaction mixture? Addition of water is a disturbance to the equilibrium system, which will respond by reacting to use up some of the added water. The reaction will proceed from the right to the left, using up both water and ethyl acetate. The amount of ethyl acetate will be less when the extra water is added. (b) Would a mixture of 0.2 mole of acetic acid and 0.1 mole of ethanol form more, less, or the same amount of ethyl acetate as a mixture of 0.1 mole of acetic acid and 0.1 mole of ethanol? Addition of acetic acid (0.2 mol instead of 0.1 mol) is a disturbance to the equilibrium system, which will respond by reacting to use up some of the added acetic acid. The reaction will proceed from the left to the right, using up both acetic acid and ethanol and forming more ethyl acetate. The amount of ethyl acetate will be more when the extra water is added. Problem 2.105. Our task in this problem is to use Le Chatelier’s principle to predict and/or clearly explain the outcome of the changes in the reaction conditions for two equilibrium reaction systems: (a) To explain why the solubility of carbon dioxide is greater in an aqueous sodium hydroxide solution, Na+(aq) + OH–(aq), than in water itself, we need to examine the reactions of dissolved carbon dioxide and see how increasing the hydroxide anion concentration would affect them. The dissolution equilibrium reactions are: CO2(g) + H2O(l) (HO)2CO(aq)

H3O+(aq) + HOCO2–(aq) (HO)2CO(aq) + H2O(l) Both of the products of the second reaction, H3O+(aq) and HOCO2–(aq), can react with OH–(aq) to “use them up”: H3O+(aq) + OH–(aq) 2H2O(l) HOCO2–(aq) + OH–(aq) H2O(l) + CO32–(aq) These two equations are also reversible and could be shown with equilibrium arrows, but the directional arrow is used to show how the H3O+(aq) and HOCO2–(aq) are used up by added OH– (aq) anion. Removing the two products of the dissolution reactions is a disturbance to the system, which reacts by forming more of these products (which are, in turn, used up until all the hydroxide anion has reacted). The direction of the response to the disturbance is shown by this series of reactions with directional arrows: CO2(g) + H2O(l) (HO)2CO(aq) (HO)2CO(aq) + H2O(l)

H3O+(aq) + HOCO2–(aq)

H3O+(aq) + OH–(aq) 2H2O(l) HOCO2–(aq) + OH–(aq) H2O(l) + CO32–(aq) The overall reaction, in the presence hydroxide anion is the sum of these four reactions: CO2(g) + 2OH–(aq) H2O(l) + CO32–(aq) 54

ACS Chemistry Chapter 2 suggested solutions

Chapter 2

Aqueous Solutions and Solubility

(b) To predict what will happen to the concentration of calcium cation, [Ca2+(aq)], if sodium sulfate (solid), Na2SO4(s), is added to a saturated aqueous solution of sparingly soluble calcium sulfate, we need know what equilibrium reaction might be disturbed and what happens to the added sodium sulfate. Sodium sulfate is a soluble ionic compound, so it will dissolve in the solution to give Na+(aq) and SO42–(aq) ions. The equilibrium reaction in a saturated solution of calcium sulfate is the equilibrium dissolution of the solid ionic compound: CaSO4(s) Ca2+(aq) + SO42–(aq) Sodium sulfate is a soluble ionic compound, so it will dissolve in the solution to give Na+(aq) and SO42–(aq) ions. The added SO42–(aq) anions are a disturbance to the equilibrium system and the system will respond by trying to use up some of the ions by reacting to form more CaSO4(s). This requires using up some of the Ca2+(aq) cation, so we predict that [Ca2+(aq)] will decrease when Na2SO4(s) is added to a saturated aqueous solution of CaSO4(s). Problem 2.106. We are to represent each of these statements as a complete balanced chemical equation: (a) Carbonic acid is formed when carbon dioxide reacts with water. CO2(g) + H2O(l) (HO)2CO(aq) (b) Calcium carbonate (limestone) reacts with carbonic acid to form an aqueous solution of calcium hydrogen carbonate. CaCO3(s) + (HO) 2CO(aq) Ca2+(aq) + 2HOCO2–(aq) (c) Calcium hydrogen carbonate reacts with calcium hydroxide to form calcium carbonate precipitate. [The implication of this statement is that one or both of the reactants is in aqueous solution. Calcium hydroxide is a sparingly soluble ionic compound, so we might write the reaction of solid calcium hydroxide with a solution of calcium hydrogen carbonate. We’ll do this as well as write the reaction for both ionic compounds in solution.] Ca2+(aq) + 2HOCO2–(aq) + Ca(OH)2(s) 2CaCO3(s) + 2H2O(l)

Ca2+(aq) + 2HOCO2–(aq) + Ca2+(aq) + 2OH–(aq)

2CaCO3(s) + 2H2O(l)

(d) Calcium hydrogen carbonate reacts with sodium hydroxide to form calcium carbonate precipitate and the water-soluble salt sodium carbonate. [Assume both ionic compounds are in aqueous solution.] Ca2+(aq) + 2HOCO2–(aq) + 2Na+(aq) + 2OH–(aq) 2CaCO3(s) + 2Na+(aq) + CO32–(aq) + 2H2O(l) (e) The mixing of aqueous solutions of sodium hydrogen carbonate and sodium hydroxide is exothermic. A reaction has occurred but there is no precipitate. Na+(aq) + HOCO2–(aq) + Na+(aq) + OH–(aq) 2Na+(aq) + CO32–(aq) + H2O(l) Since the reaction is exothermic, you may show “energy” as a product, if you wish. In this case, we are given this information, because there is no visible evidence (precipitate, gas formation, color change, etc.) that a reaction has occurred. Problem 2.107. (a) As a droplet of solution containing calcium cations and hydrogen carbonate anions evaporates from the tip of a stalactite, the concentration of these solute ions increases. The

ACS Chemistry Chapter 2 suggested solutions

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volume of solution is less, so the number of ions per unit volume is greater, that is, a higher concentration. (b) The ions present in the droplet of solution in part (a) were a result of reaction (2.40): Ca2+(aq) + 2HOCO22–(aq) CaCO3(s) + H2O(l) + CO2(g) The reaction is reversible and is shown here as an equilibrium. The increased concentrations of the products of reaction (2.40), which were noted in part (a), are a disturbance to the system. Le Chatelier’s Principle states that the system will respond by minimizing the disturbance, that is, by reacting in a way that decreases the concentrations. The reaction that does this is the reverse of reaction (2.40), which will proceed to precipitate solid calcium carbonate and release carbon dioxide gas into the surrounding air in the cave. (c) The result of the reaction in part (b) is to precipitate some solid calcium carbonate, which causes the stalactite to grow downward (very slowly). If the drop of liquid should drop off the tip of the stalactite, it will end up directly below and the same evaporation and precipitation process on the floor of the cave will build stalagmites upward toward the stalactites. If they meet, they form a column, which is another common structure in limestone caves. Problem 2.108. (a) We are asked what gas is evolved when calcium carbonate, a solid, is placed in an aqueous solution of HCl(g). We know that HCl(g) dissolves and reacts completely with water to produce hydronium cations and chloride anions (hydrochloric acid solution): HCl(g) + H2O(l) H3O+(aq) + Cl–(aq) Calcium carbonate is a sparingly soluble solid whose dissolution in water is represented by this equilibrium reaction: (i) CaCO3(s) Ca2+(aq) + CO32–(aq) The carbonate ion product of this dissolution is a Brønsted-Lowry acid that can react with the hydronium cation in the solution [from the dissolved HCl(g)] to form hydrogen carbonate anion and this anion can react further with another hydronium ion to form carbonic acid: (ii) CO32–(aq) + 2H3O+(aq) (HO)2CO(aq) + 2H2O(l) We know that carbonic acid is formed in water when carbon dioxide gas dissolves and we can write this equilibrium reaction in reverse to show how carbon dioxide gas is released by the decomposition of carbonic acid: (iii) (HO)2CO(aq) CO2(g) + H2O(l) This series of reactions shows that carbon dioxide is likely to be the gas evolved when calcium carbonate is placed in hydrochloric acid solution. (b) The net ionic equation for the reaction that produces the gas when solid calcium carbonate is placed in hydrochloric acid solution is the sum of reactions (i), (ii), and (iii) in part (a): CaCO3(s) + 2H3O+(aq) Ca2+(aq) + CO2(g) + 3H2O(l) (c) Only a tiny amount of calcium carbonate (marble, chalk) dissolves in water, which is represented by equation (i) in part (a). The addition of hydronium ion [from the dissolved HCl(g)] is a disturbance to this solubility equilibrium because the hydronium ion reacts with one of the products, carbonate anion and reduces its concentration. In response, LeChatelier’s principle, more of the solid calcium carbonate dissolves to try to increase the carbonate anion

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concentration. If there is a good deal of hydronium present, it continues to react (and the calcium carbonate continues to dissolve) forming more and more carbonic acid, because the presence of a lot of hydronium ion displaces equilibrium equation (ii) toward products. The increase in concentration of carbonic acid is a disturbance to equilibrium reaction (iii), which responds, LeChatelier’s principle, by trying to use it up to form products, one of which, carbon dioxide gas, we observe as it leaves the reaction solution. Problem 2.109. The molecular level representation of sucrose in the Web Companion Chapter 2, Section 2.2, page 3 shows that sugar molecules have several regions of positive and negative charge, which are due to the presence of a number of –OH groups that are shown in the molecular structure on this page. Thus, in its solid crystals, the sugar molecules are oriented in specific directions, as shown in the movie, that maximize their positive-negative attractions and hydrogen bonding between molecules. When you put pressure on the crystal (by trying to crush it, for example), these many oriented interactions strongly resist being reoriented and the crystal breaks into smaller pieces as some of the interactions are finally broken. The molecular level representation of grease on page 5, shows that the molecules are long chains of hydrocarbons that have essentially no polarity. Attractions of these molecules for one another are through nondirectional dispersion (induced dipole) interactions and their packing in the solid (or semisolid) form is rather random, as shown in the movie. Thus, when pressure is applied, the hydrocarbon molecules can slide over each other because their attractions are non-directional and any orientation of the molecules with respect to one another is as good as any other. Problem 2.110. (a) The dissolution of calcium sulfate, CaSO4(s), in water can be represented as: Ca2+(aq) + SO42–(aq) CaSO4(s) About 2 g of calcium sulfate dissolve in a liter of water, so we can calculate the number of moles that dissolve in a liter, using the molar mass of CaSO4(s) (= 136 g·mol–1), and from the stoichiometry of the dissolution reaction find the molarities of the ions in the saturated solution: 1 mol CaSO 4 2 (g CaSO4)·L = [2 (g CaSO4)·L ] 136 g CaSO4 –1

–1

2+

1 mol Ca 1 mol CaSO4

= [Ca2+(aq)]

= 0.015 mol·L–1 The stoichiometry of the dissolution shows that [Ca2+(aq)] = [SO42–(aq)] = 0.015 mol·L–1. Thus, a saturated solution of calcium sulfate is 0.015 M in both ions. From the table in Consider This 2.54, we find that seawater contains 0.010 M Ca2+(aq) and 0.028 M SO42–(aq). Seawater contain less calcium ion than in a saturated solution of calcium sulfate, but almost twice as much sulfate ion as in the saturated calcium sulfate solution. It’s hard to tell from these data whether seawater is saturated with calcium sulfate, because, at this point in the textbook, our analysis of solubility has not provided a way to handle a situation like this. [The remainder of the problem is designed to help get a qualitative feeling for what is going on and Chapter 9 will provide a quantitative way to analyze solubilities in a solution like seawater.] (b) Solubility equilibria, like the one represented in the equation in part (a), should respond (Le Chatelier’s principle) to the disturbance caused by adding more of the cation or anion (by adding a soluble ionic compound containing the cation or anion, for example). Since the ions are on the product side of the solubility equilibrium, the system will react by trying to remove ACS Chemistry Chapter 2 suggested solutions

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some of the added ion by formation of more solid. Removing some of the added ion also, of course, removes some of the other ion, the counterion. In the case of the calcium sulfate in this problem, addition of SO42–(aq) (by adding some sodium sulfate, Na2SO4, for example) to a saturated solution of calcium sulfate will result in formation of a little more solid calcium sulfate and a decreased concentration of Ca2+(aq). Looked at from another point of view, the amount of calcium sulfate that will dissolve in a solution that already contains SO42–(aq) will be lower, the solid will be less soluble, than in pure water. A saturated solution under these conditions will contain a lower concentration of Ca2+(aq) than a saturated solution in pure water. The ionic compound will be less soluble than it would be in pure water because Le Chatelier’s principle favors the precipitation (or decreased solubility). (c) Seawater contains more sulfate ion than is present in the saturated solution of calcium sulfate we calculated in part (a). Since seawater contains more of the anion than could have been present from CaSO4(s) alone, we can imagine that the solution is formed when CaSO4(s) dissolves in a solution that already contains some of the anion. From part (b), we find that the solubility of the CaSO4(s) will be lower in such a solution. This is the direction of the effect we observe; there is less calcium cation in seawater than would be present in a saturated solution in pure water. Although we still cannot be certain, this analysis makes it more likely that we can conclude that seawater is saturated with CaSO4(s), in the presence of extra sulfate anion from some other source. [In Chapter 9, we will find that the product of the concentrations of the cation and anion characterizes the solubility equilibrium for a saturated solution of a sparingly soluble salt like CaSO4. For the saturated solution in water, we have [Ca2+(aq)][SO42–(aq)] = (0.015 M)(0.015 M) = 2.2 10–4 M2 For seawater, we have [Ca2+(aq)][SO42–(aq)] = (0.010 M)(0.028 M) = 2.8 10–4 M2 These two products are almost the same and reinforce the conclusion that seawater is saturated with CaSO4, in the presence of extra sulfate anion.] (d) Since seawater already contains a substantial concentration of calcium anion, the solubility of calcium carbonate will be lower than it would be in pure water. The solubility of calcium carbonate is very low in pure water, so lowering it even further helps explain why seashells do not redissolve (at least not rapidly) in the sea. Problem 2.111. We are asked to find the molarity of nitrate, NO3–(aq), anion in a sample of salt water with a density of 1.02 g·mL–1 that contains 17.8 ppm (parts per million, by mass) of NO3–(aq). If we can find the mass of NO3–(aq) in a known volume of the solution, we can use the molar mass of NO3–(aq) (= 62.01 g) to get the number of moles and then the molarity. The concentration, 17.8 ppm (by mass) means that exactly one million grams of solution contain 17.8 g of NO3–(aq). We can use this information to convert the density (mass per milliliter) to grams and then to moles and molarity:

1.02 g·mL–1 sol’n = (1.02 g·mL–1 sol’n) = 2.93

58

17.8 g NO–3 1 10 6 g sol'n

–

1 mol NO3 – 62.01 g NO3

10–4 mol·L–1

ACS Chemistry Chapter 2 suggested solutions

1000 mL 1L

Chapter 2

Aqueous Solutions and Solubility

Problem 2.112. (a) We are asked to find the molarity of sodium chloride, NaCl, in a solution of salt water (representing seawater) that has a density of 1.025 g·mL–1 and contains 3.50% by weight (mass) of NaCl. If we can find the mass of NaCl in a known volume of the solution, we can use the molar mass of NaCl (= 58.44 g) to get the number of moles and then the molarity. The concentration, 3.50% by mass means that exactly one hundred grams of solution contain 3.50 g of NaCl. We can use this information to convert the density (mass per milliliter) to grams and then to moles and molarity:

1.025 g·mL–1 sol’n = (1.025 g·mL–1 sol’n)

3.50 g NaCl 1 10 2 g sol'n

1 mol NaCl 58.44 g NaCl

1000 mL 1L

= 0.614 mol·L–1 (b) In Consider This 2.54(b) we found that the sum of the concentrations of positive charges in seawater is 0.606 M (and for negative charges 0.603 M). The result in part (a) is consistent (within about 1.3%) with the answer to Consider This 2.54(b), because the solution in part (a) has a 0.614 M concentration of positive charge from the Na+ cation (and the same concentration of negative charge from the Cl– anion). Problem 2.113. We are given a good deal of information about a solution prepared by a student. The solution contains 5.15 g of a compound dissolved in 10.0 g of water, has a density of 1.34 g·mL–1, and is 2.7 M. We are asked to determine which of these ionic compounds is the solute: (NH4)2SO4, KI, CsCl, or Na2S2O3. If we can find out how many moles of solute are dissolved, we can determine its molar mass and compare the result with the molar masses of the possible solutes. The known molarity of the solution, 2.7 M, is equal to the number of moles dissolved divided by the volume (liters) of the solution. To get the volume of solution, we divide the total mass of the solution 15.15 g (= 5.15 g + 10.0 g) by its density:

15.15 g sol’n = (15.15 g sol’n)

1 mL sol'n 1.34 g sol'n

1L 1000 mL

Therefore, using the molarity of the solution, we have: 2.7 mol solute 0.0113 L sol’n = (0.0113 L sol’n) 1 L sol'n

= 0.0113 L sol’n

= 0.0305 mol solute

It is not really legitimate to carry the third significant figure in this result, since the molarity has an uncertainly of about 3% (1 part in 27), but we’ll carry it along and check whether it makes a difference later. Now we know that 5.15 g of the solute is 0.0305 mol of the solute, so we can find its molar mass: 5.15 g solute molar mass = = 169 g·mol–1 0.0305 mol solute The molar masses of the four possible compounds are: (NH4)2SO4 , 132 g·mol–1; KI, 166 g·mol– 1 ; CsCl, 168 g·mol–1; and Na2S2O3, 158 g·mol–1. The molar mass we calculated has about a 3% uncertainty from the uncertainty in the second significant figure of molarity, so the range of possibilities is about 164 to 174 g·mol–1. The possibilities, therefore, are KI or CsCl for the compound the student dissolved. The calculation fits CsCl best, but the uncertainty makes this identification unsure. We would need a better value for the molarity to be sure. ACS Chemistry Chapter 2 suggested solutions

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Problem 2.114. We are asked to find the approximate percentage of the water molecules that is protonated in aqueous solutions with different pHs. We can use Figure 2.24, or the relationship, [H3O+(aq)] = 10–pH, to find the molarity of hydronium ion (protonated water molecules) in each solution and compare this molarity to the molarity of the water, 55.5 M, to get the fraction/percentage of the water molecules that are protonated. (a) At pH 7, we have [H3O+(aq)] = 10–7 M, so the percent of molecules protonated is:

% H2O protonated =

10

–7

+

mol [H3 O (aq)] 100% = 2 55.5 mol H2 O(l )

10–7 %

(b) At pH 6, we have [H3O+(aq)] = 10–6 M, so the percent of molecules protonated is:

% H2O protonated =

10

–6

+

mol [H3 O (aq)] 100% = 2 55.5 mol H2 O(l )

10–6 %

(a) At pH 4, we have [H3O+(aq)] = 10–4 M, so the percent of molecules protonated is:

% H2O protonated =

10

–4

+

mol [H 3O (aq )] 100% = 2 55.5 mol H2 O(l )

10–4 %

Problem 2.115. (a) Figure 2.24 shows that [OH–(aq)] = 10–11 M and [H3O+(aq)] = 10–3 M in aqueous solution at pH 3. At this pH, the numeric value of the mathematical product [H3O+(aq)]·[OH–(aq)] = (10– 3 M)(10–11 M) = 10–14 M2. (b) For any pH you choose, the mathematical product [H3O+(aq)]·[OH–(aq)] = 10–14 M2. This observation certainly suggests that there must be some underlying molecular phenomenon that is responsible for the constancy of this product of concentrations. In pure water, the only source of hydronium and hydroxide ions is the reaction of water with itself: H2O(l) + H2O(l) H3O+(aq) + OH–(aq) Recall that in pure water, [H3O+(aq)] = [OH–(aq)] = 10–7 M and, therefore, as in any aqueous solution, [H3O+(aq)]·[OH–(aq)] = 10–14 M2. If the concentration of hydroxide anion is increased (by adding a tiny bit of sodium hydroxide, for example) in this system, the equilibrium is disturbed and the system will respond (Le Chatelier’s principle) by trying to use up some of the added hydroxide. This will, of necessity require reaction of some of the hydronium cation to form some water, that is, the equilibrium reaction, as written, will go toward reactants. The result will be a solution with [H3O+(aq)] < 10–7 M and [OH–(aq)] > 10–7 M. Apparently, given our observation about the product of concentrations, [H3O+(aq)]·[OH–(aq)], the quantitative decrease in [H3O+(aq)] and increase in [OH–(aq)] exactly compensate one another. Problem 2.116. (a) If the sulfur dioxide, SO2(g), molecule has a permanent dipole moment, the S–O bond dipoles must not cancel out. Sulfur dioxide must have a bent structure, similar to water. See Chapter 1, Section 1.6, for a discussion of bond dipoles and molecular dipole moments.

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(b) If a solution of SO2 in water conducts an electric current, the dissolved gas must have formed ions in the solution. From Section 2.13, we know that the oxides of nonmetals often dissolve and react with water to give a solution of an oxyacid. For sulfur dioxide, the reaction is: SO2(g) + H2O(l) (HO)2SO(aq) The product, sulfurous acid, can transfer one or both protons to water to form ions: H3O+(aq) + HOSO2–(aq) (HO)2SO(aq) + H2O(l)

H3O+(aq) + SO32–(aq) HOSO2–(aq) + H2O(l) Although sulfurous acid transfers only about 10% of its protons to water, enough ions will be present in the solution to conduct electricity. Problem 2.117. (a) The name “bicarbonate” differentiates the HCO3– ion from the carbonate ion, CO32–. It could be that the “bi” refers to the fact that it takes two HCO3– ions to be “equivalent” to one CO32– ion when forming ionic compounds and in their reactions. For example, when calcium carbonate is reacted with an acid, carbon dioxide gas is formed and one mole of gas is formed for every two moles of hydronium ion that react: CaCO3(s) + 2H3O+(aq) Ca2+(aq) + CO2(g) + 2H2O(l) However, when calcium bicarbonate is reacted with an acid two moles of carbon dioxide gas are formed for every two moles of hydronium ion that react: Ca(HOCO2)2(s) + 2H3O+(aq) Ca2+(aq) + 2CO2(g) + 4H2O(l) Ratios like these, amount of gas formed for a given amount of acid used, for example, were the only kind of data scientists had when molecular formulas and atomic masses were not known. Thus, the names for the constituents of compounds had names reflecting such observations and results. (b) Sodium ions are monopositive, Na+, and phosphate anions have a 3– charge, PO43–, so the ionic compound they form must be Na3PO4, trisodium phosphate = TSP. Problem 2.118. (a) The equivalence point in a reaction occurs when the moles of one reactant are exactly stoichiometrically equivalent to the moles of the other. In this problem, we are reacting an acid, (HO)2SO2(aq), with a base dissolved in water, KOH(aq). The reaction that occurs is between hydronium cations from the acid and hydroxide anions from the base, but we can also write the reaction in terms of the two compounds: (HO)2SO2(aq) + 2KOH(aq) 2H2O(l) + 2K+(aq) + SO42–(aq) We want to know the volume of 0.075 M sulfuric acid solution, (HO)2SO2(aq), required to reach the equivalence point of the reaction with 1.00 g of KOH(s) dissolved in 75 mL of water. Use the molar mass of KOH (= 56.1 g) to find the number of moles of KOH we have. Then determine the number of moles of (HO)2SO2 required to react with this amount of KOH and, finally, convert moles of (HO)2SO2 to volume of sulfuric acid solution, using the molarity of the sulfuric acid solution. A single equation that accomplishes these three steps is:

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1 mol KOH 56.1 g KOH

1.00 g KOH = (1.00 g KOH)

1 mol (HO)2 SO2 2 mol KOH

1 L sol'n 0.075 mol (HO)2 SO2

= 0.119 L = 119 mL sulfuric acid solution (b) In the second part of this problem, we are to find the volume of 0.075 M sulfuric acid solution required to reach the equivalence point of the reaction with 1.00 g of KOH(s) dissolved in 150 mL of water. The amount of KOH here is exactly the same as in part (a). Note that the amount of water the KOH is dissolved in was never considered in part (a), because this volume is irrelevant. All that matters for the calculation is the mass/moles of KOH. Therefore, the volume of 0.075 M sulfuric acid solution required here is the same as in part (a), 119 mL. Problem 2.119. (a) The reaction between drain cleaner, NaOH(s), and vinegar, a 0.9M aqueous solution of ethanoic acid, CH3C(O)OH(aq), actually occurs between hydroxide anions as they dissolve in the aqueous vinegar and the hydronium cations from the acid, but we can write a balanced chemical reaction equation for the reaction between the drain cleaner and vinegar in terms of the compounds: CH3C(O)OH(aq) + NaOH(s) H2O(l) +Na+(aq) + CH3C(O)O–(aq) (b) To find the minimum volume of vinegar required to react completely with one pound (454 g) of drain cleaner [that is, for the reactants to be equivalent by the stoichiometric equation in part (a)], use the molar mass of NaOH (= 40 g) to find the number of moles of NaOH we have. Then determine the number of moles of CH3C(O)OH(aq) required to react with this amount of NaOH and, finally, convert moles of CH3C(O)OH(aq) to volume of vinegar, ethanoic acid solution, using the molarity of the ethanoic acid solution. A single equation that accomplishes these three steps is:

454 g NaOH = (454 g NaOH)

1 mol NaOH 40 g NaOH

1 mol CH2 C(O)OH 1 mol NaOH

1 L sol'n 0.9 mol CH2 C(O)OH

= 13 L vinegar (ethanoic acid solution) Problem 2.120. (a) We are asked to find the number of moles of hydronium ion, H3O+(aq), in an acid-rain acidified pond with a volume of 4.5 104 m3 (1 m3 = 1000 L) and [H3O+(aq)] = 5.0 10–5 M (pH = 4.30). Use the volume and molarity to calculate number of moles:

4.5

4

3

10 m = (4.5

1000 L 10 m ) 1 m3 4

3

5.0 10 –5 mol H3 O+ (aq) 1L

= 2.25 103 mol H3O+(aq) 2.3 103 mol H3O+(aq) The experimental data (volume of the pond and its pH) do not justify three significant figures (better than 1% uncertainty) in the moles of hydronium, but we will carry the three in the next calculation and round off at the end. (b) You wish to explain how much lime, CaO(s), will be required to react with 90% of the hydronium ion in the pond by this reaction: 2H3O+(aq) + CaO(s) 3H2O(l) + Ca2+(aq)

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This equation shows us that we need one mole of lime for every two moles of hydronium ion that we want to react. We want to react 90% of the H3O+(aq): [2.25 103 mol H3O+(aq)](.90) = 2.03 103 mol H3O+(aq). Thus, the number of moles of CaO(s) needed is: 103 mol H3O+)

(2.03

1 mol CaO + 2 mol H3 O

= 1.0

103 mol CaO(s) [about 1000 mol CaO]

One mole of CaO has a mass of 56 g, so 1.0 103 mol CaO has a mass of 56 103 g = 56 kg. A mass of one kilogram is 2.2 pounds, so we need (56 kg)(2.2 pound·kg–1) = 123 pounds of CaO(s) to react with 90% of the hydronium ion in the pond. Since the lime is purchased in 40-pound bags, we will need to purchase three bags and use all of two bags and about half of the third. (c) After the lime has been added to react with 90% of the hydronium ion, 10% of the original number of moles of H3O+(aq) will remain, that is, [2.3 103 mol H3O+(aq)](0.10) = 2.3 102 mol H3O+(aq) are still present in the pond. The molar concentration, [H3O+(aq)], is: 2.3 10 2 mol H3O+ = 5.1 [H3O (aq)] = 4.5 10 7 L +

10–6 M

The pH of the pond is now about 5.3 [–log(5.1 10–6 M) = 5.29], that is, about one pH unit higher (less acidic) than before the addition of the lime. Note that reduction of the [H3O+(aq)] by a factor of 10 increases the pH by one unit. You can also see this relationship in Figure 2.24. Problem 2.121. (a) When chickens pant in warm weather, the shells of their eggs are less sturdy than usual because they contain too little calcium carbonate, which is what make an eggshell hard. The precipitation of calcium carbonate is being affected because the concentration of carbonate in the chicken’s bloodstream goes down when more than the normal amount of carbon dioxide is excreted. The directional flow of the reactions (by La Chatelier’s principle) is: CaCO3(s) Ca2+(aq) + CO32–(aq)

CO32–(aq) + H2O(l) HOCO2–(aq) + H2O(l)

HOCO2–(aq) + OH–(aq) (HO)2CO(aq) + OH–(aq)

(HO)2CO(aq) CO2(g) + H2O(l) As the chicken pants and more CO2(g) leaves, CaCO3(s) tends to dissolve, as shown, rather than precipitate to form strong eggshells. (b) To counteract the loss of CO2(g) in the chickens’ breath, the farmers give them carbonated water, seltzer water, to drink. Chickens apparently like the seltzer water better than plain water and drink even more than usual, so the method is even more effective than might have been expected. Problem 2.122. The animated movie in the Web Companion Chapter 2, Section 2.6, page 3, showing the interaction of chloride and silver ions provides one way to visualize the precipitation process that we usually write like this: Ag+(aq) + Cl–(aq) AgCl(s)

ACS Chemistry Chapter 2 suggested solutions

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The movie tries to suggest some of the complexity of the molecular/ionic interactions that result in the formation of a precipitate when appropriate cations and anions are mixed. The ions are surrounded by water molecules in their hydration layers and these interactions with the water molecules compete with the interactions of the cations with the anions. However, the cations and anions sometimes come together to form an ion pair, AgCl(aq), in which some of the waters of hydration from the individual ions are lost. When these ion pairs meet one another in a favorable orientation, larger groups of ions (aggregates) can form, Ag2Cl2(aq), with the likely loss of more waters of hydration. This process continues with the aggregates growing as more ion pairs and individual ions join, until the aggregate is no longer solvated enough to hold the mass in solution and it falls out of solution as a solid, AgxClx(s). Ion pairs and individual ions continue to join this solid and ultimately form the visible crystal that we represent as AgCl(s), with no indication of the unknown (very large) number of each ion present. All these processes are reversible, but go in the direction of precipitation until the solution reaches saturation and the concentrations of the ions are in equilibrium with the solid crystals. It is not easy to translate the molecular level animation to a symbolic representation, particularly because the number of waters of hydration on all the species is unknown. Assuming that each ion has six waters of hydration and that about half are lost by the aggregates at each step, we can write these molecular level reactions: Ag+(H2O)6 + Cl–(H2O)6 AgCl(H2O)6 + 6H2O(l) AgCl(H2O)6 + AgCl(H2O)6

Ag2Cl2(H2O)6 + 6H2O(l)

Ag2Cl2(H2O)6 + AgCl(H2O)6 Ag3Cl3(H2O)6 + 6H2O(l) And so on and on the process continues until the aggregates are too large to be held in solution by their waters of hydration and, finally, we might write: AgnCln(aq) AgCl(s) + some H2O(l) molecules leaving the solid surface Remember that these symbolic representations still simplify what is going on, since many of each of these kinds of reactions are happening simultaneously in the solution. A “simple” reaction like precipitation is really not so simple.

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