CHAPTER 15 Aqueous Equilibria: Chemistry of the Water World

CHAPTER 15 | Aqueous Equilibria: Chemistry of the Water World 15.1. Collect and Organize Figure P15.1 shows four lines to describe the possible depend...
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CHAPTER 15 | Aqueous Equilibria: Chemistry of the Water World 15.1. Collect and Organize Figure P15.1 shows four lines to describe the possible dependence of percent ionization of acetic acid with concentration. We are to choose the one that best represents the trend for this weak acid. Analyze The ionization of acetic acid is described by the following chemical reaction: CH3COOH(aq)  CH3COO–(aq) + H+ The degree of ionization is the ratio of the quantity of a substance that is ionized to the concentration of the substance before ionization. Solve According to Figure 15.10, the change in degree of ionization of a weak acid with concentration is not linear and is best described by the red line in Figure P15.1. The degree of ionization increases with decreasing acetic acid concentration. Think about It The percent ionization could be calculated for each concentration if we knew the equilibrium concentration of the acetate ion in solution and the initial concentration of acetic acid dissolved. ⎡⎣ H + ⎤⎦ equilibrium % ionization = × 100 [ acetic acid]initial 15.2. Collect and Organize The bar graph in Figure P15.2 shows the percent ionization for three aqueous solutions of HOX, where X = Cl, Br, or I. From the graph, we can compare the Ka values for these three weak acids to determine which bar represents the percent ionization of HXO. Analyze The Ka values for the acids (from Table A5.1 in the Appendix) are as follows: Ka = 2.9 × 10–8 for HClO, Ka = 2.3 × 10–9 for HBrO, Ka = 2.3 × 10–11 for HOI. From this we see that HIO is the weakest acid because it has the smallest Ka value. Solve The weakest acid has the lowest percent ionization, so the third bar (shortest one) corresponds to HOI. Notice, too, that as the electronegativity of X in HXO decreases, so too does the acidity and percent ionization. Think about It Even though all three acids are weak, the percent ionization differs greatly among them. The strongest of these acids, HClO, has a percent ionization of just under 0.5%, whereas the weakest acid, HIO, has a percent ionization of well under 0.1%, a more than fivefold difference. 15.3. Collect and Organize From Figure P15.3, we are to choose which titration curve represents a strong acid and which represents a weak acid, each of 1 M concentration at the start of the titration. Analyze A strong acid is completely ionized in solution and has a lower initial pH than a solution with the same concentration of a weak acid, which is only partially ionized in solution. Solve The blue titration curve represents the titration of a 1 M solution of strong acid. The red titration curve represents the titration of a 1 M solution of weak acid. This is because the pH of the strong acid is expected to be much lower than that of the weak acid at the start of the titration (where no base has yet been added).

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176 | Chapter 15 Think about It The equivalence point of the titration of the strong acid (pH 7) does not equal that of the weak acid (pH 10). 15.4. Collect and Organize From the titration curve shown for a weak acid (red line in Figure P15.3), we can estimate the pKa. We can then use that value to determine the pKb of the sodium salt of the weak acid to estimate the pH of a 0.5 M solution of that sodium salt. Analyze At the midpoint, pKa = pH. From the plot, we see that the equivalence point is at pH 10. At the midpoint, where half of the volume of titrant has been added, the pH is 6, so the pKa is 6, or Ka = 1 × 10–6. The Kb then is calculated by using Kw: K w 1× 10 –14 Kb = = = 1× 10 –8 K a 1× 10 – 6

We can then use this value of Kb with a RICE table to solve for the pH of the 0.5 M solution of A– (from the sodium salt of the weak acid HA).

Solve The pH of the sodium salt of HA is approximately equal to the pH at the equivalence point in the titration curve of the weak acid (pH = 10). This makes sense because at the equivalence point we have the following equilibrium: A–(aq) + H2O(ℓ)  HA(aq) + OH–(aq) This is exactly the same as the reaction of NaA with water. Think about It We can solve for the pH of a 0.5 M A– solution where Kb = 1 × 10–8 by using a RICE table: Reaction Initial   Change   Equilibrium  

A–(aq) + H2O(ℓ)  [A–]   0.5   –x   0.5 – x   ( x)( x) K b = 1 × 10 –8 = 0.5 – x x = 7.1 × 10 –5

HA(aq) [HA]   0   +x   x   x2 ≈ 0.5

+

OH–(aq)   [OH–]   0   +x   x  

pOH = –log(7.1 × 10 –5 ) = 4.15 pH = 14 – pOH = 14 – 4.15 = 9.85 Our estimate of pH 10 was close. 15.5. Collect and Organize For the red titration curve in Figure P15.3, we are to choose the indicator, according to its pKa, that would be best for the titration. Analyze The best indicator is the one with a pKa that is nearest to the end point of the titration. Solve The end point for the red curve is at approximately pH 10. Therefore, the best indicator is the one with a pKa of 9.0. Think about It The lower pKa indicators would show a color change before the end point of the titration was reached. Using these would therefore underestimate the concentration of the weak acid in the original solution.

Aqueous Equilibria | 177 15.6. Collect and Organize From the titration curve for the weak acid shown in Figure P15.3, we are to determine the pKa for the acid. Analyze In the titration curve the pH at the midpoint is equal to the pKa of the weak acid. Solve From the titration curve we see that when half the volume of strong base has been added compared to that to reach the equivalence point, the pH is 6. So the pKa for this weak acid is 6. Think about It At the midpoint, [A–] ≈ [HA] for the equilibrium reaction HA(aq) + H2O(ℓ)  A–(aq) + H3O+(aq) This describes a buffer solution in which [HA] = [A–]. By the Henderson–Hasselbalch equation, [A – ] pH = pKa + log [HA] and because log([A–]/[HA]) = log 1 = 0, then pH = pKa. 15.7. Collect and Organize We are shown two titration curves in Figure P15.7. The blue curve has one equivalence point and the red curve has two equivalence points. We are to assign each of these curves to either Na2CO3 or NaHCO3. Analyze Both bases being titrated are soluble sodium salts. The equation describing the titration of CO32– (Na+ is a spectator ion) shows CO32– to be “dibasic”; it reacts in two steps to form H2CO3. CO32–(aq) + H+(aq) → HCO3–(aq) HCO3–( aq) + H+(aq) → H2CO3(aq) HCO3–, however, reacts with acid in one step; it is “monobasic”: HCO3–(aq) + H+(aq) → H2CO3(aq) Solve The red titration curve represents the titration of Na2CO3 because it shows two equivalence points. The blue titration curve represents the titration of NaHCO3 because it shows one equivalence point. Think about It Both titration curves start at high pH. This is due to the hydrolysis of CO32– and HCO3– in water: CO32–(aq) + H2O(ℓ) → HCO3–(aq) + OH–(aq) HCO3–(aq) + H2O(ℓ) → H2CO3(aq) + OH–(aq) 15.8. Collect and Organize Figure P15.8 shows three beakers: one containing a yellow solution, one containing a very light green solution, and the last one containing a blue solution. Given that the bromthymol blue indicator in each beaker is yellow in acidic solutions and blue in basic solutions, we are to match solutions of the dissolved salts NH4Cl, NH4C2H3O2, and NaC2H3O2 to the correct beaker. Analyze We must consider the hydrolysis of the constituent cation and anion in each salt. In NH4Cl, the NH4+ ion reacts with water to give an acidic solution: NH4+(aq) + H2O(ℓ)  NH3(aq) + H3O+(aq) but Cl– does not. In NaC2H3O2 the Na+ ion does not hydrolyze, but the acetate ion does to give a basic solution: C2H3O2–(aq) + H2O(ℓ)  HC2H3O2(aq) + OH–(aq) In NH4C2H3O2 both cation and anion hydrolyze, giving a nearly neutral solution.

178 | Chapter 15 Solve NH4Cl is dissolved in the yellow solution, NaC2H3O2 is dissolved in the blue solution, and NH4C2H3O2 is dissolved in the light green solution. Think about It The relative magnitude of the Ka of NH4+ (5.7 × 10–10) compared to the Kb of C2H3O2– (5.7 × 10–10) shows that both salts hydrolyze to the same extent and that we should expect a solution of NH 4C2H3O2 to be approximately neutral. 15.9. Collect and Organize For HBr(aq) we are to identify the Brønsted–Lowry acid and base. Analyze A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry base is a proton acceptor. Solve HBr is a strong acid in water. It acts as a Brønsted–Lowry acid, donating its proton to H2O, the Brønsted–Lowry base: HBr(aq) + H2O(ℓ) → H3O+(aq) + Br–(aq) Think about It Hydrobromic acid is a strong acid in water. It completely dissociates in water to H3O+ and Br–. 15.10. Collect and Organize For HNO3(aq) we are to identify the Brønsted–Lowry acid and base. Analyze A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry base is a proton acceptor. Solve HNO3 is a strong acid in water. It acts as a Brønsted–Lowry acid, donating its proton to H2O, the Brønsted– Lowry base: HNO3(aq) + H2O(ℓ) → NO3–(aq) + H3O+(aq) Think about It Nitric acid is a strong acid in water. It completely dissociates in water to H3O+ and NO3–. 15.11. Collect and Organize For NaOH(aq) we are to identify the Brønsted–Lowry acid and base. Analyze NaOH is a soluble salt that forms Na+ and OH– in water. Na+ does not react with water, so it is a spectator ion. We need then to consider the behavior of OH– in water. A Brønsted–Lowry acid is a proton donor. A Brønsted– Lowry base is a proton acceptor. Solve OH– is a strong base in water. It acts as a Brønsted–Lowry base, removing a proton from H2O, the Brønsted– Lowry acid: OH–(aq) + H2O(ℓ)  H2O(ℓ) + OH–(aq) Think about It In Problems 15.9 and 15.10, water acted as a Brønsted–Lowry base. In this problem, it acts as an acid. This dual acid–base behavior makes water amphoteric.

Aqueous Equilibria | 179 15.12. Collect and Organize We are to compare the neutralization capacity of equimolar solutions of NaOH and Ca(OH)2. Analyze Both bases are strong and completely dissociate in water: NaOH(s) → Na+(aq) + OH–(aq) Ca(OH)2(s) → Ca2+(aq) + 2 OH–(aq) Solve No, these two compounds do not have the same capacity to neutralize strong acids. As seen from the dissolution equations, Ca(OH)2 produces two equivalents of OH– in solution and can neutralize twice as much acid as the same molar concentration of NaOH. Think about It However, equimolar solutions of NaOH and CsOH do have the same neutralizing capacity. 15.13. Collect and Organize For three acid–base reactions, we are to identify which reactant is the acid and which reactant is the base. Analyze For all these reactions that involve the transfer of a proton between the acid and base, we can apply the Brønsted–Lowry definitions of acid and base. A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry base is a proton acceptor. Solve (a) HNO3 is the acid. It transfers H+ to the base, NaOH. (b) HCl is the acid. It transfers H+ to the base, CaCO3. (c) HCN is the acid. It transfers H+ to the base, NH3. Think about It In reactions (a) and (b), Na+, Ca2+, and Cl– do not get involved in the reaction. They are spectator ions. The net ionic equations are HNO3(aq) + OH–(aq) → NO3–(aq) + H2O(ℓ) CO32–(aq) + 2 H+(aq) → CO2(g) + H2O(ℓ) 15.14. Collect and Organize For three acid–base reactions we are to identify which reactant is the acid and which reactant is the base. Analyze For all these reactions that involve the transfer of a proton between the acid and base, we can apply the Brønsted–Lowry definitions of acid and base. A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry base is a proton acceptor. Solve (a) H2O is the acid. It transfers a proton to the base, NH2–. (b) HClO4 is the acid. It transfers a proton to the base, H2O. (c) HSO4– is the acid. It transfers a proton to the base, CO32–. Think about It In reaction (a), water acts as an acid, but in reaction (b), water acts as a base. This behavior classifies water as amphoteric. Also, NH2– is a strong base in aqueous solution and HClO4 is a strong acid in aqueous solution. Both completely react with water to form NH3 and ClO4–.

180 | Chapter 15 15.15. Collect and Organize For each species listed we are to write the formula for the conjugate base. Analyze The conjugate base form of a species has H+ removed from its formula. Solve The conjugate base of HNO2 is NO2–. The conjugate base of HOCl is OCl–. The conjugate base of H3PO4 is H2PO4–. The conjugate base of NH3 is NH2–. Think about It Be sure to account for the change in charge when H+ is removed to form the conjugate base. 15.16. Collect and Organize For each species listed we are to write the formula for the conjugate acid. Analyze The conjugate acid form of a species has an H+ added to its formula. Solve The conjugate acid of NH3 is NH4+. The conjugate acid of ClO2– is HClO2 The conjugate acid of SO42– is HSO4–. The conjugate acid of OH– is H2O. Think about It Be sure to account for the change in charge when the base adds H+ to form the conjugate acid. 15.17. Collect and Organize Given that the concentration of a nitric acid solution is 1.50 M, we are to calculate the concentration of H+ ions in the solution. Analyze Nitric acid is a strong acid that completely dissociates in water, so the concentration of H+ ions is stoichiometrically related to the concentration of HNO3: HNO3(aq) → H+(aq) + NO3–(aq) Solve

[H+] = 1.50 M

Think about It The concentration of OH– in strong base solutions, likewise, is the same as the concentration of the strong base dissolved into the solution. 15.18. Collect and Organize Given a solution of the strong acid HCl that is prepared by diluting 20.0 mL of 11.6 M HCl solution to 500 mL, we are to calculate the concentration of H+ ions in the final solution. Analyze For this problem we use the relationship from Chapter 8 for diluting solutions: Vi × M i = Vf × M f where V is the volume of the solution and C is the concentration.

Aqueous Equilibria | 181 Solve

20.0 mL × 11.6 M = 500 mL × Mf Mf = 0.464 M

Think about It Because this is a solution of a strong acid, we know that the acid is completely dissociated and so [HCl] = [H +]. 15.19. Collect and Organize Given that a solution is 0.0800 M in the strong base Sr(OH)2, we are asked to calculate the concentration of OH– . Analyze Sr(OH)2, being a strong base, completely dissociates according to the equation Sr(OH)2(aq) → Sr2+(aq) + 2 OH–(aq) Therefore, [OH–] = 2 × [Sr(OH)2]. Solve

2 × 0.0800 M = 0.160 M = [OH–]

Think about It Be sure to account for both OH– ions in this strong base. 15.20. Collect and Organize Given that a solution is prepared by dissolving 5.0 g of NaOH in water and diluting to a final volume of 250 mL, we are asked to calculate the final concentration of OH– in the solution. Analyze Because NaOH is a strong base, we know that it is completely ionized in solution: NaOH(aq) → Na+(aq) + OH–(aq) Therefore, we can relate the moles of NaOH dissolved to the moles of OH– in solution and then calculate the molarity by dividing the moles of OH– by the final volume in liters. Solve

5.0 g NaOH ×

1 mol NaOH 1 mol OH – 1 × × = 0.50 M 40.00 g NaOH 1 mol NaOH 0.250 L

Think about It Other strong bases used in the laboratory include KOH, LiOH, Mg(OH)2, Ca(OH)2, Ba(OH)2, and Sr(OH)2. 15.21. Collect and Organize We are asked how to prepare 2.50 L of 0.70 M OH– by using NaOH(s). Analyze From the desired volume and concentration we first calculate the moles of OH– required for the solution. Since 1 mol of OH– is produced for every 1 mol of NaOH dissolved, this is also the moles of NaOH required. To calculate the mass of NaOH needed we multiply the number of moles by the molar mass of NaOH (40.00 g/mol).

182 | Chapter 15 Solve Mass of NaOH needed

0.70 mol OH – 1 mol NaOH 40.00 g NaOH × × = 70.0 g NaOH L 1 mol NaOH 1 mol OH – Dissolve 70.0 g of NaOH(s) in water and dilute to a total volume of 2.50 L. mass NaOH = 2.50 L ×

Think about It Remember that volume multiplied by the concentration for solutions gives us the moles of that substance in that solution. 15.22. Collect and Organize We are to calculate the volume of a 1.00 M NaOH solution that is needed to prepare 250 mL of a solution that is 0.0200 M in OH– concentration. Analyze Because NaOH is a strong base, 1.00 M NaOH is 1.00 M in OH–. We can use the relationship from Chapter 8 for dilutions to solve this problem for Vi: Vi × M i = Vf × M f Solve

Vi × 1.00 M = 250 mL × 0.0200 M Vi = 5.00 mL

Think about It Be careful in this type of calculation if the strong base is one like Ba(OH)2. In that case, a 1.00 M Ba(OH)2 solution would be 2.00 M in OH– concentration. 15.23. Collect and Organize We are asked to explain why H2SO4 is a stronger acid (greater Ka ) than H2SeO4. 1 Analyze The only difference in these acids is the central atom. Sulfur and selenium belong to the same group in the periodic table. These elements differ in size and electronegativity. Solve Sulfur is more electronegative than selenium. This higher electronegativity on the sulfur atom stabilizes the anion HSO4– more than the anion HSeO4–. Therefore, H2SO4 is a stronger acid.

Think about It We would expect this trend to continue, so we predict that H2TeO4 is a weaker acid than H2SeO4. 15.24. Collect and Organize We are to explain why H2SO4 is a stronger acid (larger Ka ) than H2SO3. 1 Analyze The only difference between these acids is the extra oxygen atom bound to sulfur in H2SO4.

Aqueous Equilibria | 183 Solve The presence of the additional electron-withdrawing oxygen atom (because of its high electronegativity) on H2SO4 compared to H2SO3 delocalizes the negative charge on HSO4– better than on HSO3–. This stabilization of the anion makes H2SO4 a stronger acid.

Think about It Even the second proton on H2SO4 is more acidic (Ka2 = 1.2 × 10–2) than that of H2SO3 (Ka2 = 6.2 × 10–8). 15.25. Collect and Organize We are to predict which acid of a pair is stronger. Analyze The more oxygen atoms bound to the central atom and the more electronegative the central atom (X) in the acid, the more acidic is the compound. Solve (a) H2SO3 is a stronger acid than H2SeO3. (b) H2SeO4 is a stronger acid than H2SeO3. Think about It The presence of oxygen atoms bound to the central atom in an oxyacid can have a dramatic effect on acidity. 15.26. Collect and Organize We are to predict which acid of a pair is stronger. Analyze The more oxygen atoms bound to the central atom and the more electronegative the central atom in the HXO acid, the more acidic is the compound. Solve (a) HBrO2 is more acidic than HBrO. (b) HClO is more acidic than HBrO. Think about It The presence of oxygen atoms bound to the central atom in an oxyacid can have a dramatic effect on acidity. The range for Ka values for HClO to HClO4 is 2.9 × 10–8 to greater than 1, as listed in Appendix 5. 15.27. Collect and Organize We are to explain why the pH value decreases for solutions as acidity increases.

184 | Chapter 15 Analyze The pH of a solution is calculated through

pH = –log[H+]

Solve Because the pH function is a –log function, as [H+] increases, the value of –log[H+] decreases. Think about It The pH scale is typically 0 –14 for concentrations of H+ from 1 M to 1 × 10–14 M, but values of pH may be negative or greater than 14. 15.28. Collect and Organize We are to determine the difference in pH between two solutions when solution A is 100 times more acidic than solution B. Analyze If solution A is 100 times more acidic than solution B, [H+]A = 100[H+]B and [H+]A/[H+]B = 100. Taking the log of both sides of this equation gives ⎛ [H + ] ⎞ log ⎜ + A ⎟ = log[H + ]A – log[H + ]B = log100 ⎝ [H ]B ⎠ + Because pHA = –log[H ]A and pHB = –log[H+]B, then pHB – pHA = log 100. Solve The difference in pH between the two solutions is pHB – pHA = log 100 = 2.000. Think about It Because pH is a logarithmic scale, one unit change in pH represents a 10-fold increase or decrease in [H+]. 15.29. Collect and Organize We are asked under what conditions the pH of a solution may be negative. Analyze The pH scale is often seen as 0 –14. This occurs when [H+] is between 1 M and 1 × 10–14 M. Solve When [H+] is greater than 1 M, the pH of the solution is negative. Think about It For example, a 3.00 M solution of HCl has a pH of pH = –log[H+] = –log(3.00 M) = – 0.48 15.30. Collect and Organize For the autoionization of ethanol we are to explain why this ionization has a K value much smaller than the autoionization constant of water. Analyze For water the autoionization is described by 2 H2O(ℓ)  H3O+(aq) + OH–(aq) For ethanol the autoionization is described by 2 CH3CH2OH(ℓ)  CH3CH2OH2+(ethanol) + CH3CH2O–(ethanol)

Aqueous Equilibria | 185 Solve In both autoionizations an O — H bond breaks and a new O—H bond forms. The O—H bond in ethanol is less polar than that of water (because it is a C—O—H bond rather than a H—O—H bond), so ethanol is a weaker base than water and the K value for the acid–base transfer of H+ for ethanol is smaller. Think about It Another species that might autoionize is ammonia: 2 NH3(ℓ)  NH4+(ammonia) + NH2–(ammonia) 15.31. Collect and Organize Given either the [OH–] or [H+] for a solution, we are asked to calculate the pH and pOH and determine whether the solution is acidic, basic, or neutral. Analyze To calculate the pH or the pOH from the [H+] or [OH–], respectively, we use pH = –log[H+] pOH = –log[OH–] To find the pOH from the pH, and vice versa, we use the relationship pH + pOH = 14 If the pH of a solution is less than 7, the solution is acidic. If the pH is equal to 7, the solution is neutral. If the pH is greater than 7, the solution is basic. Solve (a) pH = –log(3.45 × 10–8) = 7.462 pOH = 14 – pH = 6.538 This solution is basic. (b) pH = –log(2.0 × 10–5) = 4.70 pOH = 14 – pH = 9.30 This solution is acidic. (c) pH = –log(7.0 × 10–8) = 7.15 pOH = 14 – pH = 6.85 This solution is basic. (d) pOH = –log(8.56 × 10– 4) = 3.068 pH = 14 – pOH = 10.932 This solution is basic. Think about It When determining how many significant figures to include in your answers when computing the pH or pOH, remember that the first number in the pH or pOH gives the location of the decimal point. The significant digits, therefore, follow the decimal point. 15.32. Collect and Organize Given either the [OH–] or [H+] for a solution, we are asked to calculate the pH and pOH and determine whether the solution is acidic, basic, or neutral. Analyze To calculate the pH or the pOH from the [H+] or [OH–], respectively, we use pH = –log[H+] pOH = –log[OH–] To find the pOH from the pH, and vice versa, we use the relationship pH + pOH = 14 If the pH of a solution is less than 7, the solution is acidic. If the pH is equal to 7, the solution is neutral. If the pH is greater than 7, the solution is basic.

186 | Chapter 15 Solve (a) pOH = –log(7.69 × 10–3) = 2.114 pH = 14 – pOH = 11.886 This solution is basic. (b) pOH = –log(2.18 × 10–9) = 8.662 pH = 14 – pOH = 5.338 This solution is acidic. (c) pH = –log(4.0 × 10–8) = 7.40 pOH = 14 – pH = 6.60 This solution is basic. (d) pH = –log(3.56 × 10– 4) = 3.449 pOH = 14 – pH = 10.551 This solution is acidic. Think about It When determining how many significant figures to include in your answers when computing the pH or pOH, remember that the first number in the pH or pOH gives the location of the decimal point. The significant digits, therefore, follow the decimal point. 15.33. Collect and Organize Given that stomach acid is 0.155 M HCl, we are to calculate its pH. Analyze The pH is defined as pH = –log[H+]. Because HCl is a strong acid, we know that 0.155 M HCl is also 0.155 M H+. Solve

pH = –log(0.155) = 0.810

Think about It The acid in your stomach is fairly strong and the HCl is produced by parietal cells to break down food. Your stomach is protected in turn by epithelial cells that secrete a bicarbonate solution that neutralizes the acid and forms a coating to protect the stomach’s tissues. 15.34. Collect and Organize Given a 0.00500 M HNO3 solution, we are to calculate its pH. Analyze The pH is defined as pH = –log[H+]. Because HNO3 is a strong acid, we know that 0.00500 M HNO3 is also 0.00500 M H+. Solve

pH = –log(0.00500) = 2.301

Think about It This solution is about as acidic as the juice of a lemon (Figure 15.8). 15.35. Collect and Organize Given that the concentration of NaOH in a solution is 0.0450 M, we are to calculate its pH and pOH. Analyze Because NaOH is a strong base, we know that the [OH–] in this solution is 0.0450 M. The pOH is calculated from pOH = –log[OH–], with the pH being found from the pOH through the relationship pH + pOH = 14.

Aqueous Equilibria | 187 Solve

pOH = –log(0.0450) = 1.347 pH = 14 – pOH = 12.653

Think about It This solution is as alkaline (basic) as drain cleaner (Figure 15.8). 15.36. Collect and Organize Given that the concentration of KOH in a solution is 0.160 M, we are to calculate its pH and pOH. Analyze Because KOH is a strong base, we know that the [OH–] in this solution is 0.160 M. The pOH is calculated from pOH = –log[OH–], with the pH being found from the pOH through the relationship pH + pOH = 14. Solve

pOH = –log(0.160) = 0.796 pH = 14 – pOH = 13.204

Think about It This solution is as alkaline (basic) as drain cleaner (Figure 15.8). 15.37. Collect and Organize Given a solution that is 2.3 × 10–8 M in HNO3, we are to calculate its pH. Analyze Because HNO3 is a strong acid, we know that for this solution the [H+] from the HNO3 is 2.3 × 10–8 M. The calculated pH from this value, however, gives a pH of greater than 7, which would mean that this solution is basic: pH = –log(6.9 × 10–8) = 7.64 This cannot be, as we have added an acid to water. This is a case in which the autoionization of water (where [H+] = 1.00 × 10–7 M) is important. We therefore have to use include the ionization of water in the Kw expression: Kw = [H+][OH–] 1.00 × 10–14 = [H+][OH–] = (x + [H+]acid)(x) Solve Rearranging this equation, we obtain the quadratic equation where [H+]acid = 6.9 × 10–8 M x2 + [HNO3]x – 1.00 × 10–14 = 0 where for this problem [HNO3] = 2.3 × 10–8 M. x2 + (2.3 × 10–8)x – 1.00 × 10–14 = 0 Solving by the quadratic equation gives x = 1.1216 × 10–7 M for a total hydrogen ion concentration of 8.9159 × 10–8 M +2.3 × 10–8 = 1.1215 × 10–7 M pH = –log 1.1215 × 10–7 = 6.95. Think about It This pH agrees with our instinct that the solution should be slightly acidic. 15.38. Collect and Organize Given a solution that is 6.9 × 10–8 M in LiOH, we are to calculate its pH. Analyze Because LiOH is a strong base, we know that for this solution the [OH–] from the LiOH is 6.9 × 10–8 M. The calculated pOH from this value, however, gives a pOH of greater than 7, which would mean that this solution is acidic:

188 | Chapter 15 pOH = –log(6.9 × 10–8) = 7.64 pH = 14 – 7.64 = 6.36 This cannot be, as we have added an base to water. This is a case in which the autoionization of water (where [OH–] = 1.00 × 10–7 M) is important. We therefore have to use include the ionization of water in the Kw expression: Kw = [H+][OH–] 1.00 × 10–14 = [H+][OH–] = (x)(x + [OH–]base) Solve Rearranging this equation, we obtain the quadratic equation where [OH–]base = 6.9 × 10–8 M x2 + [LiOH]x – 1.00 × 10–14 = 0 where for this problem [HNO3] = 2.3 × 10–8 M. x2 + (6.94 × 10–8)x – 1.00 × 10–14 = 0 Solving by the quadratic equation gives x = 7.1284 × 10–8 M for a total hydrogen ion concentration of 7.1284 × 10–8 +6.9 × 10–8 = 1.403 × 10–7 M pOH = –log 1.1215 × 10–7 = 6.85. pH = 14 – 6.85 = 7.15 Think about It This pH agrees with our instinct that the solution should be slightly basic. 15.39. Collect and Organize For the dilution of a pH 2.0 HCl solution, we are to determine the pH once the solution’s volume is doubled by adding water. Analyze The volume of the solution doubles and therefore the acid is now diluted by a factor of 2. Solve For a pH = 2.0 solution the concentration of HCl is 1.00 × 10–2 M. Diluting to twice the volume would therefore give a solution with an HCl concentration of 5.00 × 10–2 M with pH = –log(5.00 × 10–2 M) = 2.3 Think about It We did not have to use the dilution equation in this problem because the dilution factor was easy. 15.40. Collect and Organize Given the pH of venous and arterial blood, we are asked to calculate the ratio of H+ in venous blood compared to that in arterial blood. Analyze We will first need to convert the pH of each blood sample to concentration of H+ by using [H+] = 10–pH. Solve For venous blood, the [H+] = 10–pH = 10–7.35 = 4.467 × 10–8 M. For arterial blood, the [H+] = 10–pH = 10–7.45 = 3.548 × 10–8 M. The ratio of hydrogen ion concentrations is [H + ]venous 4.467 × 10 −8 M = = 1.3 [H + ]arterial 3.548 × 10 −8 M Think about It Blood in the veins, therefore, is about 30% more acidic than blood in the arteries, as it carries CO2 from the cells to be released by the lungs.

Aqueous Equilibria | 189 15.41. Collect and Organize For 1 M solutions of CH3COOH, HNO2, HClO, and HCN, we are to rank these in order of decreasing concentration of H+. For this we need the Ka values for each acid. Analyze The Ka values for these acids are listed in Appendix 5. The greater the value of the Ka, the greater the concentration of H+ in the solution of the acid. The Ka values for the acids are as follows: CH3COOH, 1.76 × 10–5; HNO2, 4.0 × 10–4; HClO, 2.9 × 10–8; and HCN, 6.2 × 10–10. Solve In order of largest Ka (strongest acid) to smallest Ka (weakest acid), HNO2 > CH3COOH > HClO > HCN. Think about It The weak acids in this series have a wide range in acidities (about 10,000-fold, comparing their Ka values). 15.42. Collect and Organize Given the degree of ionization for four weak acids of 0.100 M concentration, we are to determine which acid has the smallest value of its acid dissociation constant, Ka. Analyze The lower the degree of ionization, the weaker the acid and the lower its Ka. Solve Acetic acid, CH3COOH, has the lowest degree of ionization and is the acid with the smallest Ka value. Think about It The acid with the largest degree of ionization, HF, has the largest Ka value. 15.43. Collect and Organize We are to explain why the electrical conductivity of 1.0 M NaNO2 is much better than that of 1.0 M HNO2. Analyze Solutions with a larger concentration of dissolved ions conduct electricity better than those with lower concentrations of dissolved ions. Solve NaNO2 is completely soluble in water, separating into Na+ and NO2– ions, each in 1.0 M concentration, for a total ion concentration of 2.0 M. HNO2, however, only weakly dissociates in water: HNO2(aq)  NO2–(aq) + H+(aq) so it produces just slightly greater than 1.0 M ions in solution. NaNO2, therefore, with more dissolved ions in solution, is a better conductor of electricity. Think about It Later in this chapter, you will learn that NO2– reacts with water to a small extent: NO2–(aq) + H2O ()  HNO2(aq) + OH–(aq) However, this produces another ion, OH–, so our analysis that a solution of 1.0 M NaNO2 is 2.0 M in ions is still correct. 15.44. Collect and Organize We are to explain why HCl, although a molecular species, when dissolved in water (also a molecular species) is a good conductor of electricity. Analyze Solutions with dissolved ions are good conductors of electricity.

190 | Chapter 15 Solve When HCl dissolves in water, it completely dissociates because it is a strong acid that produces 2 mol of ions for every mole of HCl dissolved: HCl(aq) → H+(aq) + Cl–(aq) This makes HCl(aq) a good electrical conductor. Think about It This is true for other typical strong acids and bases, such as HNO3, HBr, HClO4, KOH, and NaOH. 15.45. Collect and Organize The formula for hydrofluoric acid is HF. From this we can write the mass action expression for this weak acid. Analyze The general form of the mass action expression for weak acids, on the basis of HA  A– + H+, is

Ka = Solve

[H+ ][A – ] [HA]

HF(aq)  F–(aq) + H+(aq)

Ka =

[H+ ][F– ] [HF]

Think about It The hydrofluoric acid is transferring a proton to water in this equation, so an equivalent expression is HF(aq) + H2O ()  F–(aq) + H3O+(aq)

Ka =

[H3O+ ][F– ] [HF]

15.46. Collect and Organize Given the formula of formic acid, HCOOH, and the information that one H atom is ionizable, we are to write the mass action expression for this acid’s ionization. Analyze The general form of the mass action expression for weak acids, on the basis of HA  A– + H+, is

[H+ ][A – ] [HA] In formic acid, the H atom bonded to O (not the H bonded to C) is ionizable. Ka =

Solve

HCOOH(aq)  HCOO–(aq) + H+(aq)

Ka =

[H+ ][HCOO– ] [HCOOH]

Think about It The formic acid is transferring a proton to water in this equation, so an equivalent expression is HCOOH(aq) + H2O(ℓ)  HCOO–(aq) + H3O+(aq)

Ka =

[H3O+ ][HCOO– ] [HCOOH]

Aqueous Equilibria | 191 15.47. Collect and Organize Given that the Ka of alanine is less when it is dissolved in ethanol than it is when dissolved in water, we are to determine which solvent ionizes alanine to the larger extent and which solvent is the stronger Brønsted–Lowry base. Analyze The larger the value of Ka, the greater the extent to which a substance has ionized. The solvent in which an acid ionizes the most must be the strongest Brønsted–Lowry base toward that acid. Solve (a) Because Ka for alanine is greater in water than in ethanol, alanine in water is ionized to a greater extent than in ethanol. (b) Water is the stronger base for alanine than ethanol because alanine is ionized to a greater extent in water. Think about It This question demonstrates that acid–base strengths depend on the basicity of the solvent. 15.48. Collect and Organize Given the Ka values of proline in water, 28% aqueous ethanol, 37% and aqueous formaldehyde, we are to determine in which solvent proline is the strongest acid and rank the solvents in order of increasing basicity. Analyze The order of Ka values from weakest acid to strongest is as follows: proline in water (2.5 × 10–11) < proline in 28% ethanol (2.8 × 10–11) < proline in 37% aqueous formaldehyde (1.66 × 10–8). The larger the value of the Ka of proline in a solvent, the stronger proline’s acidity. Solve (a) Proline is the strongest acid in formaldehyde solution. (b) In order of increasing basicity toward proline, water < 28% ethanol < 37% formaldehyde. Think about It This question demonstrates that acid–base strengths depend on the basicity of the solvent. 15.49. Collect and Organize Given that CH3NH2 is slightly basic in water, we can write the equation describing its reaction with water to identify the species in the reaction that is the Brønsted–Lowry acid and the species that is the base. Analyze Acting as a base, CH3NH2 accepts H+ from surrounding water molecules. Solve The reaction describing the basicity of CH3NH2 is CH3NH2(aq) + H2O(ℓ)  CH3NH3+(aq) + OH–(aq) In this reaction, H2O is the acid and CH3NH2 is the base. Think about It In another solvent, such as diethylamine, methylamine may act as an acid: CH3NH + (CH3CH2)2NH(ℓ)  CH3NH– + (CH3CH2)2NH2+ This occurs because diethylamine (Kb = 8.6 × 10– 4) is a stronger base than methylamine (Kb = 4.4 × 10– 4). 15.50. Collect and Organize Given that an aqueous solution of H2NCH2CH2NH2 is basic, we are to write the formula of the product of its reaction with HCl.

192 | Chapter 15 Analyze Because H2NCH2CH2NH2 is a base, it will accept the H+ in solution from the dissolved HCl. Solve H2NCH2CH2NH3+ forms in the solution as described by the equation H2NCH2CH2NH2(aq) + H+(aq)  H2NCH2CH2NH3+(aq) If two molar equivalents of H+(aq) are present, the second protonation occurs: H2NCH2CH2NH3+(aq) + H+(aq)  H3NCH2CH2NH32+(aq) The Cl– ions are the counterions in the salt: (H3NCH2CH2NH3)Cl2. Think about It Because this compound has two basic sites on the nitrogen atoms, it is called a dibasic compound. 15.51. Collect and Organize Using the measured percent ionization of 1.00 M lactic acid of 2.94%, we are to calculate the Ka of this weak acid. Analyze The equilibrium equation from Appendix 5 that describes the ionization of lactic acid is CH3CHOHCOOH(aq)  H+(aq) + CH3CHOHCOO–(aq) We can set up a RICE table to solve this problem, where x = 1.00 M × 0.0294 = 0.0294. The Ka expression is [H + ][CH3CHOHCOO – ] Ka = [CH3CHOHCOOH] Solve

Reaction  

CH3CHOHCOOH(aq)  H+(aq) + CH3CHOHCOO–(aq)  

Initial   Change   Equilibrium  

[CH3CHOHCOOH]   1.00   –x = – 0.0294   0.9706  

Ka =

[H+]   0   +x = + 0.0294   0.0294  

[CH3CHOHCOO–]   0   +x = + 0.0294   0.0294  

(0.0294) 2 = 8.91× 10 – 4 0.9706

Think about It Compare this with the Ka 3F WH3>G7A8 × 10 –4 listed in Appendix 5. The difference may be attributable to 3F7?B7D3FGD76;887D7@574753GE74A6KF7?B7D3FGD7;E W 15.52. Collect and Organize Using the measured percent ionization of 0.100 M butanoic acid of 1.23%, we are to calculate the Ka of this weak acid. Analyze The equilibrium equation that describes the ionization of butanoic acid is C3H7COOH(aq)  H+(aq) + C3H7COO–(aq) We can set up a RICE table to solve this problem, where x = 0.100 M × 0.0123 = 0.00123. The Ka expression is [H + ][C3 H 7 COO – ] Ka = [C3 H 7 COOH]

Aqueous Equilibria | 193 Solve

Reaction   Initial   Change   Equilibrium  

C3H7COOH(aq)  H+(aq) + C3H7COO–(aq)   + [C3H7COOH]   [H ]   [C3H7COO–]   0.100   0   0   –x = – 0.00123   +x = +0.00123   +x = +0.00123   0.09877 0.00123 0.00123 2 (0.00123) Ka = = 1.53 × 10–5 0.09877

Think about It This value compares well with the Ka value listed in Appendix 5 of 1.5 × 10–5. 15.53. Collect and Organize We are given the [H+] for an equilibrium solution of an unknown acid with an initial concentration of 0.250 M. From this information we can calculate the degree of ionization and Ka for this weak acid. Analyze The equilibrium expression for the unknown acid is HA(aq)  H+(aq) + A–(aq) with a Ka expression of [H+ ][A – ] Ka = [HA] We can set up a RICE table to show how this weak acid ionizes; for this acid, [H +] at equilibrium is 4.07 × 10–3 M. Solve

HA(aq) H+(aq) + A–(aq)    [HA]   [H+]   [A–]   Initial   0.250   0   0   Change   –x = –0.00407   +x = +0.00407   +x = +0.00407   Equilibrium   0.24593   0.00407   0.00407   2 (0.00407) Ka = = 6.74 × 10–5 0.24593 [H + ]eq 0.00407 M Degree of ionization = = = 0.0163 or 1.63% [HA]initial 0.250 M

Reaction  

Think about It In this problem we can use the [H+] as equivalent to x in the RICE table, enabling us to calculate both the Ka and the degree of ionization for the acid. 15.54. Collect and Organize Because when HNO3 is very concentrated it is not completely ionized, we can calculate its Ka. To do this we are given that only 33% of the HNO3 is ionized when its concentration is 7.5 M. Analyze The equilibrium expression for concentrated nitric acid is HNO3(aq)  H+(aq) + NO3–(aq) with a Ka expression of [H + ][NO3 – ] Ka = [HNO3 ] We can set up a RICE table to solve this expression when x = 7.5 M × 0.33 = 2.475 M.

194 | Chapter 15 Solve

Reaction   Initial   Change   Equilibrium  

HNO3(aq)  H+(aq) + NO3–(aq)   + [HNO3]   [H ]   [NO3–]   7.5   0   0   –x = –2.475   +x = +2.475   +x = +2.475   5.025   2.475   2.475   2 (2.475) Ka = = 1.2 5.025

Think about It This Ka is still larger than 1, so the products (H+ and NO3–) are favored in this equilibrium. 15.55. Collect and Organize Given that formic acid has Ka = 1.8 × 10–4, we can calculate the pH of a 0.060 M aqueous solution of this weak acid. Analyze To solve this problem we set up a RICE table where the initial concentration of formic acid, HCOOH, is 0.060 M. We let x be the amount of formic acid that ionizes. The equilibrium and Ka expressions are HCOOH(aq)  H+(aq) + HCOO–(aq)

[H+ ][HCOO– ] = 1.8 × 10–4 [HCOOH] The pH can be calculated through the equation pH = –log[H+]. Ka =

Solve

H+(aq) + HCOO–(aq)   [H+]   [HCOO–]   Initial   0   0   Change   +x   +x   Equilibrium   x   x   2 2 x x 1.8 × 10–4 = ≈ 0.060 − x 0.060 x = 3.29 × 10–3 We should first check the simplifying assumption we made above before calculating the pH. 3.29 × 10–3 × 100 = 5.5% 0.060 This is a little over 5%, so technically we should solve this by the quadratic equation (which we do below). However, if we allow this simplifying assumption to be valid, the pH of the solution would be calculated by Reaction  

HCOOH(aq) [HCOOH]   0.060   –x   0.060 – x  



pH = –log 3.29 × 10–3 = 2.48 Now, solving by the quadratic equation gives 1.8 × 10–4(0.060 – x) = x2 2 x + 1.8 × 10–4x – 1.08 × 10–5 = 0 x = 3.20 × 10–3 pH = –log 3.20 × 10–3 = 2.49 Think about It The difference between the pH values when making the simplifying assumption and solving the equation exactly by using the quadratic equation is 2.49 – 2.48 = 0.01, which is fairly small. 15.56. Collect and Organize Given that uric acid has pKa = 3.89, we can calculate the pH of a 0.0150 M aqueous solution of this weak acid.

Aqueous Equilibria | 195 Analyze To solve this problem we can set up a RICE table where the initial concentration of uric acid is 0.0150 M. We let x be the amount of uric acid that ionizes. The equilibrium and Ka expressions are uric acid(aq)  H+(aq) + urate–(aq)

Ka = From the pKa we can calculate the Ka:

[H + ][urate – ] [uric acid]

Ka = 1×10–pKa = 1×10–3.89 = 1.29 ×10– 4

The pH can be calculated through the equation pH = –log[H+]. Solve

H+(aq) + Urate–(aq)   [H+]   [Urate–]   Initial   0   0   Change   +x   +x   Equilibrium   x   x   2 2 x x 1.29 ×10– 4 = ≈ 0.0150 − x 0.0150 x = 1.39 ×10–3 Checking the simplifying assumption shows that our simplifying assumption is not valid: 1.39 × 10–3 × 100 = 9.3% 0.0150 so we must solve this by the quadratic equation: 1.29 × 10–4(0.0150 – x) = x2 2 x + 1.29 × 10–4x – 1.935 × 10– 6 = 0 x = 1.33 × 10–3 pH = –log 1.33 × 10–3 = 2.876 Reaction  

Uric acid (aq) [Uric acid]   0.0150   –x   0.0150 – x  



Think about It Simplifying assumptions are more often valid when a large difference exists between a larger initial concentration of the weak acid or base and a smaller value of Ka. 15.57. Collect and Organize By comparing the pH of rain in a weather system in the Midwest of 5.02 with the pH of the rain in that same system when it reached New England of 4.66, we can calculate how much more acidic the rain in New England was. Analyze We want to find the ratio

[H + ]New England [H + ]Midwest

Because pH = –log[H+], the [H+] = 1 × 10–pH. Therefore, [H + ]New England +

[H ]Midwest Solve

[H + ]New England

=

1 × 10 –4.66 1 × 10 –5.02

1 × 10 –4.66 2.19 × 10 –5 M = = 2.3 [H + ]Midwest 1 × 10 –5.02 9.55 × 10 –6 M The rain in New England in this weather system was 2.3 times more acidic than the rain in the Midwest. =

196 | Chapter 15 Think about It Among the causes of the acidity of the rain in New England are the coal-burning electricity power plants in the Midwest, which expel SO2 and SO3 into the air that make H2SO3 and H2SO4. 15.58. Collect and Organize We are given that the pH of a water sample in 1998 was 200 times lower than that of the lowest pH taken in the preceding 23 years. We are to calculate what this difference is in pH units. Analyze The ratio is expressed mathematically as

⎡⎣ H + ⎤⎦ 1 today = + 200 ⎡⎣ H ⎤⎦ previous

Solve In pH units this difference is –log 200 = –2.301 pH units. So the measurement in 1998 was 2.301 pH units more acidic than any reading in the previous 23 years. Think about It Because the pH scale is a log scale, a large change in [H+] gives just a small change in pH. 15.59. Collect and Organize Given the Kb for dimethylamine of 5.9 × 10– 4, we are to find the pH of a solution of dimethylamine that is 1.20 × 10–3 M. Analyze The equilibrium and Kb expressions for dimethylamine are (CH3)2NH(aq) + H2O ()  (CH3)2NH2+(aq) + OH–(aq)

[OH – ][(CH3 )2 NH 2 + ] [(CH3 ) 2 NH] We can set up a RICE table to solve this problem, where x is the amount of dimethylamine that ionizes. By solving for x we can calculate [OH–], from which we can determine the pOH and pH: pOH = –log[OH–] pH = 14 – pOH Kb =

Solve

Reaction   Initial   Change   Equilibrium  

(CH3)2NH(aq) [(CH3)2NH]   1.20 × 10–3   –x   1.20 × 10–3 – x  



OH–(aq) [OH–]   0   +x   x  

+

(CH3)2NH2+(aq) [(CH3)2NH2+]   0   +x   x  

x2 x2 ≈ 1.2 × 10 –3 − x 1.2 × 10–3 x = 8.41 × 10 –4 Checking the simplifying assumption shows that our simplifying assumption is not valid: 8.41 × 10–4 × 100 = 70% 1.20 × 10–3 so we must solve this by the quadratic equation: 5.9 × 10– 4 (1.20 × 10–3 – x) = x2 x2 + 5.9 × 10– 4x – 7.08 × 10–7 = 0 x = 5.966 × 10– 4 5.9 × 10–4 =

Aqueous Equilibria | 197 pOH = –log 5.97 × 10– 4 = 3.22 pH = 14 – 3.22 = 10.77 Think about It Our answer of a pH > 7 is consistent with dimethylamine’s behavior as a base in aqueous solution. 15.60. Collect and Organize Given the pKb of morphine, a weak base, we are asked to calculate the pH of a 0.115 M solution. Analyze From the pKb we can calculate the Kb

Kb = 1×10–pKb = 1×10–5.79 = 1.62 ×10– 6 The equilibrium and Kb expressions for the ionization of morphine are morphine(aq) + H2O ()  morphineH+(aq) + OH–(aq)

[OH – ][morphineH + ] = 1.62 ×10– 6 [morphine] We can set up a RICE table to solve this problem, where x is the amount of morphine that ionizes. By solving for x we can calculate [OH–], from which we can determine the pOH and pH: pOH = –log[OH–] pH = 14 – pOH Kb =

Solve

OH–(aq) + MorphineH+(aq)   [OH–]   [MorphineH+]   Initial   0   0   Change   +x   +x   Equilibrium   x   x   2 2 x x 1.62 × 10 –6 = ≈ 0.115 − x 0.115 x = 4.32 × 10–4 Checking the simplifying assumption shows that our simplifying assumption is valid: 4.32 × 10–4 × 100 = 0.38% 0.115 – From [OH ] we can calculate the pH: pOH = –log 4.32 × 10–4 = 3.365 pH = 14 – 3.365 = 10.635 Reaction  

Morphine(aq) [Morphine]     0.115 –x   0.115 – x  



Think about It Our calculation of the pH of this solution as greater than 7 is consistent with morphine’s behavior as a base in aqueous solution. 15.61. Collect and Organize Given the pKa of the conjugate acid of codeine (pKa = 8.21), we are to calculate the pH of a 3.42 × 10– 4 M solution of codeine, a weak base. Analyze We first need to determine the value of the Kb from the pKa for codeine:

Ka = 1 × 10– pKa = 1 × 10–8.21 = 6.17 × 10–9 Kb =

Kw Ka

=

1× 10 –14 = 1.62 × 10 – 6 6.17 × 10 –9

198 | Chapter 15 The equilibrium and Kb expressions for the ionization of codeine are codeine(aq) + H2O(ℓ)  codeineH+(aq) + OH–(aq)

[OH – ][codeineH + ] = 1.62 ×10– 6 [codeine] We can set up a RICE table to solve this problem, where x is the amount of codeine that ionizes. By solving for x we can calculate [OH–], from which we can determine the pOH and pH: pOH = –log[OH–] pH = 14 – pOH Kb =

Solve

Codeine(aq) + H2O(ℓ)  OH–(aq) + CodeineH+(aq)   [Codeine]   [OH–]   [CodeineH+]   Initial   0   0   3.42 × 10–4   Change   –x   +x   +x   Equilibrium   x   x   3.42 × 10–4 – x   2 2 x x ≈ 1.62 × 10 – 6 = 3.42 × 10– 4 − x 3.42 × 10 – 4 x = 2.35 × 10 –5

Reaction  

Checking the simplifying assumption shows that our simplifying assumption is not valid: 2.35 × 10–5 × 100 = 6.9% 3.42 × 10–4 so we must solve this by the quadratic equation: 1.62 × 10–6(3.42 × 10–4 – x) = x2 2 x + 1.62 × 10–6x – 5.54 × 10–10 = 0 x = 2.27 × 10–5 pOH = –log 2.27 × 10–5 = 4.644 pH = 14 – 4.644 = 9.356 Think about It Comparing the pKb of codeine to that of morphine in Problem 15.60, we see that codeine and morphine are identical in their basicity. 15.62. Collect and Organize Given that the Kb of pyridine is 1.7 × 10–9 in Appendix 5, we are to calculate the pH of a 0.125 M solution of this weak base. Analyze The equilibrium and Kb expressions for the ionization of pyridine are C5H5N(aq) + H2O(ℓ)  C5H5NH+(aq) + OH–(aq)

[OH – ][C5 H5 NH + ] = 1.7 × 10 –9 [C5 H5 N] We can set up a RICE table to solve this problem, where x is the amount of pyridine that ionizes. By solving for x we can calculate [OH–], from which we can determine the pOH and pH: pOH = –log[OH–] pH = 14 – pOH Kb =

Aqueous Equilibria | 199 Solve

OH–(aq) + C5H5NH+(aq)   – [OH ]   [C5H5NH+]   Initial   0   0   Change   +x   +x   Equilibrium   x   x   2 2 x x 1.7 × 10–9 = ≈ 0.125 − x 0.125 x = 1.46 × 10–5 Checking the simplifying assumption shows that our simplifying assumption is valid: 1.46 × 10 –5 × 100 = 0.01% 0.125 From [OH–] we can calculate the pH: pOH = –log 1.46 × 10–5 = 4.84 pH = 14 – 4.84 = 9.16 Reaction  

C5H5N(aq) + H2O(ℓ) [C5H5N]     0.125 –x   0.125 – x  



Think about It This solution of pyridine is more basic than a solution of baking soda. 15.63. Collect and Organize Given that the pKa of aspirin is 3.50, we are to calculate the percentage of aspirin that is present as the anion if 325 mg of aspirin is in 1 L of an aqueous solution that has a pH of 2.0. Analyze We first will have to calculate the concentration of aspirin that is present in the 1 L volume of the stomach. 1 mol 1 0.325 g × × = 1.804 × 10 −3 M 180.157 g L The equilibrium and Ka expressions for the ionization of aspirin are aspirin(aq) + H2O(ℓ)  H+(aq) + aspirin–(aq)

[H + ][Aspirin – ] = 10 –3.50 = 3.162 × 10−4 [Aspirin] In the stomach, the pH is 2.0, so the initial concentration of H+ is 1.00 × 10–2 M. We can set up a RICE table to solve this problem, where x is the amount of aspirin that ionizes. By solving for x we can calculate [Aspirin–]. For the calculation of the percent of aspirin that ionizes in the stomach at pH 2.0 we use x × 100 = percent ionization of aspirin 1.804 × 10 −3 M Ka =

Solve

Reaction   Initial   Change   Equilibrium  

Aspirin(aq) H+(aq) +  [Aspirin]   [H+]   1.804 × 10–4   1.00 × 10–2   –x   +x   1.804 × 10–4 – x   1.00 × 10–2 + x   –4

3.162 × 10 =

(

x 1.0 × 10−2 + x

Aspirin–(aq)   [Aspirin–]   0   +x   x  

) ≈ x (1.0 × 10 ) −2

1.804 × 10−3 − x 1.804 × 10−3 x = 5.70 × 10 –5 M Checking the simplifying assumption shows that our simplifying assumptions are valid: 5.70 × 10 –5 5.70 × 10 –5 × 100 = 0.5% and × 100 = 3.0% 0.010 1.804 × 10 –3

200 | Chapter 15 Therefore the percentage of aspirin that is ionized in the stomach at pH 2.0 is 5.70 × 10 −5 M × 100 = 3% 1.804 × 10 −3 M Think about It If the pH of the environment increases (higher than 2.0), then more aspirin will ionize. 15.64. Collect and Organize Given that the pKa of pseudoephedrine hydrochloride is 9.22, we are to calculate the pH of a 5 mL solution containing 30.0 mg. Analyze We first will have to calculate the concentration of pseudoephedrine hydrochloride that is in the 5 mL volume of the cough syrup. 1 mol 1 0.0300 g × × = 0.02975 M 201.70 g 0.005 L The equilibrium and Ka expressions for the ionization of codeine are pseudoephedrine⋅HCl(aq)  H+(aq) + pseudoephedrine⋅Cl–(aq)

[H + ][Pseudoephedrine ⋅Cl– ] = 10−9.22 = 6.026 × 10−10 [Pseudoephedrine ⋅ HCl] We can set up a RICE table to solve this problem, where x is the amount of pseudoephedrine hydrochloride that ionizes. By solving for x we can calculate [H+], from which we can determine the pH: pH = –log[H+] Ka =

Solve

H+(aq) + Pseudoephedrine ⋅ Cl–(aq) [H+]   [Pseudoephedrine ⋅ HCl]   [Pseudoephedrine ⋅ Cl–]   Initial   0.02975 0   0   Change   –x   +x   +x   Equilibrium   0.02975 – x   x   x   2 2 x x ≈ 6.026 × 10 –10 = 0.02975 − x 0.02975 x = 4.234 × 10 –6 Checking the simplifying assumption shows that our simplifying assumption is valid: 4.234 × 10−6 M × 100 = 0.0142% 0.02975M From x = [H+] we can calculate the pH: pH = –log 4.234 × 10–6 = 5.37 Reaction  

Pseudoephedrine ⋅ HCl (aq)



Think about It The use of the hydrochloride salt of ephedrine helps it to dissolve in the cough syrup’s ethanol and water solution. 15.65. Collect and Organize We are to explain why for H3PO4 Ka1 > Ka2 > Ka3. Analyze The equations describing these acid dissociation constants are as follows: H3PO4(aq)  H2PO4–(aq) + H+(aq) H2PO4–(aq)  HPO42–(aq) + H+(aq) HPO42–(aq)  PO43–(aq) + H+(aq)

K a1 K a2 K a3

Aqueous Equilibria | 201 Solve With each successive ionization, it becomes more difficult to remove H+ from a species that is negatively charged. Therefore it is harder to remove H+ from HPO42– than from H2PO4– than from H3PO4. This is reflected in decreasing Ka values as H3PO4 is ionized. Think about It From Appendix 5 we can compare the Ka values for phosphoric acid: Ka1 = 7.11 × 10–3, Ka2 = 6.32 × 10–8, Ka3 = 4.5 × 10–13. These values span 10 orders of magnitude. 15.66. Collect and Organize We are to explain why we can ignore the second ionization of H2SO3 but not that in H2SO4 in calculating the pH of 1.0 M aqueous solutions of these acids. Analyze The equations describing the ionization reactions of these two acids, with Ka values from Appendix 5, are as follows: Ka1= 1.7 × 10–2 H2SO3(aq)  HSO3–(aq) + H+(aq) – 2– + Ka2 = 6.2 × 10–8 HSO3 (aq)  SO3 (aq) + H (aq) H2SO4(aq)  HSO4–(aq) + H+(aq) HSO4–(aq)  SO42–(aq) + H+(aq)

Ka1 >> 1 Ka2 = 1.2 × 10–2

Solve Because there is a large difference (six orders of magnitude) in Ka2 and Ka1 for H2SO3, the ionization of HSO3– occurs to only a very small extent compared to the ionization of H2SO3, so the second ionization can be ignored. For H2SO4, Ka2 is sufficiently large (10–2) that it cannot be ignored. Think about It For most calculations involving polyprotic acids, the second ionization can be ignored. 15.67. Collect and Organize We have to use the Ka2 of H2SO4 to calculate the pH of a solution of 0.300 M H2SO4. Analyze The first H+ is completely removed from the H2SO4 and the initial concentrations of the species in solution before the second ionization are [H+] = 0.300 M, [H2SO4] = 0.0 M, and [HSO4–] = 0.300 M. The equation describing the second ionization is HSO4–(aq)  SO42–(aq) + H+(aq)

Ka = Solve

Reaction   Initial   Change   Equilibrium  

HSO4–(aq) [HSO4–]     0.300 –x   0.300 – x  

[H + ][SO4 2– ] = 1.2 × 10–2 [HSO4 – ] 

H+(aq) [H+]   0.300   +x   0.300 + x  

+

Plugging equilibrium concentrations into the Ka expression gives

1.2 × 10–2 = Solving this by the quadratic equation gives

x(0.300 + x) 0.300 − x

x2 + 0.312x – 0.0036 = 0 x = 0.0111

SO42–(aq)   [SO42–]   0   +x   x  

202 | Chapter 15 [H+] = 0.300 + 0.0111 = 0.311 pH = –log 0.311 = 0.51 Think about It If we did not take into consideration the second ionization of H2SO4, we would have underestimated the acidity of the solution by 0.016 pH units. 15.68. Collect and Organize Given a 0.150 M H2SO3 solution and Ka1 = 1.7 × 10–2 and Ka2 = 6.2 × 10–8 for this weak acid, we are to calculate the pH. Analyze Because Ka2 is so much smaller than Ka1 for this acid, we can say that the second ionization of H2SO3 contributes little to the [H+] in the solution. Therefore, we can solve this by examining only the first ionization. Solve

Reaction  

H2SO3(aq) [H2SO3]     0.150 –x   0.150 – x  

Initial   Change   Equilibrium  



HSO3–(aq) [HSO3–]   0   +x   x  

+

H+(aq)   [H+]   0   +x   x  

Substituting into the K a expression gives 1

1.7 × 10–2 = Solving this by the quadratic equation gives

x2 0.150 − x

x = 0.0427 pH = –log 0.0427 = 1.37

Think about It A simplifying assumption would not be valid in this problem because the initial concentration of H2SO3 is close to the value of Ka . 1

15.69. Collect and Organize Given a 0.250 M solution of ascorbic acid and fact that it is a weak triprotic acid (1.0 × 10–5 and 5 × 10–12, respectively), we are to calculate the pH. Analyze Because K a1 (1.0 × 10–5) for ascorbic acid is so much larger than the second or third ionization constants, we can say that the second and third ionizations contribute little to the [H+] in the solution. Therefore, we can solve this by examining only the first ionization. Solve

Reaction   Initial   Change   Equilibrium  

Ascorbic acid  Ascorbate– [Ascorbic acid]   [Ascorbate–]     0.250 0   –x   +x   0.250 – x   x  

x2 x2 ≈ 0.250 − x 0.250 x = 1.58 × 10–3

1.0 × 10–5 =

+

H+   [H+]   0   +x   x  

Aqueous Equilibria | 203 Checking the simplifying assumption shows that it is valid: 1.58 × 10–3 × 100 = 0.63% 0.250 The pH is found from the [H+]: pH = –log 1.58 × 10–3 = 2.80 Think about It This solution is about as acidic as vinegar. 15.70. Collect and Organize Given a 0.0288 M solution of oxalic acid (HO2CCO2H) and the pKa1 and pKa2 (from Appendix 5) for this weak diprotic acid (1.23 and 4.19, respectively), we are to calculate the pH. Analyze The K a1 and K a 2 are calculated from the pKa1and pKa2: Ka1 = 1 × 10–1.23 = 5.89 × 10–2 Ka2 = 1 × 10–4.19 = 6.46 × 10–5 Because the pKa2 value is so much smaller than the Ka1 for oxalic acid, we can say that the second ionization contributes little to the [H+] in the solution. Therefore, we can solve this by examining only the first ionization. Solve

HO2CCO2H(aq)  HO2CCO2–(aq) [HO2CCO2H] [HO2CCO2–]     Initial   0.0288 0   Change   –x   +x   Equilibrium   0.0288 – x   x   2 x 5.89 × 10–2 = 0.0288 − x Solving this by the quadratic equation gives x = 0.0212 pH = –log 0.0212 = 1.67 Reaction  

+ H+(aq)   [H+]   0   +x   x  

Think about It For a weak acid, oxalic acid with a pKa1 of 5.89 × 10–2 is fairly strong. Uses for this acid include bleaches, rustproofing chemicals, wood restorers, and rust removers. 15.71. Collect and Organize Given a 0.00100 M solution of nicotine and the pKb1 and pKb2 (from Appendix 5) for this weak dibasic compound, we are to calculate the pH. Analyze The pKb2 (1.3 × 10–11) is so much smaller than the pKb1 (1.0 × 10–6) for nicotine that we may ignore the contribution of the second ionization of nicotine to the [OH–] in the solution. We can therefore solve this problem by examining only the first ionization. Solve

Reaction   Initial   Change Equilibrium  

Nicotine(aq) + H2O(ℓ)  NicotineH+(aq) + OH–(aq)   [Nicotine]   [NicotineH+]   [OH–]     0.00100 0   0   –x   +x   +x   0.00100 – x   x   x  

204 | Chapter 15

x2 x2 ≈ 0.00100 − x 0.00100 x = 3.16 ×10 –5 Checking the simplifying assumption shows that it is valid: 3.16 × 10–5 × 100 = 3.2% 0.00100 1.0 ×10 – 6 =

We can calculate the pOH and pH from the [OH–]: pOH = –log 3.16 × 10–5 = 4.500 pH = 14 – 4.500 = 9.500 Think about It The relatively high pH of this dilute solution of nicotine shows that this compound is a strong weak base. 15.72. Collect and Organize Given a 2.50 × 10–4 M solution of H2NCH2CH2NH2 and the pK b and pKb for this weak dibasic compound 2 1 (3.29 and 6.44, respectively), we are to calculate the pH. Analyze The Kb1 and Kb2 values are calculated from the pKb1 and pKb2:

K b1 = 1 × 10–3.29 = 5.13 × 10–4

Kb2 = 1 × 10–6.44 = 3.63 × 10–7 Because K b is so much smaller than K b for H2NCH2CH2NH2, we can say that the second ionization 2 1 contributes little to the [OH–] in the solution. Therefore, we can solve this by examining only the first ionization. Solve

Reaction   Initial   Change   Equilibrium  

H2NCH2CH2NH2(aq) + H2O(ℓ)  H2NCH2CH2NH3+(aq) + OH–(aq)   [H2NCH2CH2NH2]  

[H2NCH2CH2NH3+]  

2.50 × 10–4   –x   2.50 × 10–4 – x  

0   +x   x  

5.13 × 10 – 4 = Solving this by the quadratic equation gives

[OH–]   0   +x   x  

x2 2.50 × 10− 4 − x

x = 1.84 × 10–4 pOH = –log 1.84 × 10–4 = 3.735 pH = 14 – 3.735 = 10.265

Think about It We must solve this problem by using the quadratic equation because pKb1 is about the same magnitude as the initial concentration of the weak base. In these cases, the usual simplifying assumption is often invalid. 15.73. Collect and Organize Given a 0.01050 M solution of quinine and the K b and K b for this weak dibasic compound, we are to 1 2 calculate the pH. Analyze The K b2 (1.4 × 10–9) is so much smaller than pKb1 (3.3 × 10–6) for quinine that we may ignore the contribution of the second ionization of quinine to the [OH–] in the solution. We can therefore solve this problem by examining only the first ionization.

Aqueous Equilibria | 205 Solve

Reaction Initial Change   Equilibrium  

Quinine(aq) + H2O(ℓ)  QuinineH+(aq) + OH–(aq)   [Quinine]   [QuinineH+]   [OH–]     0.01050 0 0 –x   +x   +x   0.01050 – x   x   x  

x2 x2 ≈ 0.01050 − x 0.01050 x = 1.86 × 10– 4 Checking the simplifying assumption shows that it is valid: 1.86 × 10 – 4 × 100 = 1.8% 0.01050 The pOH and pH can be calculated from the [OH–]: pOH = –log 1.86 × 10–4 = 3.730 pH = 14 – 3.730 = 10.270 3.3 × 10– 6 =

Think about It Quinine has a very complicated molecular structure that includes aromatic rings, an alcohol, an amine, an alkene, and an ether as functional groups.

15.74. Collect and Organize Given a 0.0133 M solution of piperazine and the pKb1 and pKb2 for this weak dibasic compound, we are asked to calculate the pH and then draw the structure of piperazine that would be present in 0.15 M HCl (stomach acid). Analyze (a) The pKb2 (2.15 × 10–9) is so much smaller than pKb1 (5.38 × 10–5) for piperazine that we may ignore the contribution of the second ionization of piperazine to the [OH–] in the solution. We can therefore solve this problem by examining only the first ionization. (b) Stomach acid is quite strong, so we would expect both nitrogen atoms to be protonated. Solve (a)

Reaction Initial   Change   Equilibrium  

Piperazine(aq) + H2O(ℓ)  PiperazineH+(aq) + OH–(aq)   [Piperazine]   [PiperazineH+]   [OH–]     0.0133 0   0   –x   +x   +x   0.0133 – x   x   x  

x2 x2 ≈ 0.0133 − x 0.0133 x = 8.46 ×10– 4 Checking shows that our simplifying assumption is not valid: 8.46 ×10 – 4 × 100 = 6.4% 0.0133 5.38 ×10–5 =

206 | Chapter 15 so we must solve this by the quadratic equation: x2 + 5.38 × 10–5x – 7.155 × 10–7 = 0 x = 8.19 × 10–4 pOH = –log 8.19 × 10–4 = 3.087 pH = 14 – 3.087 = 10.913 (b) H H

CH2

H2C

N

N H2C

CH2

H

2+

H

Think about It Piperazine is one of the stronger of the weak nitrogen-containing bases and a little more basic than ammonia (Kb = 1.76 × 10–5). 15.75. Collect and Organize Given the pKa values of conjugate acids of three pyridine derivatives where methyl groups are added, we are to determine whether more methyl groups increase or decrease the basicity of pyridine. Analyze The pKa is a measure of the conjugate acid’s acidity. From the equation pKa = –log Ka we see that the larger the pKa, the weaker the acid. The weaker the acid, the stronger the conjugate base. Solve As methyl groups are added, the pKa increases, so the acidity decreases. If the acidity decreases, the basicity of the conjugate base increases. Therefore, more methyl groups on the parent pyridine increases the base strength. Think about It Our prediction is true. The Kb values of the pyridine bases show that more methyl groups lead to increased basicity.

HC HC

H C

N

CH3

CH

HC

CH

HC

Kb = 1.51 × 10–9

C

N

CH3

CH C

HC CH3 H3C

Kb = 9.77 × 10–8

C

C

N

CH C

CH3

Kb = 2.69 × 10–7

15.76. Collect and Organize We are to explain why we do not need a table of Kb values if we already have the base’s conjugate acid Ka values. Analyze The acid and conjugate base are related to each other. Their respective Ka and Kb values are related by Kw: Kw = Ka × Kb and their pKa and pKb values are related through pKwI:;5:;E 3F W 14 = pKa + pKb

Aqueous Equilibria | 207 Solve If we have the Ka value of the conjugate acid, we can easily compute the Kb value of the base: Kw Kb = Ka Think about It Whether a species is listed in a Kb table or a Ka table usually depends on the neutral species’ dominant acid– base character. For example, H2SO3 acts as an acid, so it is listed in a Ka table, but NH3 acts as a base, so it is found in a Kb table. 15.77. Collect and Organize Of the three salts given, we are to determine which, when dissolved in water, gives an acidic solution. Analyze To give an acidic solution, the cation of the salt must donate a proton to water without the anion reacting with water, or if the anion hydrolyzes, then the pKa of the cation must be lower than the pKb of the anion. Solve Both NH4+ and CH3COO– of ammonium acetate hydrolyze: pKa = 9.25 NH4+(aq) + H2O(ℓ)  NH3(aq) + H3O+(aq) pKb = 9.25 CH3COO–(aq) + H2O(ℓ)  CH3COOH(aq) + OH–(aq) Because pKa = pKb, this salt’s solution is neutral. Only NH4+ of ammonium nitrate hydrolyzes. This gives an acidic solution: NH4+(aq) + H2O(ℓ)  NH3(aq) + H3O+(aq) – Only HCOO of sodium formate hydrolyzes. This gives a basic solution: HCOO–(aq) + H2O(ℓ)  HCOOH(aq) + OH–(aq) Therefore, of the three salts, only ammonium nitrate dissolves to give an acidic solution. Think about It Remember that neither Na+ nor NO3– hydrolyzes because it would form either a strong base or a strong acid, both of which are always 100% ionized. Na+(aq) + H2O(ℓ) ⎯⎯ → NaOH(aq) + H+(aq) NO3–(aq) + H2O(ℓ) ⎯⎯ → HNO3(aq) + OH–(aq) 15.78. Collect and Organize Of the three salts, NaF, KCl, and NH4Cl, we are to determine which would give a basic solution when dissolved in water. Analyze All these salts are soluble, so we need to examine how and whether the cation and anion of each salt react with water (hydrolyze). If the hydrolysis forms a strong acid or base, then that cation or anion will not react with water to change the pH of the solution. Solve For NaF: Na+(aq) + H2O(ℓ)  NaOH(aq) + H+(aq) F–(aq) + H2O(ℓ)  HF(aq) + OH–(aq) For KCl: K+(aq) + H2O(ℓ)  KOH(aq) + H+(aq) Cl–(aq) + H2O(ℓ)  HCl(aq) + OH–(aq)

NaOH is a strong base, so Na+ does not change the pH. HF is a weak acid, so F– does make this solution basic. KOH is a strong base, so K+ does not change the pH. HCl is a strong acid, so Cl– does not change the pH.

208 | Chapter 15 For NH4Cl: NH4+(aq) + H2O(ℓ)  NH3(aq) + H+(aq) Cl–(aq) + H2O(ℓ)  HCl(aq) + OH–(aq) Therefore, only NaF produces a basic solution.

NH3 is a weak base, so NH4+ does make this solution acidic. HCl is a strong acid, so Cl– does not change the pH.

Think about It When both cation and anion hydrolyze, the relative values of the Ka and Kb for the hydrolysis reactions determine whether the solution is acidic, basic, or neutral. 15.79. Collect and Organize We consider why lemon juice is used to reduce the fishy odor due to the presence of (CH3)3N in not-so-fresh seafood. Analyze Trimethylamine is a weak base, and lemon juice contains citric acid, which is a weak acid. Solve The citric acid in the lemon juice neutralizes the volatile trimethylamine to make a nonvolatile dissolved salt that neutralizes the fishy odor: HOC(CH2)2(COOH)3(aq) + (CH3)3N(aq)  HOC(CH2)2(COOH)2COO–(aq) + (CH3)3NH+(aq) Think about It Because the pKb of trimethylamine of 4.19 and the pKa of citric acid of 3.13 are lower than the pKa of trimethylammonium (9.81) and the pKb of citrate (10.87), this equilibrium lies to the right, favoring the products. 15.80. Collect and Organize We consider the acid–base properties of the calcium salt of malonic acid, Ca(OOCCH2COO). Analyze The conjugate base of malonic acid, –OOCCH2COO–, reacts with water, but Ca2+ does not. Solve The hydrolysis of the anion of calcium malonate gives a basic solution, as described by the equation – OOCCH2COO–(aq) + 2 H2O(ℓ)  HOOCCH2COOH(aq) + 2 OH–(aq) The pH of the beets is raised because of the presence of the calcium salt of malonic acid because of this hydrolysis reaction. Think about It Ca2+ does not hydrolyze because if it did, the strong base Ca(OH)2 would be formed. 15.81. Collect and Organize Given Ka = 2.1 × 10–11 for the conjugate acid of saccharin, we are asked to calculate the value of pKb for saccharin. Analyze From the Ka we can calculate the pKa:

pKa = –log Ka From the pKa we can calculate the pKb because pKI 

3F W: pKb = 14.00 – pKa

Aqueous Equilibria | 209 Solve

pKa = –log(2.1 × 10–11) = 10.68 pKb = 14.00 – 10.68 = 3.32

Think about It Alternatively, we could calculate the Kb from Ka by using

Kb = and then calculate pKb by using

Kw Ka

pKb = –log Kb

15.82. Collect and Organize Given the Ka1 and Ka2values for H2CrO4 (0.16 and 3.2 × 10–7, respectively), we are to calculate the value of Kb1 and Kb2 for CrO42–. Analyze The relevant equilibrium equations are CrO42–(aq) + H2O(ℓ)  HCrO4–(aq) + OH–(aq) HCrO4–(aq) + H2O(ℓ)  H2CrO4(aq) + OH–(aq) Solve The Kb for the first reaction is

K b1 = The Kb for the second reaction is

K b2 =

Kw Ka2 Kw K a1

=

1 × 10 –14 = 3.1 × 10 –8 3.2 × 10 –7

=

1 × 10 –14 = 6.3 × 10 –14 0.16

Think about It The first equilibrium contributes the most to the basicity of a solution of CrO42–, as its Kb is much larger than that of the second equilibrium equation. 15.83. Collect and Organize From Appendix 5 we know that the Ka of HF is 6.8 × 10–4. Using this, we are to calculate the pH of a solution that is 0.00339 M in NaF. Analyze When NaF dissolves in water, the F– ion hydrolyzes to give a basic solution: F–(aq) + H2O(ℓ)  HF(aq) + OH–(aq) The Kb for this reaction is Kw 1× 10 –14 Kb = = = 1.47 × 10 –11 K a 6.8 × 10 – 4

We can solve for [OH–] by using a RICE table and then compute the pH. Solve

Reaction   Initial Change   Equilibrium  

F–(aq) + H2O(ℓ) [F–]     0.00339 –x   0.00339 – x  



HF(aq) [HF]   0   +x   x  

+

OH–(aq)   [OH–]   0   +x   x  

210 | Chapter 15

x2 x2 ≈ 0.00339 − x 0.00339 x = 2.23 × 10–7 Checking the simplifying assumption shows that it is valid: 2.23 × 10–7 × 100 = 0.0066% 0.00339 – The pH is calculated from the [OH ]: pOH = –log 2.23 × 10–7 = 6.65 pH = 14 – 6.65 = 7.35 1.47 × 10–11 =

Think about It Because HF is a moderately strong weak acid, F–, its conjugate base, is a fairly weak base. 15.84. Collect and Organize Given the Kb of ephedrine (3.86), we are to calculate the pH of a 1.25 × 10–2 M solution of its hydrochloride salt. Analyze When ephedrine ⋅ HCl dissolves in water, it produces an acidic solution: ephedrineH+(aq)  ephedrine(aq) + H+(aq) The Ka for this reaction is K w 1 × 10 –14 = = 7.24 × 10 –11 Ka = –3.86 K b 1 × 10 We can solve for [H+] by using an RICE table.

Solve

EphedrineH+(aq)  Ephedrine(aq) + H+(aq)   [EphedrineH+]   [Ephedrine]   [H+]     Initial   0.0125 0   0   Change   –x   +x   +x   Equilibrium   0.0125 – x   x   x   2 2 x x 7.24 × 10–11 = ≈ 0.0125 − x 0.0125 x = 9.51 × 10–7 Checking the simplifying assumption shows that it is valid: 9.57 × 10–7 × 100 = 0.007% 0.0125 The pH is found from the [H+]: pH = –log 9.51 × 10–7 = 6.02 Reaction  

Think about It EphedrineH+ is quite a weak acid, and for this solution the percent ionization is only 0.007%. 15.85. Collect and Organize We are to explain why a solution of CH3COOH with CH3COONa is a better pH buffer than a solution containing NaCl and HCl. Analyze A buffer is composed of a weak acid and its conjugate base.

Aqueous Equilibria | 211 Solve Because the weak acid CH3COOH is combined with its weak conjugate base, CH3COO–, this buffer can absorb added H+ or OH–. The other mixture, HCl with Cl–, is a strong acid paired with its very, very weak conjugate base. This pairing cannot absorb added H+ or OH–. Think about It That both conjugates be weak for the acid–base pair in a buffer system is a key idea. 15.86. Collect and Organize We consider why a solution composed of a weak base with its conjugate acid is a better buffer than a solution of only the weak base. Analyze A buffer acts to absorb both OH– and H+ added to the solution. Solve The presence of both the base and the conjugate acid in a buffer means that the system can absorb both added H+ and OH–. Because the concentration of the conjugate acid (present only because of a small degree of hydrolysis of the base) is too small, a buffer composed only of a weak base cannot absorb much OH–. Think about It A good buffer has relatively high and roughly equal concentrations of the weak base and its conjugate acid. 15.87. Collect and Organize For a buffer that is 0.244 M in acetic acid and 0.122 M in sodium acetate, we can use the Henderson–Hasselbalch 7CG3F;A@FA53>5G>3F7F:7B A8F:;E4G887D3F WKa = 1.76 × 10–53@63F WKa = 1.64 × 10–5). Analyze The Henderson–Hasselbalch equation is

[base] [acid] For this problem, [base] = [acetate] = 0.122 M and [acid] = [acetic acid] = 0.244 M. Because the Ka values are different for the two temperatures, the pH of these solutions will differ. pH = pK a + log

Solve

0.122 = 4.453 0.244 0.122 F W pH = –log(1.64 × 10–5 ) + log = 4.484 0.244 F W pH = –log(1.76 × 10–5 ) + log

Think about It At the lower temperature the pH of this buffer is less acidic than at the higher temperature. 15.88. Collect and Organize For a buffer that is 0.100 M in pyridine and 0.275 M in pyridinium chloride, we can use the Henderson– Hasselbalch equation to calculate the pH of this buffer (pKb for pyridine = 8.8). Analyze We are given the pKb for pyridine, and from this we can calculate the pKa for use in the Henderson–Hasselbalch 7CG3F;A@3F W pKa = 14 – pKb = 14 – 8.8 = 5.2

212 | Chapter 15 The Henderson–Hasselbalch equation is

[base] [acid] For this problem, [base] = [pyridine] = 0.100 M and [acid] = [pyridinium] = 0.275 M. pH = pK a + log

Solve

pH = 5.2 + log

0.100 = 4.8 0.275

Think about It If we were to increase the concentration of base in relation to the acid, the pH of the solution would increase because the value of log([base]/[acid]) in the Henderson–Hasselbalch equation becomes more positive. 15.89. Collect and Organize For a buffer that is 0.225 M in both HPO42– and PO43– (with the Ka of HPO42– equal to 4.5 × 10–13), we can use the Henderson–Hasselbalch equation to calculate the pH of the buffer. From the pH we can also calculate the pOH. Analyze The Henderson–Hasselbalch equation is

pH = pK a + log For this problem, [base] = [acid] = 0.225 M. Solve

[base] [acid]

0.225M = 12.34 0.225M pOH = 14 – pH = 14 –12.34 = 1.65

pH = –log(4.5 × 10–13 ) + log

Think about It When the concentrations of the acid and its conjugate base are equal, pH = pKa because log 1 = 0. 15.90. Collect and Organize For a buffer that is 0.0200 M in boric acid and 0.0250 M in sodium borate (with the pKa of boric acid equal to 9.27), we can use the Henderson–Hasselbalch equation to calculate the pH of the buffer. From the pH we can also calculate the pOH. Analyze The Henderson–Hasselbalch equation is

pH = pK a + log For this problem, [base] = 0.0250 M and [acid] = 0.0200 M.

[base] [acid]

Solve

0.0250 = 9.37 0.0200 pOH = 14 – pH = 14 – 9.37 = 4.63 pH = 9.27 + log

Think about It Because the ratio [base]/[acid] is greater than 1, the calculated pH is greater than the pKa of the acid.

Aqueous Equilibria | 213 15.91. Collect and Organize Given the pH of an acetic acid–acetate buffer solution (pH = 3.56) where the Ka of acetic acid is 1.76 × 10–5, we can use the Henderson–Hasselbalch equation to calculate the ratio of acetate ion to acetic acid in the solution. Analyze Rearranging the Henderson–Hasselbalch equation to solve for the ratio gives [acetate] pH = pK a + log [acetic acid]

log

[acetate] = pH – pK a [acetic acid]

[acetate] ( pH–pKa ) = 1× 10 [acetic acid] Solve The pKa = –log(1.76 × 10–5) = 4.754, so pH – pKa = 3.56 – 4.754 = –1.194 and the ratio of acetate to acetic acid is [acetate] = 1 × 10 –1.194 = 0.064 [acetic acid] Think about It Because [acetic acid] > [acetate], the pH of this buffer is less than the pKa of acetic acid. 15.92. Collect and Organize Given the pH of a lactic acid–lactate buffer solution (pH = 4.00) where the Ka of lactic acid is 1.4 × 10– 4, we can use the Henderson–Hasselbalch equation to calculate the ratio of lactic acid to lactate in the solution. Analyze Rearranging the Henderson–Hasselbalch equation to solve for the ratio gives [lactate] pH = pK a + log [lactic acid]

log

[lactate] = pH – pK a [lactic acid]

1 [lactic acid] = pH–pKa ) [lactate] 1× 10( Solve The pKa = –log(1.4 × 10–4) = 3.85, so pH – pKa = 4.00 –3.85 = 0.15 and the ratio of lactic acid to lactate is [lactic acid] 1 = = 0.71 [lactate] 1 × 100.15 Think about It Because [lactate] > [lactic acid], the pH of this buffer is higher than the pKa of lactic acid. 15.93. Collect and Organize We can use the Henderson–Hasselbalch equation to determine the pH of a solution prepared by mixing a volume of 0.05 M NH3 with an equal volume of 0.025 M HCl.

214 | Chapter 15 Analyze Mixing equal volumes of solutions dilutes them both. Therefore, after mixing and before reaction, [NH3] = 0.025 M and [HCl] = 0.0125 M in the combined solution. When HCl reacts with NH3, NH4+ is produced according to the equation NH3(aq) + H+(aq) → NH4+(aq) Stoichiometrically, this would give [NH3] = 0.0125 M and [NH4+] = 0.0125 M after complete reaction with H+. Because this is a solution of an acid (NH4+) and its conjugate base (NH3), we can use the Henderson– Hasselbalch equation to calculate the pH of the solution. To do so we need the pKa of NH4+ (9.25) from Appendix 5 (Table A5.1). Solve

pH = pKa + log

[NH3 ] 0.0125 = 9.25 + log = 9.25 0.0125 [NH4 + ]

Think about It Because the HCl added to the NH3 in this solution converts exactly half of the NH3 to NH4+, the ratio of the concentrations equals 1 and the pH of the solution equals the pKa of NH4+. 15.94. Collect and Organize We can use the Henderson–Hasselbalch equation to determine the pH of a solution prepared by mixing a volume of 0.05 M acetic acid with an equal volume of 0.025 M NaOH. Analyze Mixing equal volumes of solutions dilutes them both. Therefore, after mixing and before reaction, [CH3COOH] = 0.025 M and [NaOH] = 0.0125 M in the combined solution. When NaOH reacts with CH3COOH, CH3COO– is produced according to the equation CH3COOH(aq) + OH–(aq) → CH3COO–(aq) + H2O () Stoichiometrically, this would give [CH3COOH] = 0.0125 M and [CH3COO–] = 0.0125 M after complete reaction with OH–. Because this is a solution of an acid (CH3COOH) and its conjugate base (CH3COO–), we can use the Henderson–Hasselbalch equation to calculate the pH of the solution. To do so we need the pKa of CH3COOH from Appendix 5 (4.75). Solve

pH = pK a + log

[CH3 COO – ] 0.0125 = 4.75 + log = 4.75 [CH3 COOH] 0.0125

Think about It Because the NaOH added to the CH3COOH in this solution converts exactly half of the CH3COOH to CH3COO–, the ratio of the concentrations equals 1 and the pH of the solution equals the pKa of CH3COOH. 15.95. Collect and Organize We are to calculate how much of the strong acid HNO3 (10 M) we would add to a buffer solution that is 0.010 M acetic acid and 0.10 M sodium acetate to give a solution of pH 5.00. Analyze We can use the Henderson–Hasselbalch equation to first calculate the ratio [acetate]/[acetic acid] needed to give a solution of pH 5.00. We then use that ratio to determine how much HNO3 to add to convert the acetate in solution to acetic acid to give that ratio. The pKa of acetic acid is 4.75.

Aqueous Equilibria | 215 Solve

[acetate] [acetic acid] [acetate] 5.00 = 4.75 + log [acetic acid] [acetate] = 1.78 [acetic acid] When we add HNO3 to the solution, acetate is converted to acetic acid and the amounts of each are [acetate] = 0.10 – x [acetic acid] = 0.10 + x The ratio of these is 1.78. Solving for x gives [acetate] 0.10 − x = 1.78 = [acetic acid] 0.10 + x 0.10 − x = 0.0178 + 1.78 x pH = pK a + log

x = 0.0296 The value of x is the moles of HNO3 that we need to add to the solution. In milliliters of 10 M HNO3, then, 1000 mL 0.0296 mol × = 3.0 mL for 1.00 L of the buffer solution 10 mol Think about It The small volume addition of HNO3 does not appreciably change the concentration of the acid and the base, so we do not need to account for it. 15.96. Collect and Organize For this problem we are to calculate how much of the strong base NaOH (6.0 M) we would add to 0.500 L of a buffer solution that is 0.0200 M acetic acid and 0.0250 M sodium acetate to give a solution that is pH 5.75. Analyze We can use the Henderson–Hasselbalch equation to first calculate the ratio of [acetate]/[acetic acid] needed to give a solution of pH 5.75. We then use that ratio to determine how much NaOH to add to convert the acetate in solution to acetic acid to give that ratio. The pKa of acetic acid is 4.75. Solve

[acetate] [acetic acid] [acetate] 5.75 = 4.75 + log [acetic acid] [acetate] = 10 [acetic acid] When we add NaOH to the solution, acetic acid is converted to acetate and the amounts of each are [acetate] = 0.0250 + x [acetic acid] = 0.0200 – x The ratio of these will be 10. Solving for x gives [acetate] 0.0250 + x = 10 = [acetic acid] 0.0200 – x 0.0250 + x = 0.200 –10 x x = 0.0159 pH = pK a + log

216 | Chapter 15 The value of x is the moles of NaOH that we need to add to 1 L of the buffer solution. In milliliters of 6.0 M NaOH, then, 1000 mL 0.0159 mol × = 2.7 mL for 1 L of the buffer solution 6.0 mol For 500 mL of the buffer solution, we need only one-half of this amount, or 1.3 mL of 6 M NaOH solution. Think about It The small volume addition of NaOH does not appreciably change the concentration of the acid and the base, so we do not need to account for it. 15.97. Collect and Organize We are to compare the pH of 1.00 L of a buffer that is 0.120 M in HNO2 and 0.150 M in NaNO2 before and after 1.00 mL of 12.0 M HCl is added. Analyze We can use the Henderson–Hasselbalch equation in both cases. After the addition of HCl, however, the amounts (calculated in moles) of HNO2 and NO2– have to be adjusted before using the Henderson–Hasselbalch equation. The pKa of HNO2 is 3.40. Solve Without added HCl, the pH of the buffer solution is

pH = 3.40 + log

0.150 = 3.50 0.120

Because we have 1.00 L of the buffer solution, we originally have 0.120 mol HNO2 and 0.150 mol NO2– in the solution. Adding 1.00 mL of 12.0 M HCl adds

1.00 mL ×

12.0 mol = 0.0120 mol H+ 1000 mL

This will increase the moles of HNO2 to 0.120 mol + 0.0120 mol = 0.132 mol and decrease the moles of NO2– to 0.150 mol – 0.0120 mol = 0.138 mol. Because the volume of the solution is 1.00 L, the concentrations of these species are [HNO2] = 0.132 M and [NO2–] = 0.138 M. Using the Henderson–Hasselbalch equation to calculate the pH gives

pH = 3.40 + log

0.138 = 3.42 0.132

Think about It The pH of the buffer changed very little. The change in pH of 1.00 L of water after adding 0.0120 mol of H + would be from 7.00 to 1.92. 15.98. Collect and Organize We are to compare the pH of 100.0 mL of a buffer that is 0.100 M in NH4Cl and 0.100 M in NH3 before and after the addition of 1.0 mL of 6 M HNO3. Analyze We can use the Henderson–Hasselbalch equation in both cases. After the addition of HNO3, however, the amounts (calculated in moles) of NH4+ and NH3 have to be adjusted before using the Henderson–Hasselbalch equation. The pKa of NH4+ is 9.25. Solve Without added HNO3 the pH of the buffer solution is

pH = 9.25 + log

0.100 = 9.25 0.100

Aqueous Equilibria | 217 Because we have 100.0 mL of the buffer solution, we originally have 0.0100 mol NH4+ and 0.0100 mol NH3 in the solution. Adding 1.0 mL of 6 M HNO3 adds

1.00 mL ×

6.0 mol = 0.0060 mol H+ 1000 mL

This will increase the amount of NH4+ to 0.0100 mol + 0.0060 mol = 0.0160 mol and decrease the amount of NH3 to 0.0100 mol – 0.0060 mol = 0.0040 mol. Because the volume of the solution is 0.1000 L, the concentrations of these species are [NH4+] = 0.160 M and [NH3] = 0.040 M. Using the Henderson–Hasselbalch equation to calculate the pH gives

pH = 9.25 + log

0.040 = 8.65 0.160

Think about It Our answer shows that adding acid to the buffer lowers the pH (slightly), as we would expect. 15.99. Collect and Organize We are to describe the difference between a titration curve for a strong acid and that of a weak acid. Analyze A strong acid is completely ionized in aqueous solution, whereas a weak acid is only partially hydrolyzed. This affects the pH of the solution at the start of the titration for equal concentrations of the acids. The equivalence or end point of the titration is where equal moles of OH– have been added to the acid. The species formed at the end point for a strong acid and a weak acid differ. This affects the pH of the solution at the equivalence point. Solve The weak acid titration curve has an initial pH that is higher (less acidic) than that of an equimolar solution of a strong acid (lower pH, more acidic). The pH at the equivalence point in the titration of a strong acid is 7.00 because the species formed in the titration are water and a nonhydrolyzing anion, such as Cl–: HCl(aq) + OH–(aq) → H2O(ℓ) + Cl–(aq) The pH at the equivalence point for a weak acid is basic because of the formation of a hydrolyzing anion, such as NO2– in the titration of HNO2: HNO2(aq) + OH–(aq) → NO2–(aq) + H2O(ℓ) NO2–(aq) + H2O(ℓ)  HNO2(aq) + OH–(aq)

Think about It Titration of a weak acid always gives an end point of pH > 7. 15.100. Collect and Organize We are asked whether the pH at the equivalence point in all titrations of a strong base with a strong acid is the same. Analyze A strong base such as OH– (for example, from NaOH or KOH), when titrated with strong acid, produces water plus a dissolved salt.

218 | Chapter 15 Solve Yes. The pH at the equivalence point of strong bases during titrations with strong acids is always pH 7.00, as seen by the formation of H2O and salts that do not hydrolyze, as shown in the following examples: NaOH(aq) + HCl(aq) → H2O(ℓ) + Na+(aq) + Cl–(aq) Ca(OH)2(aq) + 2 HNO3(aq) → 2 H2O(ℓ) + Ca2+(aq) + 2 NO3–(aq) Na2O(aq) + 2 HClO4(aq) → H2O(ℓ) + 2 Na+(aq) + 2 ClO4–(aq)

Think about It Likewise, the equivalence point is at pH 7.00 for all titrations of strong acids with strong bases. 15.101. Collect and Organize We are asked whether the pH at the equivalence point is the same for the titration of all weak acids with strong base. Analyze When a weak acid is titrated, the species present in solution at the equivalence point is the conjugate base of the weak acid. This conjugate base is a weak base and will hydrolyze in water according to the equation A–(aq) + H2O(ℓ)  HA(aq) + OH–(aq) Solve No. Because the extent to which A– (the conjugate base) hydrolyzes depends on the base strength of A–, the pH at the equivalence point in the titration for weak acids is not expected to be the same.

Think about It Likewise, the pH values at the equivalence point for weak bases differ according to the strength of the conjugate acid.

Aqueous Equilibria | 219 15.102. Collect and Organize We are to list the properties of an acid–base indicator. Analyze An indicator is a visual indicator of the pH of a solution. Solve An indicator must have a different color in its base from that in its acid form. Think about It In using an indicator in a titration we have to be careful to choose an indicator that changes color in the pH range of the expected equivalence point. 15.103. Collect and Organize We are to calculate the pH along various points of the titration curve when 25.0 mL of 0.100 M acetic acid (Ka = 1.76 × 10–5) is titrated with 0.125 M NaOH. Analyze For each step of the titration we have to consider the moles of NaOH added that react with the moles of CH3COOH initially present:

25.0 mL ×

0.100 mol = 0.00250 mol CH3COOH 1000 mL

For each point in the titration curve, the moles of OH– (from NaOH) added are as follows: 0.125 mol 10.0 mL × = 0.00125 mol OH – 1000 mL 0.125 mol 20.0 mL × = 0.00250 mol OH – 1000 mL 0.125 mol 30.0 mL × = 0.00375 mol OH – 1000 mL The OH– reacts with the CH3COOH in solution to give CH3COO–, the conjugate base of acetic acid. Our strategy for the problem is to first react as much of the added OH – with acetic acid as possible and then use the equilibrium Ka expression to calculate the pH: CH3COOH(aq)  CH3COO–(aq) + H+(aq)

[CH3COO – ][H + ] [CH3COOH] Or we can use the equivalent equilibrium Kb expression: CH3COO–(aq) + H2O(ℓ)  CH3COOH(aq) + OH–(aq) Ka =

[CH3COOH][OH – ] [CH3COO – ] For that calculation we have to be careful to determine the molarity of the species in solution by remembering that the volume in a titration increases through the addition of the titrant. Kb =

Solve When 10.0 mL of OH– is added, the 0.00125 mol of OH– reacts with the 0.00250 mol of CH3COOH to produce 0.00125 mol of CH3COO– and leaves 0.00125 mol of CH3COOH unreacted. Because the total volume of the solution is now 25.0 + 10.0 = 35.0 mL, the molarity of these species is 0.00125 mol = 0.0357 M 0.0350 L

220 | Chapter 15 Using a RICE table to calculate the pH of this solution gives the following: CH3COO–(aq) Reaction   CH3COOH(aq)  [CH3COOH]   [CH3COO–]     Initial   0.0357 0.0357   Change   –x   +x   Equilibrium   0.0357 – x   0.0357 + x  

+

H+(aq)   [H+]   0   +x   x  

x(0.0357 + x) x(0.0357) ≈ 0.0357 − x 0.0357 x = 1.76 × 10 –5

1.76 × 10 –5 =

The simplifying assumption is valid (0.05%), so pH = –log 1.76 × 10–5 = 4.754 – When 20.0 mL of OH is added, we have added an equal number of moles of OH–, as moles of CH3COOH are initially present. This reaction produces 0.00250 mol of CH3COO–, so using the Kb expression to calculate the pH of the solution makes sense here. Because the total volume of the solution is now 45.0 mL, the molarity of these species is 0.00250 mol = 0.0556 M 0.0450 L Using a RICE table to calculate the pH of this solution gives the following: Reaction   CH3COO–(aq) + H2O(ℓ)  CH3COOH(aq) + OH–(aq)   [CH3COO–]   [CH3COOH]   [OH–]     Initial 0.0556 0   0   Change   –x   +x   +x Equilibrium   0.0556 – x   x   x  

Kb =

Kw Ka

=

1× 10−14 x2 x2 –10 = 5.68 × 10 = ≈ 0.0556 − x 0.0556 1.76 × 10 –5 x = 5.62 × 10 – 6

The simplifying assumption is valid (0.01%), so pOH = –log 5.62 × 10–6 = 5.250 pH = 14 – 5.250 = 8.750 When 30.0 mL of OH– is added, we convert all the CH3COOH to 0.00250 mol CH3COO– and have 0.00125 mol OH– remaining. Because the total volume of the solution is now 55.0 mL, the molarity of these species is 0.00250 mol = 0.0455 M CH3COO – 0.0550 L 0.00125 mol = 0.0227 M OH – 0.0550 L Using a RICE table to calculate the pH of this solution gives the following: Reaction CH3COO–(aq) + H2O(ℓ)  CH3COOH(aq) + OH–(aq)   [CH3COO–]   [CH3COOH] [OH–]     Initial   0.0455 0   0.0227   Change –x   +x   +x   Equilibrium   0.0455 – x   x   0.0227 + x   x(0.0227 + x) x(0.0227) ≈ 5.68 × 10 –10 = 0.0455 − x 0.0455 –9 x = 1.14 × 10 The simplifying assumption is valid, so pOH = –log 0.0227 = 1.644 pH = 14 – 1.644 = 12.356

Aqueous Equilibria | 221

Think about It When exactly half the moles of strong base are added in the titration of a weak acid (as in this problem, where 10 mL of the titrant was added), this point is the midpoint of the titration. At this point pH = pKa of the weak acid. 15.104. Collect and Organize We are to calculate the pH along various points of the titration curve when 25.0 mL of 0.100 M trimethylamine (Kb = 6.46 × 10–5) is titrated with 0.125 M HCl. Analyze For each step of the titration we have to consider the moles of HCl added that react with the moles of (CH3)3N initially present:

25.0 mL ×

0.100 mol = 0.00250 mol (CH3 )3 N 1000 mL

For each point in the titration curve the moles of H+ (from HCl) added are as follows: 0.125 mol 10.0 mL × = 0.00125 mol H + 1000 mL 0.125 mol 20.0 mL × = 0.00250 mol H + 1000 mL 0.125 mol 30.0 mL × = 0.00375 mol H + 1000 mL The H+ reacts with the (CH3)3N in solution to give (CH3)3NH+, the conjugate acid of trimethylamine. Our strategy for the problem is to first react as much of the added H + with trimethylamine as possible and then use the equilibrium Kb expression to calculate the pH: (CH3)3N(aq) + H2O(ℓ)  (CH3)3NH+(aq) + OH–(aq)

[(CH3 )3 NH + ][OH – ] [(CH3 )3 N] Or we can use the equivalent equilibrium Ka expression: (CH3)3NH+(aq)  (CH3)3N(aq) + H+(aq) Kb =

[(CH3 )3 N][H + ] [(CH3 )3 NH + ] For that calculation we have to be careful to determine the molarity of the species in solution by remembering that the volume in a titration increases through the addition of the titrant. Ka =

222 | Chapter 15 Solve When 10.0 mL of H+ is added, the 0.00125 mol of H+ reacts with the 0.00250 mol of (CH3)3N to produce 0.00125 mol of (CH3)3NH+ and leave 0.00125 mol of (CH3)3N unreacted. Because the total volume of the solution is now 25.0 + 10.0 = 35.0 mL, the molarity of these species is 0.00125 mol = 0.0357 M 0.0350 L Using a RICE table to calculate the pH of this solution gives Reaction   (CH3)3N(aq) + H2O(ℓ)  (CH3)3NH+(aq) + OH–(aq)   [(CH3)3N]   [(CH3)3NH+]   [OH–]     Initial   0.0357 0.0357   0   Change   –x   +x   +x   Equilibrium   0.0357 – x   0.0357 + x   x   x(0.0357 + x) x(0.0357) ≈ 6.46 × 10–5 = 0.0357 − x 0.0357 –5 x = 6.46 × 10 The simplifying assumption is valid (0.2%), so pOH = –log 6.46 × 10–5 = 4.190 pH = 14 – 4.190 = 9.810 When 20.0 mL of H+ is added, we have added an equal number of moles of H+ as there are moles of (CH3)3NH initially present. This reaction produces 0.00250 mol of (CH3)3NH+, so using the Ka expression to calculate the pH of the solution makes sense here. Because the total volume of the solution is now 45.0 mL, the molarity of this species is 0.00250 mol = 0.0556 M 0.0450 L Using a RICE table to calculate the pH of this solution gives Reaction   (CH3)3NH+(aq)  (CH3)3N(aq) + H+(aq)   [(CH3)3NH+]   [(CH3)3N]   [H+]     Initial   0.0556 0   0   Change   –x   +x   +x   Equilibrium   0.0556 – x   x   x   −14 2 2 Kw 1× 10 x x Ka = = = 1.55 × 10 –10 = ≈ –5 K b 6.46 × 10 0.0556 − x 0.0556

x = 2.94 × 10 – 6 The simplifying assumption is valid (0.005%), so pH = –log 2.94 × 10– 6 = 5.532 + When 30.0 mL of H is added, we convert all the (CH 3)3N to 0.00250 mol of (CH3)3NH+ and have 0.00125 mol of H+ remaining. Since the total volume of the solution is now 55.0 mL, the molarity of these species is 0.00250 mol = 0.0455 M (CH3 )3 NH + 0.0550 L 0.00125 mol = 0.0227 M H + 0.0550 L Using a RICE table to calculate the pH of this solution gives Reaction   (CH3)3NH+(aq)  (CH3)3N(aq) + H+(aq)   + [(CH3)3NH ]   [(CH3)3N]   [H+]     Initial   0.0455 0   0.0227   Change   –x   +x   +x   Equilibrium   0.0455 – x   x   0.0227 + x  

Aqueous Equilibria | 223

x(0.0227 + x) x(0.0227) ≈ 0.0455 − x 0.0455 x = 3.11 × 10 –10

1.55 × 10 –10 = The simplifying assumption is valid, so

pH = –log 0.0227 = 1.644

Think about It In the titration of a weak base with a strong acid, the pH at the midpoint is equal to the pKa of the conjugate acid. In this problem, we calculated the midpoint to be at pH = 9.810. This is equal to –log Ka = –log 1.55 × 10–10. 15.105. Collect and Organize In the titration of a 100.00 mL NH3 solution with 0.1145 M HCl, it takes 22.35 mL to reach the equivalence point. From this information we are to calculate the concentration of ammonia in the solution. Analyze Because at the equivalence point the moles of HCl added as a titrant equal the moles of NH3 in the solution, we can calculate the amount of NH3 through 1 mol NH 3 moles NH 3 = mL HCl used as titrant × molarity of HCl solution × 1 mol HCl Because we know the volume of the original solution of NH3, the molarity of the sample is mol NH3

volume of sample in L Solve

0.1145 mol 1 mol NH 3 × = 2.559 ×10 –3 mol 1 mol HCl 1000 mL 2.559 ×10 –3 mol molarity of NH 3 = = 0.02559 M 0.100 L

moles NH 3 = 22.35 mL HCl ×

Think about It Remember that at the equivalence point, what is equal is the moles, not the volume or the concentration of acid and base. 15.106. Collect and Organize For the titration of a 100.0 mL sample from a hot spring containing CO32– and HCO3–, we are to calculate the concentrations of each species, knowing that the first equivalence point is reached when 2.56 mL of a 0.0355 M solution of HCl has been added and the second equivalence point is reached when 10.42 mL of the HCl has been added.

224 | Chapter 15 Analyze As the titration proceeds, the CO32– ion is converted into the HCO3– ion: CO32–(aq) + H+(aq) → HCO3–(aq) This then adds to the concentration of HCO3– already present and titrated to reach the second equivalence point: HCO3–(aq) + H+(aq) → H2CO3(aq) We therefore have to subtract the moles of CO32– found in the first equivalence point from the total moles of HCO3– found from the second equivalence point to obtain the moles of HCO3– present originally in the sample. Solve From the first equivalence point,

0.0355 mol 1 mol CO32– × = 9.09 × 10–5 mol CO32– 1000 mL 1 mol H+ 9.09 ×10 –5 mol ⎡CO32– ⎤ in sample = = 9.09 ×10 – 4 M ⎣ ⎦ 0.100 L From the second equivalence point, 0.0355 mol 1 mol HCO3 – 10.42 mL HCl × × = 3.70 × 10 – 4 mol HCO3 – 1000 mL 1 mol H + moles of HCO3– present in original sample = 3.70 ×10 – 4 mol – 9.09 ×10 –5 mol = 2.79 ×10 – 4 mol 2.56 mL HCl ×

– ⎡ ⎤ ⎣ HCO3 ⎦ in sample =

2.79 × 10 – 4 mol = 2.79 × 10–3 M 0.100 L

Think about It In this sample the pH is such that [HCO3– ] > [CO32–]. At higher pH, less HCO3– would be expected. 15.107. Collect and Organize We are to calculate the volume of 0.100 M HCl required to titrate 250 mL of 0.0100 M Na2CO3 to the first equivalence point. Analyze Before setting out to do a lot of calculations here, let’s stop and think. The concentration of the titrant (HCl) is one-tenth the concentration of the base we are titrating. We don’t need, therefore, to do any calculations because the one tenth of the volume of HCl is needed to neutralize the base. Solve Titration of 250 mL of 0.0100 M Na2CO3 to the first equivalence point requires 25.0 mL of HCl. To reach the second endpoint we would need twice that amount, or 50.0 mL. Think about It To reach the first endpoint we need equimolar amount of acid and base. In this problem we hve 2.5 × 10–3 moles of CO32–. The frist equivlance point is reached, therefore, when we have added 2.5 × 10–3 moles HCl (25 mL) and the second equivalence point is reached when we have added two equivalents of acid, or 5.0 × 10–3 moles HCl (50 mL). 15.108. Collect and Organize We are to determine the volume of 0.0100 M HCl needed to titrate 250 mL of 0.0100 M Na2CO3 and 250 mL of 0.0100 M HCO3–. Analyze Before setting out to do a lot of calculations here, let’s stop and think. The concentration of the titrant (HCl) is equal to the concentration of the Na2CO3 we are titrating. We don’t need, therefore, to do any calculations here.

Aqueous Equilibria | 225 We do have to keep in mind that Na2CO3 is a dibasic compound and, therefore, two equivalents of acid are needed to titrate it. HCO3– is monobasic and requires only one equivalent of HCl to reach the equivalence point. Solve We need 500 mL to titrate the Na2CO3 and 250 mL to titrate the HCO3–. Think about It To reach the third end point for 250 mL of 0.0100 M Na3PO4 solution, 750 mL of the 0.0100 M HCl titrant is required. 15.109. Collect and Organize In comparing the titration of a weak acid in which the amount of NaOH titrant needed to reach the equivalence point is double that for another titration, we are asked what the pH halfway to the equivalence point is for the second titration if the pH at that point for the first titration is 4.44. Analyze The midpoint is where half the weak acid has been converted into its conjugate base. It is at this point, where [acid] = [conjugate base], that the pH = pKa. Solve The pH at the midpoint in the second titration would be the same as it is in the first titration, 4.44. Think about It Certainly the volume of the added titrant at which the midpoint is reached for the second titration would be twice that as in the first titration, but the pH at those points would be the same. 15.110. Collect and Organize Using the data of pH versus the volume of OH– added to a 125.0 mg sample of a weak acid dissolved in 100.0 mL of water, we can plot the titration curve to determine the Ka of this acid. Analyze From the titration curve we first determine the volume of OH– to reach the equivalence point. When one-half of that amount was added, the midpoint is reached and at that point pH = pKa. By reading the pH off the titration curve at this volume of OH–, we can calculate the Ka of the acid. Solve

The equivalence point is at 22.5 mL; therefore, the midpoint in the titration is when 11.25 mL of OH– has been added. The pH at this point is 4.2, so the Ka = 6.31 × 10–5.

226 | Chapter 15 Think about It Because we know the moles of OH– added to reach the equivalence point in this titration 0.050 mol OH – 22.5 mL × = 1.125 × 10–3 mol 1000 mL we also know that 1.125 × 10–3 mol of the monoprotic acid is present in the sample. Because the sample was weighed, we can calculate the molar mass of the acid:

0.125 g = 111.1 g/mol 1.125 × 10–3 mol

15.111. Collect and Organize To sketch the titration curve for the titration of 50.0 mL of a solution of the weak acid HNO2 with 1.00 M NaOH, we must use the information given to calculate the pH of the HNO2 solution before any NaOH is added and at the equivalence point. Analyze The initial pH of 0.250 M HNO2 is calculated from the value of Ka = 4.0 × 10– 4 after setting up a RICE table where the initial concentrations of H+ and NO2– are 0.00 M. To calculate the pH of the solution at the equivalence point, we need to recognize that at that point all the HNO2 is converted to NO2–. We use Kb, therefore, for the RICE table calculation. We also need to take into account the added volume of the solution when NaOH titrant is added. Solve The initial pH of 0.250 M HNO2 is found as follows: Reaction   HNO2(aq)  [HNO2]     Initial   0.250 Change   –x   Equilibrium   0.250 – x  

H+(aq) + [H+]   0.00   +x   x   2 x x2 ≈ K a = 4.0 ×10– 4 = 0.250 − x 0.250 x = 0.0100

NO2–(aq)   [NO2–]   0.00   +x   x  

pH = –log 0.0100 = 2.00

At the equivalence point, moles of OH– added = moles of HNO2 present in the initial solution: 0.250 mol moles HNO 2 = 50.0 mL × = 0.0125 mol 1000 mL – Because all the HNO2 is converted to NO2 at the equivalence point, the amount of NO2– at equivalence point is initially 0.0125 mol. The NO2– produced hydrolyzes by the equation NO2–(aq) + H2O(ℓ)  HNO2(aq) + OH–(aq) to give a slightly basic solution at the equivalence point. We need to determine the volume of NaOH titrant added so that we can use the [NO 2–] in a RICE table to calculate the pH at the equivalence point. The 0.0125 mol of HNO2 requires 0.0125 mol of OH– to neutralize it. The volume of NaOH needed to supply this 0.0125 mol of OH– is 1 mol NaOH – 1000 mL 0.0125 mol OH – × × = 12.5 mL 1.00 mol 1 mol OH – The total volume of the solution at the equivalence point is 12.5 + 50.0 mL = 62.5 mL. The molarity of NO2– at the equivalence point is 0.0125 mol NO2 = 0.200 M 0.0625 L

Aqueous Equilibria | 227

Reaction  

NO2–(aq) + H2O(ℓ)  HNO2(aq) + OH–(aq)  

[NO2–]   [HNO2]     Initial   0.200 0   Change   –x   +x   Equilibrium   0.200 – x   x   This is a Kb expression, so we must calculate Kb from Ka: Kw 1× 10−14 Kb = = = 2.5 × 10 –11 K a 4.0 × 10 – 4

[OH–]   0   +x   x  

Solving the equilibrium constant expression for x gives

x2 x2 ≈ 0.200 − x 0.200 x = 2.24 × 10− 6 pOH = –log 2.24 × 10– 6 = 5.65 pH = 14 – 5.65 = 8.35

K b = 2.5 ×10–11 =

Think about It The titration curve has a relatively flat region before the equivalence point. Here, the pH does not change much despite the addition of more and more titrant. This area is often called the buffer region. 15.112. Collect and Organize To determine the color of the red cabbage juice at the equivalence point in the titration of 0.10 M acetic acid with 0.10 M NaOH, we need to consider whether the pH at the equivalence point is acidic or alkaline (basic). Analyze We could calculate the pH of the solution at the equivalence point, but that is not necessary to answer the question. At the equivalence point, all the acetic acid, CH3COOH, is converted to the acetate anion, CH3COO–. The acetate anion reacts with water. Solve The equation describing acetate’s reaction with water shows that we have a basic (alkaline) solution at the equivalence point: CH3COO–(aq) + H2O(ℓ)  CH3COOH(aq) + OH–(aq) The color of red cabbage juice at the equivalence point, then, is yellow.

228 | Chapter 15 Think about It To calculate the actual pH, solve for x in the following Kb expression: x2 x2 K b = 5.68 ×10–10 = ≈ 0.050 − x 0.050 x = 5.33 × 10− 6

pOH = –log 5.33 ×10− 6 = 5.27 pH = 14 – 5.27 = 8.73 15.113. Collect and Organize For the titration of quinine, a dibasic malaria drug, we need to calculate the initial pH and the position of each of the equivalence points in order to sketch the titration curve. Analyze Because the second base ionization constant (Kb2 = 1.35 × 10–9) is much smaller than the first (K b1 = 3.31 × 10 – 6), we can ignore it in calculating the initial pH of the solution. Because the concentration of the quinine and HCl titrant are equal (both solutions are 0.100 M), we know that the first equivalence point will be at the point where 40.0 mL of HCl has been added and the second equivalence point is at 80.0 mL of HCl added. Solve The initial pH is calculated as follows: Reaction Quinine(aq) + H2O(ℓ)  QuinineH+(aq) + OH–(aq) [Quinine] [QuinineH+] [OH–] Initial 0.100 0 0 Change –x +x +x Equilibrium 0.100 – x x x

x2 x2 ≈ 0.100 − x 0.100 x = 5.75 × 10− 4

K b = 3.31×10– 6 = 1

pOH = –log 5.75 ×10− 4 = 3.240 pH = 14 – 3.240 = 10.760 The first equivalence point is where 40.0 mL of HCl has been added to give 0.100 mol quinine 1 mol quinineH + 40.0 mL HCl × × = 4.00 × 10–3 mol 1000 mL 1 mol quinine The molarity of quinineH+ is 4.00 × 10–3 mol = 0.0500 M 0.080 L + At this point, quinineH can act both as an acid and as a base: quinineH+(aq) + H2O(ℓ)  quinineH22+(aq) + OH–(aq) quinineH+(aq)  quinine(aq) + H+(aq)

Ka2 =

K b1 =1.35 × 10–9 Kw K b2

=

1× 10 –14 = 3.02 × 10 –9 3.31× 10 – 6

This calculation is beyond the scope of general chemistry study, but we can see that Ka2 > Kb1, so we expect this equivalence point to be slightly acidic. In fact, it is, at approximately pH 6.8. The second equivalence point is where 80.0 mL of HCl has been added. This solution contains 4.00 × 10–3 mol quinineH22+ with a molarity of 4.00 × 10–3 mol = 0.0333 M 0.120 L

Aqueous Equilibria | 229 because the total volume is now 40.0 mL of quinine solution + 80.0 mL of HCl titrant. The pH at this equivalence point is dependent on the equilibrium Kw 1× 10 –14 quinineH2+(aq)  quinineH+(aq) + H+(aq) K a1 = = = 7.41× 10 – 6 K b2 1.35 × 10 –9

The pH at this equivalence point is as follows: Reaction QuinineH2+(aq)  quinineH+(aq) + H+(aq) [QuinineH2+] [QuinineH+] [H+] Initial 0.0333 0 0 Change –x +x +x Equilibrium 0.0333 – x x x 2 2 x x Ka = 7.41×10– 6 = ≈ 1 0.0333 − x 0.0333 x = 4.97 × 10− 4

pH = –log 4.97 ×10− 4 = 3.304

Think about It At the two midpoints (20 mL and 60 mL HCl added), the pH equals pKa2 (8.52) and pKa1 (5.13). 15.114. Collect and Organize For the titration of 100 mL of a solution of ascorbic acid, a diprotic acid, we need to calculate the initial pH and the position of each equivalence point to sketch the titration curve. The molarities of the ascorbic acid and NaOH solutions are 1.25 × 10–2 M and 1.00 × 10–2 M, respectively. We are also asked how many equivalence points the curve should have and what color indicator(s) could be used. Analyze Because the second acid ionization constant (Ka2 = 1.6 × 10–12) is much smaller than the first (Ka1 = 8.0 × 10–5), we can ignore it in calculating the initial pH of the solution. Solve The initial pH is Reaction Initial Change Equilibrium

H2C6H6O6(aq)  [H2C6H6O6] 1.25 × 10–2 –x 1.25 × 10–2 – x

HC6H6O6–(aq) [HC6H6O6–] 0 +x x

+

H+(aq) [H+] 0 +x x

230 | Chapter 15

Ka 1 = 8.0 × 10–5 =

x2 1.25 × 10–2 − x

Expanding and solving by the quadratic equation gives x 2 + 8.0 × 10 –5 x − 1.00 × 10 – 6 = 0 x = 9.61× 10− 4

pH = –log 9.61×10− 4 = 3.02 The first equivalence point is where moles of OH– added = moles H2C6H6O6: 1.25 × 10 –2 mol = 1.25 × 10 –3 mol moles H 2C6 H 6O6 = 100 mL × 1000 mL The volume of NaOH solution required to reach this equivalence point is

1.25 × 10–3 mol NaOH ×

1000 mL = 125 mL 1.00 × 10–2 mol

At the first equivalence point ascorbate can act either as an acid or as a base: HC6H6O6–(aq)  C6H6O62–(aq) + H+(aq) Ka 2 = 1.6 × 10–12 HC6H6O6–(aq) + H2O(ℓ)  H2C6H6O6(aq) + OH–(aq)

K b2 =

Kw K a1

=

1 × 10 –14 = 1.25 × 10 –10 8.0 × 10 –5

This calculation is beyond the scope of general chemistry study, but we can see that Kb2 > Ka2, so we expect this equivalence point to be slightly basic. The second equivalence point is where 250 mL of NaOH solution has been added. This solution contains 1.25 × 10–3 mol of ascorbate2– with a molarity of 1.25 × 10–3 mol = 3.57 × 10–3 M 0.350 L because the total volume is now 100 mL of ascorbic acid solution + 250 mL of NaOH titrant. The pH at this equivalence point is dependent on the equilibrium C6 H 6O6 2– (aq) + H 2O()  HC6 H 6O6 – (aq) + OH – (aq)

Kw

1 × 10 –14 = 6.25 × 10 –3 K a 2 1.6 × 10 –12 The pH at this second equivalence point is calculated as follows: Reaction C6H6O62–(aq) + H2O(ℓ)  HC6H6O6–(aq) [C6H6O62–] [HC6H6O6–] –3 Initial 0 3.57 × 10 Change –x +x Equilibrium x 3.57 × 10–3 – x 2 x 6.25 × 10–3 = 3.57 × 10–3 − x Expanding and solving by the quadratic equation gives x 2 + 6.25 × 10 –3 x − 2.23 × 10 –5 = 0 x = 2.54 × 10 −3 pOH = –log 2.54 × 10−3 = 2.60 pH = 14 − 2.60 = 11.40 K b1 =

=

+

OH–(aq) [OH–] 0 +x x

Aqueous Equilibria | 231

Two equivalence points exist for the titration of the diprotic acid ascorbic acid. The color indicators that could be used (Figure 15.15) are bromthymol blue for the first equivalence point, near pH 7, and Alizarin yellow R for the second equivalence point, near pH 11. Think about It In performing this experiment in the lab, we can use the pH at the half-equivalence points (62.5 mL and 187.7 mL) to determine pKa1 and pKa2 for this acid. 15.115. Collect and Organize / Analyze We are asked to compare the terms molar solubility and solubility product. Solve Molar solubility is the mole of a substance that dissolves in a solvent. The solubility product is the equilibrium constant for the dissolution of a substance. Think about It Solubility has units of grams or moles per volume of solution but, like other equilibrium constants, the solubility product is unitless. 15.116. Collect and Organize Using the definition of the common-ion effect, we are to give an example of how addition of a common ion to a solution saturated in a slightly soluble salt would limit that salt’s solubility. Analyze The common-ion effect says that the addition of an ion common to that of the sparingly soluble salt shifts the equilibrium to the left, according to Le Châtelier’s principle. This then decreases the amount of that salt that dissolves. Solve For the sparingly soluble salt BaSO4, if we increase the concentration of sulfate anions in the solution by adding soluble sodium sulfate (Na2SO4), some of the Ba2+ in solution will precipitate to form BaSO4. Think about It The Ksp expression for a sparingly soluble salt with singly charged ions may be expressed as K sp = [M + ][X – ] MX(s)  M + (aq) + X – (aq) If either X– or M+ is added to the solution, then Q > Ksp, so the equilibrium shifts to the left and less of the salt dissolves. 15.117. Collect and Organize By comparing the Ksp values of MgCO3, CaCO3, and SrCO3 we can identify which cation (Mg2+, Ca2+, or Sr2+) precipitates first as carbonate mineral.

232 | Chapter 15 Analyze From Appendix 5, the Ksp values are as follows:

MgCO3

Ksp = 6.8 × 10–6

CaCO3

Ksp = 5.0 × 10–9

SrCO3

Ksp = 5.6 × 10–10

Solve Because SrCO3 has the lowest Ksp, the cation Sr2+ precipitates first as a carbonate mineral. Think about It The order of solubility from least to most soluble for these carbonates is SrCO3 < CaCO3 < MgCO3. 15.118. Collect and Organize We are asked whether an increase in solubility as the temperature increases means that the Ksp value increases or decreases. Analyze As a substance becomes more soluble, the Ksp value increases. Solve Ksp increases for substances that have increased solubility as temperature increases. Think about It A substance is not always more soluble at higher temperatures. For some substances, solubility decreases as the temperature is raised. 15.119. Collect and Organize For SrSO4, whose Ksp increases as the temperature increases, we are to determine whether the dissolution is exothermic or endothermic. Analyze We can include heat as a reactant (for an endothermic reaction) or as a product (for an exothermic reaction) and apply Le Châtelier’s principle: SrSO4(s) + heat  Sr2+(aq) + SO42–(aq) SrSO4(s)  Sr2+(aq) + SO42–(aq) + heat Solve Applying Le Châtelier’s principle, we see that the reaction shifts to the right and more SrSO4 dissolves as the temperature is increased. The dissolution is endothermic. Think about It The opposite effect of temperature occurs for an exothermic dissolution: less solid dissolves as the temperature is increased. 15.120. Collect and Organize We are asked how the addition of concentrated NaOH affects the solubility of an Al3+ salt. Analyze Let’s use AlPO4 as an example. The solubility equilibrium is AlPO4(s)  Al3+(aq) + PO43–(aq) The reaction of Al3+ with excess OH– has the overall reaction Al3+(aq) + 4 OH–(aq)  Al(OH)4–(aq)

Aqueous Equilibria | 233 Solve The solubility of an Al3+ salt increases in a concentrated solution of OH– because the formation of soluble Al(OH)4– shifts the solubility equilibrium to the right. Think about It Likewise, because of the formation of Cr(OH)4–, “insoluble” salts such as Cr2(SO4)3 are expected to dissolve in concentrated OH–. 15.121. Collect and Organize By writing the equation for the dissolution of hydroxyapatite, we can explain why acidic substances erode tooth enamel. Analyze The solubility of hydroxyapatite is described by Ca5(PO4)3OH(s)  OH–(aq) + 3 PO43–(aq) + 5 Ca2+(aq) Solve Acidic substances react with the OH– released upon dissolution of hydroxyapatite. The equilibrium is shifted to the right, dissolving more hydroxyapatite. Think about It The equilibrium would be shifted in the opposite direction (to the left) in an alkaline environment. 15.122. Collect and Organize Using the formula given for the product for the conversion of hydroxyapatite into fluorapatite, we can explain why fluorapatite is less susceptible to erosion by acids. Analyze The dissolution of fluorapatite is described by the equation Ca5(PO4)3F(s)  5 Ca2+(aq) + 3 PO43–(aq) + F–(aq) Solve The F– produced in the dissolution of fluorapatite, unlike the OH– produced in the dissolution of hydroxyapatite, is a weak base. It does not react completely with H+ and does not drive the solubility equilibrium to the right. Think about It Also, the Ksp of fluorapatite is estimated to be about 10 times lower than the Ksp of hydroxyapatite. 15.123. Collect and Organize Given the [Ba2+] in a saturated solution of BaSO4 (1.04 × 10–5 M Ba2+), we are to calculate the value of Ksp for BaSO4: BaSO 4 (s)  Ba 2+ (aq) + SO 4 2− (aq) Analyze The Ksp expression is

Ksp = [Ba2+][SO42–] Because for every mole of BaSO4 that dissolves we get 1 mol of Ba2+ and 1 mol of SO42–, the molarities of Ba2+ and SO42– are the same for this saturated solution of BaSO4. Solve

Ksp = (1.04 × 10–5)(1.04 × 10–5) = 1.08 × 10–10

234 | Chapter 15 Think about It Because BaSO4 is quite insoluble, we can add SO42– to a solution of dissolved Ba2+ to quantitatively precipitate the barium out of solution. After weighing the dried precipitate, we can then calculate how much Ba2+ was present in the original solution. 15.124. Collect and Organize Given the [F–] in a saturated solution of BaF2 (1.5 × 10–2 M F–), we can calculate the value of Ksp for BaF2: BaF2 (s)  Ba 2+ (aq) + 2 F− (aq) Analyze The Ksp expression is

Ksp = [Ba2+][F–]2 Because for every mole of BaF2 that dissolves we get 1 mol of Ba2+ and 2 mol of F–, the molarity of Ba2+ in the saturated solution is half that of F–. Solve

Ksp = (1.5 × 10–2/2)(1.5 × 10–2)2 = 1.7 × 10–6

Think about It Don’t forget to square the concentration of F– to calculate the Ksp for BaF2. 15.125. Collect and Organize Given that Ksp = 1.02 × 10–6, we are to calculate [Cu+] and [Cl–] for a saturated solution of CuCl. Analyze The solubility equation and Ksp expression for CuCl are CuCl(s)  Cu + (aq) + Cl− (aq)

K sp [Cu + ][Cl– ]

The [Cu+] = [Cl–] in this solution because for every particle of CuCl that dissolves we get one particle of Cu + and one particle of Cl–. Solve Let [Cu+] = [Cl–] = x. The Ksp expression becomes Ksp = 1.02 × 10–6 = (x)(x) x = 1.01 × 10–3 + – –3 Therefore, [Cu ] = [Cl ] = 1.01 × 10 M. Think about It We do not need to know how much CuCl is originally placed into the solution because it does not appear in the Ksp expression as a pure solid. 15.126. Collect and Organize Given that Ksp = 3.2 × 10–8, we are to calculate [Pb2+] and [F–] for a saturated solution of PbF2. Analyze The solubility equation and Ksp expression for PbF2 are PbF2 (s)  Pb2+ (aq) + 2 F− (aq)

K sp = [Pb2+ ][F – ]2

In this solution [F–] = 2 × [Pb2+] because for every particle of PbF2 that dissolves we get one particle of Pb2+ and two particles of F–. Solve Let [Pb2+] = x and [F–] = 2x. The Ksp expression becomes Ksp = 3.2 × 10–8 = (x)(2x)2 x = 2.0 × 10–3 M 2+ –3 – Thus, [Pb ] = x = 2.0 × 10 M and [F ] = 2x = 4.0 × 10–3 M.

Aqueous Equilibria | 235 Think about It We do not need to know how much PbF2 is originally placed into the solution because it does not appear in the Ksp expression as a pure solid. 15.127. Collect and Organize Given the Ksp of CaCO3 (9.9 × 10–9), we are to calculate the solubility of this substance in units of grams per milliliter. Analyze The solubility equation and Ksp expression for CaCO3 are CaCO3 (s)  Ca 2+ (aq) + CO32− (aq)

K sp = [Ca 2+ ][CO32– ]

In this solution, [Ca2+] = [CO32–] because for every CaCO3 that dissolves one Ca2+ and one CO32– are produced. We can then say that Ksp = x2 and we can solve for x, which is the molar solubility (mol/L) of CaCO3. To convert this to grams per milliliter, we multiply by the molar mass of CaCO3 (100.09 g/mol) and divide by 1000 mL/L. Solve

K sp = 9.9 × 10 –9 = x 2 x = 9.95 × 10 –5 9.95 × 10 –5 mol 100.09 g 1L × × = 9.96 × 10 –6 g/mL L 1 mol 1000 mL

Think about It The value of x that we calculate in the Ksp expression is the molar solubility of the solid because it is that amount (“x”) that dissolves into the solution. 15.128. Collect and Organize Given the Ksp of AgI (1.50 × 10–16), we are to calculate the solubility of this substance in units of grams per milliliter. Analyze The solubility equation and Ksp expression for AgI are AgI(s)  Ag + (aq) + I − (aq)

K sp = [Ag + ][I − ]

In this solution, [Ag+] = [I–] because for every AgI that dissolves one Ag+ and one I– are produced. We can then say that Ksp = x2 and we can solve for x, which is the molar solubility (moles per liter) of AgI. To convert this to grams per milliliter, we multiply by the molar mass of AgI (234.77 g/mol) and divide by 1000 mL/L. Solve

K sp = 1.50 × 10 –16 = x 2 x = 1.225 × 10 –8 1.225 × 10 –8 mol 234.77 g 1L × × = 2.88 × 10 –9 g/mL L 1 mol 1000 mL

Think about It The value of x that we calculate in the Ksp expression is the molar solubility of the solid because it is that amount (“x”) that dissolves into the solution.

236 | Chapter 15 15.129. Collect and Organize Given the Ksp for the dissolution of AgOH (1.52 × 10–8) in Appendix 5, we are to calculate the pH of a saturated solution. Analyze The solubility equation and Ksp expression for AgOH are AgOH(s)  Ag + (aq) + OH − (aq)

K sp = [Ag + ][OH − ]

Letting [Ag+] = [OH–] = x (because the stoichiometry is 1:1), we can solve for x by using the Ksp expression. The pH of the solution will then be pH = 14 – (–log x) Solve

K sp = 1.52 ×10 – 8 = x 2

(

x = 1.233 ×10 – 4

)

pH = 14 – – log 1.233 ×10 – 4 = 10.091 Think about It Even though the Ksp of AgOH is not high, this solution is quite basic. 15.130. Collect and Organize Using the Ksp for the dissolution of Mg(OH)2 from Appendix 5 (5.6 × 10–12), we can calculate the pH of a saturated solution. Analyze The solubility equation and Ksp expression for MgOH are Mg OH 2(s)  Mg 2+ (aq) + 2 OH − (aq)

( )

K sp = [Mg 2+ ][OH − ]2

If we let [Mg2+] = x, then [OH–] = 2x at equilibrium. We can then solve for x by using the Ksp expression, and the pH of the solution will be pH = 14 – (–log 2x) Solve

K sp = 5.6 × 10 –12 = x(2 x) 2 x = 1.12 × 10 – 4 pH = 14 – [– log(2 ×1.12 ×10 – 4 )] = 10.35

Think about It Be careful to notice here that [OH–] = 2x and that the [OH–] must be squared in the Ksp expression. 15.131. Collect and Organize Using the common-ion effect, we can determine in which 0.1 M solution (NaCl, Na2CO3, NaOH, or HCl) the most CaCO3 dissolves. Analyze Whenever a common ion is already present in the solution, the CaCO3 is less soluble. Any solution, therefore, with Ca2+ or CO32– would have lower solubility of CaCO3 than water. We also should look for solutions that might react with either Ca2+ or CO32– and shift the solubility equilibrium to the right. Solve (a) NaCl(aq) has neither an ion common to CaCO3 nor ions that react with either Ca2+ or CO32–. CaCO3 has the same solubility in this NaCl solution as in water. (b) The 0.1 M Na2CO3 solution is 0.1 M in CO32–. This decreases the solubility of CaCO3.

Aqueous Equilibria | 237 (c) NaOH(aq) has neither an ion common to CaCO3 nor reacts with either Ca2+ or CO32–. CaCO3 has the same solubility in this NaOH solution as water. (d) A solution of HCl reacts with CO32– to form H2CO3, which then decomposes to H2O and CO2. This reaction shifts the solubility equilibrium to the right, so more CaCO3 dissolves. The solution of (d) 0.1 M HCl dissolves the most CaCO3. Think about It A higher concentration of acid dissolves even more CaCO3 as the equilibrium shifts to the right: CaCO3(s) + 2 H+(aq)  Ca2+(aq) + H2CO3(aq) 15.132. Collect and Organize Using the common-ion effect, we can determine in which solution, already containing either Ca2+ or F–, the solid CaF2 will be most soluble. Analyze The presence of either Ca2+ or F– in the solution causes less CaF2 to dissolve than in pure water. The CaF2 is most soluble, then, in the solution that has the lowest concentration of either Ca2+ or F–. For each solution, the ion concentration is (a) 0.010 M Ca2+ (b) 0.01 M F– (c) 0.001 M F– (d) 0.10 M Ca2+ Solve The solution that has the highest solubility for CaF2 is (c) 0.001 M NaF. Think about It The solution that has the lowest solubility for CaF2 is the one highest in concentration of the common ions. That is the case for (d) 0.10 M Ca(NO3)2. 15.133. Collect and Organize Given the average concentrations of SO4–2 and Sr2+ in seawater (0.028 M and 9 × 10–5 M, respectively) and the Ksp of SrSO4 (3.4 × 10–7), we are to determine whether the concentration of Sr2+ is controlled by the relative insolubility of SrSO4. Analyze The solubility equation and Ksp expression for SrSO4 are SrSO 4 (s)  Sr 2+ (aq) + SO 4 2– (aq)

K sp = [Sr 2+ ][SO 4 2– ]

Solve 2–

K sp = 3.4 × 10 –7 = [Sr 2+ ][SO 4 ] = [Sr 2+ ] ( 0.028 M ) [Sr 2 + ] = 1.21× 10 –5 M This is the expected concentration of Sr in seawater with a known [SO42–] of 0.028 M. This [Sr2+] is lower than the average [Sr2+] of 9 × 10–5 M, so some other process must be controlling the [Sr2+]. 2+

Think about It As the [SO42–] increases, the solubility of SrSO4 decreases because of the common-ion effect. 15.134. Collect and Organize Given the Ksp for Fe(OH)3 (1.1 × 10–36), we are to calculate the [Fe3+] in seawater that has a pH of 8.1.

238 | Chapter 15 Analyze The seawater has a [OH–] that is found from the pH K 1×10–14 [OH – ] = w+ = = 1.26 ×10– 6 M [H ] 1×10–8.1 This is the initial concentration of [OH–] before the Fe(OH)3 dissolves, so the RICE table is Fe3+(aq) + 3OH–(aq) Fe(OH)3(s)  Reaction [Fe3+] Initial Change Equilibrium

0 +x x

[OH–] 1.26 × 10–6 +3x 1.26 × 10–6 + 3x

Because Ksp is so small, we may assume that 3x Ksp, then CaF2 will precipitate. Solve [Ca2+]initial × [F–]2initial = 0.127 × (0.169)2 = 3.63 × 10–3. This is greater than the value of Ksp for CaF2, so it will precipitate from the mixed solution. Think about It Because CaF2 has a small solubility product constant, we expect that most of the ions will precipitate as CaF2. In this solution the Ca2+ ions are in excess (0.0469 mol) compared to that of F– (0.0625 mol), so the final solution will have only a small amount of F– in solution. 15.136. Collect and Organize Given 345 mL of solution that is 0.0100 M in Sr2+(aq), we are asked whether SrSO4 will precipitate when 75 mL of a 0.175 M K2SO4 solution is added.

Aqueous Equilibria | 239 Analyze The Ksp for SrSO4 from Appendix 5 is 3.4 × 10–7. When the two solutions are mixed, the total volume is 420 mL and the [Sr2+] and [SO42–] in the mixed solution is 345 mL × 0.0100 M = 420 mL × [Sr 2 + ]

[Sr 2 + ] = 8.21×10−3 M 2−

75 mL × 0.175 M = 420 mL × [SO 4 ] 2−

[SO 4 ] = 3.13 ×10 −2 M

From the Ksp expression,

Ksp = [Sr2+][SO42–] = 3.4 × 10–7 If the [Sr2+]initial × [SO42–]initial > Ksp, then SrSO4 will precipitate. Solve [Sr2+]initial × [SO42–]initial = (8.21 × 10–3 M) × (3.13 × 10–2 M) = 2.56 × 10– 4. This is greater than the value of Ksp for SrSO4, so it will precipitate from the mixed solution. Think about It Because SrSO4 has a small solubility product constant, we expect that most of the ions will precipitate as SrSO4. In this solution the SO42– ions are in excess (0.0131 mol) compared to that of Sr2+ (0.00345 mol), so the final solution will have only a small amount of Sr2+ in solution. 15.137. Collect and Organize When a 0.250 M solution of Pb2+(aq) is added to a solution that is 0.010 M in Br– and SO42– we can use the values of Ksp for PbBr2 and PbSO4 to determine which anion is the first to precipitate. We are then asked to calculate the concentration of the anion that precipitates first at the moment that the second ion starts to precipitate. This will be when the solution is saturated in the lead salt of the first anion to precipitate. Analyze The Ksp value shows that PbSO4 has a smaller solubility product constant (1.8 × 10–8) than PbBr2 (6.6 × 10– 6). Solve (a) PbSO4, with a smaller solubility product constant, will precipitate first from the solution, so the SO42– anion will precipitate first. (b) When Br– begins to precipitate, the maximum amount of Pb2+ that could be present that will not cause Br– to precipitate is

Ksp = 6.6 × 10−6 = [Pb 2+ ] [Br – ]2 [Pb 2+ ] =

6.6 × 10−6 = 6.6 × 10−2 M (0.010)2

The [SO42–] in the solution when [Pb2+] = 6.6 × 10–2 M is

Ksp = 1.8 × 10−8 = [Pb2+ ] [SO4 2 − ] [Pb 2+ ] =

1.8 × 10−8 = 2.7 × 10−7 M 6.6 × 10−2

Think about It We do not need a RICE table to solve this problem. 15.138. Collect and Organize Considering a solution (A) that is 0.0250 M Ag+ and 0.0250 M Pb2+, we are to determine how to separate these ions by using either a 0.500 M solution of NaCl (solution B) or a 0.500 M solution of NaBr (solution C).

240 | Chapter 15 Analyze Here we must consider the values of Ksp for the possible salts formed in the reaction. From Appendix 5 these are: Ksp(AgCl) = 1.8 × 10–10 Ksp(AgBr) = 5.4 × 10–13 Ksp(PbCl2) = 1.6 × 10–5 Ksp(PbBr2) = 6.6 × 10–6 To separate Ag+ from Pb2+ we want to employ the anion solution (Cl– or Br–) with the greatest difference in the Ksp for the Ag+ and Pb2+ salts of that anion. Solve (a) Because the solubility product constants for both Cl– and Br– salts of Ag+ are smaller than those of Pb2+, Ag+ will precipitate first from the solution whether we add solution B or solution C. The difference in the solubility product constant is approximately 107 for AgBr versus PbBr2, which is greater than the approximately 105 difference in solubility product constants between AgCl and PbCl2. Therefore, we should choose solution C (NaBr) to separate Ag+ from Pb2+ in the mixed solution. (b) When Pb2+ begins to precipitate, the maximum amount of Br– that can be present to not cause Ag+ to precipitate is

K sp = 6.6 × 10−6 = [Pb 2+ ] [Br – ]2 [Br − ] =

6.6 × 10−6 = 1.62 × 10−2 M 0.0250

The [Ag+] in the solution when [Br–] is 1.62 × 10–2 M is

Ksp = 5.4 × 10−13 = [Ag + ] [Br − ] 5.4 × 10−13 = 3.3 × 10−11 M 1.62 × 10−2 In relation to the original concentration of Ag+ in the solution, this is 3.3 ×10−11 M × 100 = 1.3 × 10−7 % 0.0250 M [Ag + ] =

This is less than 0.10%, so the two cations are completely separated. Think about It We do not need a RICE table to solve this problem. 15.139. Collect and Organize We are asked to describe the changes in bonding and intermolecular forces when the weak base CH 3NH2 is dissolved in water. Analyze Methylamine is a gas; when dissolved in water, the individual methylamine molecules are surrounded with water. CH3NH2 also reacts with water, according to the equation CH 3NH 2 (aq) + H 2O()  CH 3NH 3+ (aq) + OH – (aq) Solve The hydrogen bonds between some of the water molecules must break and re-form around the species CH3NH2. Also, the amine hydrolyzes and forms CH3NH3+ and OH–; thus, ion–dipole forces are added when these ions are surrounded by water molecules.

Aqueous Equilibria | 241

Think about It Depending on the strength of the forces formed versus those broken, this dissolution may be either exothermic or endothermic. 15.140. Collect and Organize High-sulfur fuels, when burned, produce acid rain that erodes marble. We are to describe those chemical reactions. Analyze Sulfur burned in air produces sulfur oxides such as SO3. This nonmetal oxide is an acid anhydride. The parent acid forms upon reaction with water in clouds to produce acid rain. The acid, then, erodes (dissolves) CaCO3 in marble. Solve In burning S-containing fuels: 4 O2 ( g ) S8(s) + 12 O2(g) → 8 SO2(g) ⎯⎯⎯→ 8 SO3(g)

In the presence of water, these produce a weak acid, H2SO3, and a strong acid, H2SO4:

SO 2 (g) + H 2O() → H 2SO3 (aq) SO3 (g) + H 2O() → H 2SO 4 (aq) These acids dissolve CaCO3:

2 H+(aq) + CaCO3(s) → H2CO3(aq) + Ca2+

Think about It H2CO3 decomposes to H2O and CO2 according to the equilibrium expression H 2CO3 (aq)  H 2O() + CO 2 (g) 15.141. Collect and Organize For the buffer pH values of 3.00, 5.00, 7.00, and 12.00, we are to choose appropriate acid/base pairs to prepare buffers at these pHs. Analyze The acid/base conjugate pair that we choose for each buffer should have their pKa values close to our target pH values.

242 | Chapter 15 Solve For pH 3.00 buffer: citric acid/citrate (pKa = 3.13). For pH 5.00 buffer: ascorbic acid/ascorbate (pKa = 5.00). For pH 7.00 buffer: H2PO4–/HPO42– (pKa = 7.19). For pH 12.00 buffer: HPO42–/PO43– (pKa = 12.32). Think about It For an effective buffer, the acid/conjugate base pair concentrations should be roughly equal to each other and in high enough concentrations to absorb any anticipated amount of strong acid or base. 15.142. Collect and Organize We are asked to identify the ionizable protons in the structure of H3PO3 (Figure P15.142). Analyze The ionizable protons will be those bonded to the electronegative oxygen atoms. Solve

Ionizable H atoms

O H H

P

O

H

O

Think about It The H atom bonded to P in H3PO3 is not ionizable. The electronegativity difference between P and H is not great enough to make this H atom acidic. 15.143. Collect and Organize We are asked whether the pH of the solution changes when a cook adds more baking soda to water used in a recipe and to explain why or why not. Analyze Baking soda is a soluble sodium salt with the formula NaHCO3. In solution, this salt forms Na+(aq) + HCO3–. Na+ does not react with water, but HCO3– does, which changes the pH of the solution. Solve Yes, the pH of the solution increases because of the increase in hydrolysis of HCO3– according to the equation HCO3– (aq) + H 2O()  H 2CO3 (aq) + OH – (aq) Think about It H2CO3 decomposes at baking temperatures to give CO2(g) and H2O(ℓ). 15.144. Collect and Organize For each ingredient in antacid tablets, we are to write a net ionic equation for its reaction with HCl(aq) and then explain how substances such as MgCO3, CaCO3, and Mg(OH)2 can act as antacids even though they are insoluble. Analyze Each substance acts as a base to absorb the added HCl. The Cl– anion and the Na+ cation of the soluble salts are not involved in the net ionic reaction. Solve (a) HCO3–(aq) + H+(aq) → H2CO3(aq) MgCO3(s) + 2 H+(aq) → Mg2+(aq) + H2CO3(aq)

Aqueous Equilibria | 243 CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2CO3(aq) Mg(OH)2(s) + 2 H+(aq) → Mg2+(aq) + 2 H2O(ℓ) (b) Insoluble antacids are effective because even as solids a small amount dissolves and reacts with H+ to neutralize the acid and raise the pH. Once that small amount is consumed in the reaction, more of the solid dissolves to reestablish equilibrium. Think about It Carbonic acid, H2CO3, decomposes in aqueous solution to give H2O and CO2. 15.145. Collect and Organize Given the change in pH of a lake from 6.1 to 4.7 when 400 gallons of 18 M H2SO4 was added to the lake, we are to calculate the volume of the lake. Analyze To make this calculation easier, we can assume that we ionize only the first proton on H2SO4. First, we must calculate the moles of H2SO4 added to the lake and then determine the increase in the concentration of H+ in the lake in going from pH 6.1 to pH 4.7. Knowing that we simply divide the moles of acid added by the molarity change of H+ in the lake, we can obtain the size of the lake. Solve The amount of H2SO4 added is

400 gal × The change in lake [H+] is

The size of the lake is

3.78 L 18 mol × = 27, 216 mol 1 gal L

Initial [H+] = 1 × 10– 6.1 = 7.94 × 10–7 M Final [H+] = 1 × 10– 4.7 = 2.00 × 10–5 M ∆[H+] = 1.92 × 10–5 M increase

27, 216 mol ×

1L = 1.4 × 109 L 1.92 × 10–5 mol

Think about It This volume is equivalent to 1.4 × 106 m3. If the lake were 10 m deep, it would cover an area of 1.4 × 105 m2. If thought of as a square, that is 374 m on a side. 15.146. Collect and Organize We are asked to explain why the pH of the water increases when Na3PO4 is added. Analyze Na3PO4 is a soluble salt that forms Na+ and PO43– in aqueous solution. The Na+ ion does not hydrolyze, but PO43– does. Solve The reaction with PO43– produces OH–, as seen in the following equations, which increases the pH: PO43–(aq) + H2O(ℓ)  HPO42–(aq) + OH–(aq) HPO42–(aq) + H2O(ℓ)  H2PO4–(aq) + OH–(aq) H2PO4–(aq) + H2O(ℓ)  H3PO4(aq) + OH–(aq) The first reaction contributes to nearly all the increase in the pH of the water because Kb2 (1.58 × 10–7) and K b 3 (1.41 × 10–12) are negligible compared to K b (2.2 × 10–2). 1

Think about It The ultimate acid formed in this reaction is a weak acid, not strong, and therefore we write all the acid equilibrium species through to the parent acid, H3PO4.

244 | Chapter 15 15.147. Collect and Organize For the drug Zoloft we are to use Figure P15.147 to determine which form is for the acid salt and whether solutions of Zoloft are acidic or basic. Analyze Remember that this drug is sold as the HCl salt. Solve (a) The acid salt form has H on the amine (R2NH) moiety with Cl– as a counter ion. This structure is shown on the right of Figure P15.147. (b) Because this drug is sold as the HCl salt, solutions of this drug are acidic. Think about It Many drugs are sold as HCl salts to render the drugs more soluble in aqueous solution. 15.148. Collect and Organize Given the structure of Prozac in Figure P15.148, we are asked about its acid–base character and the solubility of its HCl salt. Analyze The functional group that is likely to show acid–base character is the amine group (–NHCH3) on one end of the Prozac molecule. Solve (a) In water, amines show basic character. They pick up a proton from water to form ammonium cations and OH–. Therefore, dissolving Prozac in water gives a slightly basic solution. (b) The N atom on Prozac is more likely to react with HCl than the O atom. (c) The HCl salt of Prozac is more soluble because it is charged and water molecules form stronger ion–dipole forces around the molecule than the dipole–induced dipole forces between the neutral molecule and water. Think about It Being soluble in water also helps to deliver the drug to the body. 15.149. Collect and Organize By examining the equilibrium reactions of HF in water and aqueous F–, we are to calculate the equilibrium constant for the combination of the two equilibrium equations and then calculate the pH and [HF2–]eq when [HF] is 0.150 M. Analyze (a) The overall equation is the sum of the two equilibrium reactions, so the K for the combined reaction is the product of the two K values for the individual reactions. (c) We can use a RICE table and the value of the overall K calculated in part b to determine the pH and [HF2–]eq. Solve (a) Koverall = Ka × K = (1.1 × 10–3) × (2.6 × 10–1) = 2.86 × 10– 4 (b) We must tackle this problem in two steps. First, we consider the hydrolysis of HF (aq): Reaction HF(aq) H+(aq) + F–(aq)  [HF] [H+] [F–] Initial 0.150 0 0 Change –x +x +x Equilibrium 0.150 – x x x

Aqueous Equilibria | 245

1.1×10 –3 =

x2 ( 0.150 − x )

x 2 + 1.1×10 –3 x − 1.65 ×10 – 4 = 0 x = 0.0123 –

So [F ] = 0.0123 M and [HF] = 0.138 M Now, we consider the second equilibrium: Reaction

F–(aq) + HF(aq)  HF2–(aq) [F–] [HF] [HF2–] Initial 0.0123 0.138 0 Change –x –x +x Equilibrium 0.0123 – x 0.138 – x x x x 0.26 = ≈ (0.0123 − x)(0.138 − x) (0.0123)(0.138)

x = 4.4 ×10– 4 Therefore,

pH = –log (0.0123) = 1.91 [HF2–] = 4.4 × 10– 4 M

Think about It Be careful in making simplifying assumptions. For the first equilibrium, we must solve by using the quadratic equation. 15.150. Collect and Organize We are to explain why C5F5H is such a strong acid compared to other organic acids, after we draw the structure of its conjugate base. Analyze The presence of many nearby electronegative atoms increases the acidity of a proton. Solve (a)

_

F F

C C F

C

F

C C F

(b) C5F5H is very acidic because the presence of five very electronegative F atoms on the carbon ring stabilizes the anion formed when the proton is lost. Think about It The formation of a stable aromatic ring upon deprotonation enhances the stability of the conjugate base above the stability imparted by the presence of F atoms on the ring. 15.151. Collect and Organize Given the structure of naproxen (Figure P15.151), we are to draw the structure of the sodium salt, explain whether a solution of the salt is acidic or basic, and explain why the salt is more soluble than naproxen itself. Analyze The ionizable functional group in naproxen is the carboxylic acid (– COOH) group.

246 | Chapter 15 Solve (a)

HC CH3O

C

H C

C H

C C

H C

C H

CH3 C

CH

C

CH

O – Na+

O

(b) A solution of the salt of naproxen is basic because the ionized – COO– group reacts with water, giving – COOH + OH–: HC CH3O

C

H C

C H

C C

H C

C H

CH3

C

CH

CH

C

O–

HC

+ H2O

O

CH3O

C

H C

C H

C C

H C

C H

CH3

C CH

CH

C

OH + OH –

O

(c) The salt is more soluble because it is charged and water molecules form stronger ion – dipole forces around the molecule than the dipole–induced dipole forces between the neutral molecule and water. Think about It Being soluble in water also helps to deliver the drug to the body. 15.152. Collect and Organize For a drop in the pH of the oceans by 0.77 pH units due to increased CO2 in the atmosphere, we are asked to explain with chemical equations how CO2 might cause this decrease in pH. We are also to calculate the extent of the increase in acidity this change would cause. Finally, we are to explain how coral reef survival could be affected by the decrease in pH. Analyze Carbon dioxide gas dissolves in water to form carbonic acid, a weak diprotic acid. Coral reefs are composed of CaCO3, which reacts with acids. Solve (a) When the partial pressure of CO2 in the atmosphere increases, the following equilibria are shifted to the right (Le Châtelier’s principle): CO2(g)  CO2(aq) CO2(aq) + H 2O()  H2CO3(aq)

H2CO3(aq)  H+(aq) + HCO3–(aq) HCO3–(aq)  H+(aq) + CO32–(aq) The acidity of the oceans increases (lower pH) because of increased ionization of the weak acid carbonic acid. (b) pH f – pH i = –0.77

– log[H + ]f – ( –log[H + ]i ) = –0.77

log[H + ]f − log[H + ]i = 0.77 log

[H + ]f = 0.77 [H + ]i

[H + ]f = 100.77 = 5.9 [H + ]i [H + ]f = 5.9[H + ]i The acidity of the ocean will be about six times greater.

Aqueous Equilibria | 247 (c) Because coral reefs are composed of CaCO3, they are likely to dissolve to a greater extent as the pH decreases. CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2CO3(aq) Think about It Our answer to part b is consistent with our knowledge that one pH unit represents a 10-fold difference in [H+].