THE SOLUBILITY PRODUCT CONSTANT OF CALCIUM HYDROXIDE The solubility product constant of a substance is a special type of equilibrium constant whose value is used to express the solubility of slightly soluble compounds. The larger the value of the solubility product constant is, the greater the solubility of the compound. Note the adjacent table. Silver iodide, AgI, has the -17 lowest solubility with a Ksp of 1.8 x 10 , and silver chloride, AgCl , has the greatest solubility with a Ksp of -10 1.8 x 10 . You will also observe the order of decreasing solubility to be: AgCl > AgSCN > AgBr > AgI.
Substance
Ksp
AgBr
5.0 x 10
-13
AgCl
1.8 x 10
-10
AgI
8.3 x 10
-17
AgSCN
1.0 x 10
-12
How are solubility product constant expressions written? Given the general equation y+
x-
MxBv(s) ↔ xM (aq) + yB (aq) the equilibrium constant expression, Ksp, becomes, y+
x
x-
y
Ksp = [M (aq)] [B (aq)]
Using the general formula above one can write the equilibrium expression for the solubility product constants of some common substances at 25°C, such as: +
-
-10
Li2CO3 Ksp = [Li ] [CO3 ] = 2.5 x 10
2+
- 2
-9
As2S3
AgCl
Ksp = [Ag ][Cl ] = 1.8 x 10
PbI2
Ksp = [Pb ][I ] = 7.1 x 10
+ 2
3+ 2
2-
-2
2- 3
-22
Ksp = [As ] [S ] = 2.1 x 10
Solubility product constants are also important as the basis for quantitative analysis. Solubility product constants can be used to predict the conditions under which a precipitate will form. Consider the example of + -10 silver chloride, AgCl, above. The solubility product constant expression, Ksp= [Ag ][Cl ] = 1.8 x 10 , tells + us that when the product of the silver ion concentration, [Ag ], times the chloride ion concentration [Cl ], -10 equals 1.8 x 10 at 25°C, the compound has reached its limit of solubility, and precipitation will occur. What must the chloride ion concentration be to cause a precipitate to occur in a 0.010 M silver nitrate -8 solution? From the following calculation, we find that the chloride ion concentration must be 1.8x10 M for a precipitate to form. + -10 [Ag ][Cl ] = 1.8 x 10 -
:
(0.010M)[Cl ] = 1.8x 10 -
-10
-8
[Cl ] = 1.8x 10 M How can we calculate the solubility product constant, molar solubility, and solubility in grams/100 mL from concentration data? In a similar experiment designed to determine the solubility product constant of lead
(lI) iodide, PbI2, pure solid lead iodide was shaken with distilled water until equilibrium was achieved. The resulting solution was then filtered, and the iodide ion concentration was determined by titration with a solution of silver nitrate. Silver nitrate is a perfect substance for the determination of the iodide ion concentration, as the silver iodide produced has a very low solubility at room temperature as shown by the -17 very small solubility product constant, Ksp = 8.3 x 10 . In this experiment, 0.97ml of 0.001 M AgNO3 (aq) was required to just precipitate the iodide ion from 0.80 mL of the saturated lead iodide solution. With this information, we can find the Ksp, molar solubility, and solubility in grams/100 mL of lead iodide. +
-
+
-
NOTE: Ag (aq) + I (aq) AgI(s) Ksp = [Ag ][I ] = 8.3 x 10
-17
Since one mole of silver ions with a +1 charge reacts stiochiometrically with one mole of -1 charged iodide ions, the concentration of iodide in the sample simplifies to: Vsilver ion x Msilver ion = Viodide ion X Miodide ion +
0.97ml x 0.001 M Ag = 0.80 ml x Miodide ion Now that we know the concentration of the iodide ion, we can find the concentration of the lead ion. We know from the formula that in pure lead(lI) iodide, Pbl2, one mole of lead ions is combined with two moles of iodide ions. Therefore, if the iodide ion concentration is 0.0012M, then the lead ion concentration must be 0.006 M, which is one-half of 0.0012 M. 2+
-
PbI2 (s) Pb (aq)+ 2l (aq)
[I ]
1 mole Pb 2 [ Pb 2 ] 2 moles I
0.0012M I
1mole Pb 2 0.0060M Pb 2 2 moles I
We are ready to determine the Ksp of lead (ll) iodide from the lead ion and iodide ion concentrations determined above. 2+
- 2
2
-9
Ksp = [Pb ][I ] = (0.0060 M)(0.0012 M) = 8.6 x 10
The next step in the calculation is to find the solubility of lead (ll) iodide at the temperature of the experiment. The solubility of a substance is usually expressed as grams of solute per 100 grams of solution. Given the very low concentration of lead (ll) iodide in the solution, it is safe to assume that the density of the solution is close to the density of water or 1.0 gram per milliliter. Therefore, the solubility can be expressed as grams of solute, lead (II) iodide in this case, per 100ml of solution (0.100 liter). Since one mole of solid lead (lI) iodide forms one mole of lead (ll) ion, then the lead ion concentration, 0.00060 M, mirrors the concentration of lead (ll) iodide. This means that the molar solubility of lead (II) -4 iodide is 6.0 x 10 moles per liter. Using this value, we can find the solubility in grams/100 mL of lead (ll) iodide in two steps. First we are going to find the number of moles of Pbl2 in 100 ml (0.100 L) and then the grams of PbI2 in 100 ml.
molarity , M
number of moles of solute volume of the solution in liters
Number of moles of solute = volume of the solution (liters) x molarity (moles/liter) -5
Number of moles of PbI2 = 0.100 liter x 0.00060 M = 6.0 x 10 moles PbI2 -5
Since 100 ml of the saturated lead(ll) iodide solution contains 6.0 x 10 moles, and lead(ll) iodide has a formula mass of 461.0 grams per mole (Pb = 207.2 g/mol+ 2I @ 126.9 g/mol = 461.0 g/mol), we can finish the calculation by finding the solubility in grams of lead (ll) iodide per 100 ml of solution. number of grams of solute = number of moles of PbI2 x formula mass of PbI2 (grams/mole) -5
number of grams of solute = 6.0 x 10 moles Pbl2 x 461grams/mole = 0.028 grams/100ml In this experiment you are going to determine the solubility of a slightly soluble compound, calcium hydroxide, by titration of a saturated solution of this substance with a standard solution of hydrochloric acid. From the concentration of the unknown solution you will calculate the molar solubility, solubility in grams/100 mL and the solubility product constant as outlined above. PRE-LAB QUESTIONS: -2
1. Lithium carbonate, Li2CO3, has a solubility product constant, Ksp = 2.5 x 10 . a. Calculate the molar solubility of lithium carbonate. b. Using this information, calculate the solubility of lithium carbonate in grams/100 mL. c. Calculate the mass of Na2CO3 that must be added to 100 mL of a 0.10 M solution of LiNO 3 in order for a precipitate to form. 2. Powdered MgO was added to a beaker filled with deionized water. The mixture was covered with parafilm, and allowed to stir overnight. The mixture was filtered, and the filtrate was collected. A 100 mL sample of the solution was titrated using 0.0010 M HCl. The endpoint of the titration was reached after the addition of 22.5 mL HCl. a. Calculate the number of moles of HCl needed to reach the endpoint. b. Determine the molar solubility, and the solubility of Mg(OH) 2 in grams/100 mL. c. Calculate the Ksp of Mg(OH)2. HINT: metal oxide plus water yield metal hydroxide. MATERIALS: Calcium metal or calcium oxide 50 ml Erlenmeyer flask Para film Two 1-mL syringes (with tip extenders if needed) 1 10-mL syringe with luer-lok tip 1 0.45 µ syringe filter
0.050M hydrochloric acid Phenolphthalein in dropper bottle Graduated cylinder 2 30-mL beakers 25 mL Erlenmeyer flask Thermometer
PROCEDURE: Day 1 1. Add a piece of solid calcium metal about the size of this "O" into 25 mL of deionized or distilled water in a 50 mL Erlenmeyer flask. Calcium metal reacts with water producing calcium hydroxide and hydrogen gas. When the reaction ceases, cover the flask with parafilm and swirl to produce a homogenous mixture. Mark the flask with your initials and class, and place it in the hood until day 2. OR Add a small amount of calcium oxide shaving into 25 ml of deionized or distilled water in a 50 ml Erlenmeyer flask. Calcium oxide reacts with water producing calcium hydroxide. Cover the flask with parafilm to minimize the absorption of carbon dioxide from the air and swirl to produce a homogenous mixture. Mark the flask with your initials and class, and place it in the hood until day 2.
Day 2 1. Record the temperature of the room. This should also be the temperature of the solution in your flask. 2. Remove the plunger from the 10 mL syringe and attach the 0.45 μ filter. Fill the syringe to about 8 mL with the saturated solution of calcium hydroxide. Be careful when filling the syringe to not disturb the solid that has settled to the bottom of the flask. Replace the plunger, and filter the solution by pressing the plunger to force the liquid through the syringe. Collect the filtrate the small beaker labeled ‘base’ – make sure that the beaker is clean and dry. 3. If necessary, attach tip extenders to both 1-ml syringes. It is important to rinse and prepare the syringes as directed by your instructor. Fill the acid syringe with the 0.050M hydrochloric acid solution, 0.050 M HCl. Note that both the acid and base syringes are color coded. Fill the syringe designated for the base with the filtered, saturated Ca(OH)2 solution. Record the initial volumes of the solutions in both syringes to the nearest 0.01 mL in your data table. 3. Add 0.80-0.85 mL of the calcium hydroxide solution along with one or two milliliters of distilled or deionized water, and one drop of the phenolphthalein indicator to a clean, 25-mL Erlenmeyer flask. Add 0.050 M hydrochloric acid to the calcium hydroxide solution drop wise until one drop of acid just turns the solution colorless. Swirl the mixture after each addition, being certain to rinse the sides of the flask with water as necessary. Back titrate with the calcium hydroxide one drop at a time until the mixture turns pink and stays pink for 20 seconds or so. If you add too much base, simply add the acid drop wise again until the solution turns colorless again, and then add another drop of base. When you have reached the endpoint of the titration correctly, read the final volume of solution remaining in each syringe to the nearest 0.010 mL and record these values on your data table. 4. Refill the syringes and repeat the experiment as time permits (you should have time for at least 3 trials). 5. When the experiment is complete, rinse both the syringes with water. Thoroughly rinse and dry all other glassware, returning the common materials to their storage area.
ANALYSIS AND DISCUSSION: -
The hydroxide ion concentration, [OH ], in each of your trials can be calculated from the concentration of the hydrochloric acid, 0.050 M HCl, and the volume of both the HCl and Ca(OH)2 used in the titration. Write the net ionic equation for the dissociation of calcium hydroxide, Ca(OH) 2 (s), in water. 2+
According to this net ionic equation, determine the ratio of moles of calcium ion, Ca , to moles of hydroxide ion, OH , in the solution of calcium hydroxide, Ca(OH)2. Using this ratio, determine the molarity of calcium ion in the solution for each trial. 2+
-
Calculate the average concentration for both the calcium ion, [Ca ], and the hydroxide ion, [OH ]. Discuss the precision of your concentration values. Write the equilibrium expression for the solubility product constant, Ksp, of calcium hydroxide, Ca(OH)2(s). 2+
-
Calculate the average solubility product constant, Ksp, of calcium hydroxide from [Ca ] and [OH ] and the equilibrium expression. The table below gives the solubility of calcium hydroxide in grams/100 mL of water at various temperatures. Temperature (°C) Solubility (g/100mL) Temperature (°C) Solubility (g/100mL) 0 0.185 60 0.116 20 0.165 80 0.094 40 0.141 100 0.077 Plot the solubility of calcium hydroxide vs. temperature. What effect does an increase in temperature have on the solubility of Ca(OH)2 in water? What effect does an increase in temperature have on the solubility product constant, Ksp, for calcium hydroxide? Given the following reaction for the dissolution of calcium hydroxide in water, Ca(OH)2 (s) Ca(OH)2 (aq) + 16.3 kJ using LeChatlier’s principle as a guide, explain the effect of an increase in temperature on solubility. Based on your data, calculate the experimental solubility of Ca(OH) 2 in g/100 mL of solution. Plot this point on the graph, and circle it. How does the experimental solubility of calcium hydroxide compare to the predicted solubility at the same temperature? Using the graph, extrapolate the theoretical solubility of Ca(OH) 2 at the temperature of your solution. Use this value to calculate the theoretical molar solubility, and the theoretical solubility product constant for Ca(OH)2. Determine the % error for the experimental values for the solubility in grams/100 mL, the molar solubility, and the solubility product constant. Discuss any sources of error.
