THE SOLUBILITY PRODUCT CONSTANT OF CALCIUM HYDROXIDE The solubility product constant of a substance is a special type of equilibrium constant whose values are used to express the solubility of slightly soluble compounds. The larger the solubility product constant, the greater the solubility. Note the adjacent table. Silver iodide, AgI, has the lowest solubility with a Ksp of 1.8 x 10-17, and silver chloride, AgCl, has the greatest solubility with a Ksp of 1.8 x 10-10. You will also observe the order of decreasing solubilities 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

M x By (s)

xM y+ (aq) yBx- (aq)

the equilibrium constant expression, Ksp, becomes,

K sp = [M y+ (aq)]x [Bx- (aq)]y Using the general formula above we can write the equilibrium expression for the solubility product constants of some common substances at 25°C such as: AgCl Ksp = [Ag+][Cl-] = 1.8 x 10-10 PbI2

Ksp = [Pb2+][I-]2 = 7.1 x 10-9

Li2CO3

Ksp = [Li+]2[CO32-] = 2.5 x 10-2

As2S3

Ksp = [As3+]2[S2-]3 = 2.1 x 10-22

Solubility products 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 silver chloride, AgCl, above. The solubility expression [Ag+][Cl-] = 1.8 x 10-10 tells us that when the product of the silver ion concentration, [Ag+], times the chloride ion concentration, [Cl-], equals 1.8 x 10-10 at 25°C. What must be the chloride ion concentration to cause a precipitate to occur in a 0.010 M silver nitrate solution? From the following calculation, we find that the chloride ion concentration must be 1.8 x 10-8 M for a precipitate to form.

[Ag + ][Cl- ] = 1.8 x 10-10 (0.010 M)[Cl- ] = 1.8 x 10-10 [Cl- ] = 1.8 x 10-8 M How can we calculate the solubility product constant ans solubility from concentration data? In a similar experiment designed to determine the solubility product constant of lead (II) iodide, PbI2, pure solid lead iodide was shaken with distilled water for a few minutes. The resulting solution was then filtered and the iodide ion concentration determined using silver nitrate. Silver nitrate is a perfect substance for the determination of iodide ion as the silver iodide produced has a very low solubility at room temperature, Ksp = 8.3 x 10-17.

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0.97 ml of 0.001 M AgNO3(aq) is 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 and solubility 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 react stoichiometrically with one mole of -1 charged iodide ions, the concentration of iodide in the sample simplifies to:

Vsilver ion x M silver ion = Viodide ion x M iodide ion 0.97 ml x 0.001 M Ag + = 0.80 ml x M iodide ion M iodide ion = 0.0012 M I Now that we know the concentration of iodide ion, we can find the concentration of the lead ion. We know from the formula that in pure lead (II) iodide, PbI2, one mole of lead ion is combined with two moles of iodide ion. Therefore, if the iodide ion concentration is 0.0012 M, then the lead ion concentration must be 0.006 M, one-half of 0.0012 M.

PbI 2 (s)  Pb2  (aq) + 2I - (aq) -

[I ] x -

0.0012 M I x

1 mole Pb2+ 2 moles I

-

1 mole Pb2+ 2 moles I

-

= [Pb2+ ] = 0.0060 M Pb 2+

We are ready to determine the Ksp of lead (II) iodide from the lead ion and iodide ion concentrations determined above.

K sp = [Pb2+ ][I  ]2 = (0.0060 M)(0.0012 M) 2 = 8.6 x 10-9 The next step in the calculation is to find the solubility of lead (II) 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 (II) 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 100 ml of solution (0.100 liter). Since one mole of solid lead (II) iodide forms one mole of lead (II) ion, then the lead ion concentration, 0.00060 M, mirrors the concentration of lead (II) iodide. Using this value, we can find the solubility of lead (II) iodide in two steps. First we are going to find the number of moles of PbI2 in 100 ml (0.100 L) and then the grams of PbI2 in 100 ml.

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molarity, M =

number of moles of solute volume of the solution, liters

 moles  number of moles of solute = volume of the solution (liters) x molarity    liter  number of moles of PbI 2 = 0.100 liter x 0.00060 M = 6.0 x 10-5 moles PbI 2 Since 100 ml of the saturated lead (II) iodide solution contains 6.0 x 10-5 moles, and lead (II) iodide has a formula mass of 461 grams per mole (Pb = 207.2 g/mol + 2 I @ 126.9 g/mol = 461.0 g/mol), we can finish the calculation finding the solubility in grams of lead (II) iodide per 100 ml of solution.

 grams  number of grams of solute = number of moles of PbI 2 x formula mass of PbI2    mole  grams number of grams of solute = 6.0 x 10-5 moles PbI 2 x 461 = 0.028 grams/100 ml mole In this experiment you are going to determine the solubility of a slightly soluble compound, calcium hydroxide, by titrating a saturated solution of this substance against standard hydrochloric acid. From the concentration of the unknown solution you will calculate both its solubility and its solubility product constant as outlined above.

MATERIALS:

PROCEDURE: 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, stopper the flask and swirl to produce a homogenous mixture. Place the stoppered flask in a beaker of cold tap water allowing it to cool while you are preparing the syringes. 2. Cut the very top of a jumbo polyethylene transfer pipet and pack it with about 0.5 - 1.0 cm of cotton. This can be achieved by pushing small pieces of cotton into the bottom of pipet with a glass rod. Fill the top of the pipet filter with saturated solution of calcium hydroxide into the filter. Collecting the filtrate in a dry, 13 x 100 mm 3

Pipet filter with a cotton plug

Calcium metal, 50 ml Erlenmeyer flask w/stopper to fit, two 1-ml syringes with tip extenders, 0.050 M hydrochloric acid, phenolphthalein, 50-ml beaker, microtip polyethylene transfer pipet, small test tube and stopper, two jumbo polyethylene transfer pipets, thermometer, cotton, scissors

test tube. It may be necessary to discard the first few drops of solution until a clear filtrate is obtained. Record the temperature of the solution remaining in the flask to the nearest 1°C. Stopper both the test tube and the flask to minimize the absorption of carbon dioxide from the air. 3. If necessary, attach tip extenders to both 1-ml syringes hypodermic syringes. It is important to rinse and prepare the syringes as directed by your instructor. Fill the acid syringe with the 0.050 M hydrochloric acid solution, 0.05 M HCl. Note that both the acid and base syringes are coded. Fill the syringe designated for the base with the filtered saturated solution Ca(OH)2 solution. Partly fill a microtip polyethylene transfer pipet with phenolphthalein solution, the acid-base indicator in this experiment. Record the initial volumes of the solutions in both syringes to the nearest 0.01 ml on your data table. 4. 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 clean, 50-ml beaker. Add 0.050 M hydrochloric acid to the calcium hydroxide solution dropwise until one drop of acid just turns the solution colorless. Swirl the mixture after each addition being certain to rinse the sides of the vial or 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 dropwise 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 0.01 ml and record these values on your data table. 6. Refill the syringes and repeat the experiment as time permits. 7. When the experiment is complete, rinse both the syringes and the extender with water. Throughly rinse all other glassware returning the common materials to their storage area.

CALCULATIONS AND QUESTIONS: Q1. Calculate the hydroxide ion concentration, [OH-], in each of your trials from the concentration of the hydrochloric acid, 0.050 M HCl, and the volume of both solutions reacted. Q2. (a) Write the ionic equation for the dissociation of calcium hydroxide, Ca(OH)2(s), in water. (b) According to the ionic equation you wrote in question Q2a, what is the ratio of moles of calcium ion, Ca2+, 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?

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(c) Calculate the average concentration for both the calcium ion, [Ca2+(aq)], and the hydroxide ion, [OH-(aq)], in moles/liter (molarity). Q3. (a) Write the equilibrium expression for the solubility product constant, Ksp, of calcium hydroxide, Ca(OH)2(s). (b) Calculate the average solubility product constant, Ksp, of calcium hydroxide from the [Ca2+(aq)] and [OH-(aq)] calculated in question Q2c and your equilibrium expression given in question Q3a. Q4. Given the solubility of calcium hydroxide in grams of solute/100 ml of water at various temperatures Temperature

Solubilit y

Temperature

Solubility

0°C

0.185 g

60°C

0.116 g

20°C

0.165 g

80°C

0.094 g

40°C

0.141 g

100°C

0.077 g

(a) Plot the solubility of calcium hydroxide vs the temperature in the space provided on the same page as your data table. What effect does an increase in the temperature have on the solubility of calcium hydroxide in water? (b) According to the solubility data, what effect does an increase in the temperature have on the solubility product constant, Ksp, for calcium hydroxide? (c) Given the formula equation for the formation of a solution of calcium hydroxide in water:

Ca(OH)2 (s)

Ca(OH) 2 (aq) + 16.3 kJ

Using LeChâtelier's principle as a guide, how do you account for the effect of an increase in the temperature on the solubility of calcium hydroxide? Q5. (a) Assuming the average [Ca2+] and the molarity of Ca(OH)2(aq) are the same, calculate the mass of calcium hydroxide dissolved in 100 ml (0.100 liter) of solution. This value is your experimental solubility for Ca(OH)2. (b) Plot your experimental solubility of Ca(OH)2 at the temperature measured in this experiment on the graph prepared on your data table. Circle this point. Which term best describes your results, saturated, unsaturated, or supersaturated? Explain! Q6. Write the equation for the reaction of calcium metal with water. What other metals will also react rapidly with water producing a soluble hydroxide? 5

DATA:

Temperature of the filtrate: __________________ °C Trial #1

Trial #2

Trial #3

Initial volume of the HCl solution

ml

ml

ml

Final volume of the HCl solution

ml

ml

ml

Volume of 0.050 M HCl reacted

ml

ml

ml

Initial volume of Ca(OH)2 solution

ml

ml

ml

Final volume of Ca(OH)2 solution

ml

ml

ml

Volume of saturated Ca(OH)2 reacted

ml

ml

ml

Molarity of OH-(aq)

M.

M.

M.

Molarity of Ca2+(aq)

M.

M.

M.

Average hydroxide ion concentration, [OH-(aq)] =

M.

Average calcium ion concentration, [Ca2+(aq)] =

M.

Average Ksp of Ca(OH)2(s) = Solubility of Ca(OH)2(s) =

g/100 ml H2O

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