Solubility Lab #2 - The Solubility Product of Calcium Hydroxide

Calcium metal reacts vigorously with water thusly:   

Ca(s) + 2 H2O -------> Ca(OH)2(s) + H2(g)

The Ca(OH)2(s) is slightly soluble in water, dissolving as follows:  

Ca(OH)2(s) <==> Ca2+(aq) + 2 OH-(aq)

By measuring the concentration of the hydroxide ion, [OH-], the concentration of calcium ion, [Ca2+], can be calculated and the Ksp value for Ca(OH)2 at 25oC obtained.

The acceptable values are: 1.3 x 10-6 (no temperature given, Seinko & Plane); 7.8 x 10-5 (Merck Index); 6.25 x 10-5 at 0oC as calculated from solubility data.

1.  One partner should prepare some filtrate of the dissociated Ca(OH)2 (s). At the same time the other partner should prepare a burette, cleaned and rinsed with some of the filtrate. 
2. Three samples of potassium hydrogen phthalate of approximately 0.25 g should be weighed out.

Record these masses accurately. 

Deliver each to a 125 mL Erlenmeyer flask and add 25 to 30 mL of water. (Record the exact amounts used.) 
The potassium hydrogen phthalate will dissolve as titration proceeds. 
3. Fill the burette with saturated aqueous Ca(OH)2 and take the initial reading. (Burettes read from the top). 
4.  Add 2 to 3 drops of phenolphthalein to the Erlenmeyer. 
5. Slowly titrate. (The OH- + H+ ----> H2O) When the OH- has neutralized the hydrogen of the hydrogen phthalate ion, the pink of the phenolphthalein appears. 

Stop the addition of Ca(OH)2 when only the faintest perceptible pink appears and remains for three minutes. Record the final volume of Ca(OH)2 solution in the burette. 
6. Repeat step (5) with the other potassium hydrogen phthalate samples. Do not forget to add phenolphthalein or take burette readings.
Trials  Grams of potassium hydrogen phthalate used   Initial Burette Reading  Final Burette Reading 

Processing The Data
(1) The molar mass of potassium hydrogen phthalate is 204.23 g/mole. Calculate the number of moles of Hphth- in each sample weighed.
                   KHphth(s) <========> K+(aq) + Hphth-(aq)

(2) At the equivalence point, (i.e. neutrality), the number of moles of OH- is equal to the number of moles of Hphth-. Knowing the volume of hydroxide used, the molarity of the hydroxide can be calculated.

(3) From the dissociation equation:   

Ca(OH)2(s) <=======>  Ca2+(aq) + 2 OH-(aq)
         x                                  x                2x

At equivalence it should be obvious that [Ca2+] = 2 [OH-]

(4) Substituting values for [OH-] and [Ca2+] in the Ksp expression we can calculate the Ksp experimentally.
                                 Ksp = [Ca2+][OH-]2 at 25oC

(5) How does your value compare with the accepted experimental values given in the introduction?

Possible Sources of Errors
(1) The solubility of Ca(OH)2 varies with temperature.
(2) The CO2 in the air would interfere with the equilibrium at the neutral point.

Some Sample Student Data:
Sample Set 1
trial 1 [OH-] = 3.56 x 10-2
trial 2 [OH-] = 3.58 x 10-2
Ksp Ca(OH)2 = 2.25 x 10-5 (trial 1), 2.29 x 10-5 (trial 2)

Sample Set 2
trial 1 [OH-] = 4.27 x 10-2
trial 2 [OH-] = 4.16 x 10-2
trial 3 [OH-] = 4.32 x 10-2
average [OH-] = 4.25 x 10-2
average Ksp = 3.8 x 10-5

This compares favourably with the data in other experiments.

1. Why do you have to filter the Ca(OH)2 solution before loading it into the burette?
2. Why don't you have to accurately record the amount of Ca(s) used?