Calcium metal reacts vigorously with water
Ca(s) + 2 H2O -------> Ca(OH)2(s) + H2(g)
The Ca(OH)2(s) is slightly soluble in water, dissolving as
<==> 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.
||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.
||Three samples of
potassium hydrogen phthalate of approximately
0.25 g should be weighed out.
|Record these masses
|Deliver each to a 125
mL Erlenmeyer flask and add 25 to 30 mL of water. (Record the exact
The potassium hydrogen phthalate will dissolve as
||Fill the burette with
saturated aqueous Ca(OH)2 and
take the initial reading. (Burettes read from the top).
||Add 2 to 3 drops of
phenolphthalein to the Erlenmeyer.
||Slowly titrate. (The OH-
+ H+ ----> H2O) When the OH- has
neutralized the hydrogen of the hydrogen phthalate ion, the pink of the
|Stop the addition of
Ca(OH)2 when only the faintest
perceptible pink appears and remains for three minutes. Record the
volume of Ca(OH)2 solution in the burette.
||Repeat step (5) with
the other potassium hydrogen phthalate samples. Do not forget to add
phenolphthalein or take burette readings.
||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
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
(3) From the dissociation equation:
Ca(OH)2(s) <=======> Ca2+(aq)
+ 2 OH-(aq)
At equivalence it should be obvious that [Ca2+]
(4) Substituting values for [OH-] and [Ca2+]
in the Ksp expression we can calculate the Ksp
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
(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)