Exercise 10 | Solubility and Precipitation Reactions

General Chemistry 3 - Exercise 10

Calculate the solubility of Ca(OH)2 (s) in water in g.L-1.


Data: pH of a saturated Ca(OH)2 (aq) solution = 12.45.

Chemical equation:

Ca(OH)2 (s) ⇄ Ca2+ (aq) + 2 HO- (aq)


Pure water: the only source of Ca2+ and HO- is Ca(OH)2

From the reaction stoichiometry: [HO-] = 2 [Ca2+]

s = solubility of Ca(OH)2 (s) in M = [Ca2+] = [HO-] / 2


Kw = 10-14 = [H3O+][HO-]

⇒ [HO-] = 10-14 / [H3O+] = 10-14 / 10-12.45

⇒ [HO-] = 2.82 x 10-2 M

⇒ s = 1.41 x 10-2 M


S = solubility of Ca(OH)2 (s) in g.L-1 = s x MCa(OH)2

⇒ S = 1.04 x 10-1 g.L-1