Chemistry 102: Acid Equilibrium: Find Ksp for calcium hydroxide at a temperature

Question

The solubility of an ionic compound in water can be expressed in terms of the mass that will dissolve in a given volume of water, or in terms of the solubility product, K_sp.

If only 0.303 g of Ca(OH)₂ dissolves in enough water to give 0.100 L of aqueous solution at a given temperature, what is the K_sp value for calcium hydroxide at this temperature?

Answer 1

Ca(OH)₂(s) <==> Ca²⁺(aq) + 2OH^-(aq)

Ksp = [Ca²⁺][OH-]²

Now we need to calculate [Ca²⁺] and [OH-] (note there are 2 OH^- for each 1 Ca²⁺)

molar mass Ca(OH)₂ = 74.1 g / mol

mols Ca(OH)₂ dissolved = 0.303 g x 1 mol / 74.1 g = 0.004089 mols Ca(OH)₂

This is in 0.100 L so...

[Ca²⁺] = 0.004089 mol / 0.1 L = 0.04089 M

[OH-] = 2 x 0.04089 = 0.0818 M

Ksp = (0.0409)(0.0818)²

Ksp = 2.74x10^-4