Question

Calculate the molar solubility of calcium hydroxide in a solution buffered at each of the following pH's.

1) pH=3

2) pH=10

Answer 1

Calcium hydroxide dissolves and dissociates according to:
Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2 OH⁻(aq)
So the ionic molarities in a saturated solution satisfy the following equilibrium equation
Ksp = [Ca²⁺] ∙ [OH⁻]²
The solubility constant at 25°C is [1]
Ksp = 5.02×10^–6
Let s be the molar solubility of Ca(OH)₂. When you dissolve s moles per liter, all the salt dissolves and dissociates, forming one calcium ion per salt molecule. Hence the calcium ion molarities is:
[Ca²⁺] = x
The hydroxide ion concentration does not change due to buffering. It is fixed to a level, which is specified by the given pH:
[OH⁻] = 10^(pH - 14)

Therefore
Ksp = x ∙ (10^(pH - 14))² = x 10^(2∙pH - 28)
x = Ksp 10^(2∙pH - 28) = Ksp 10^(28 - 2∙pH)


at pH=3
x = 5.02×10^–6 *10^(28 - 2*3) = (5.02×10^–6 * 10^22) = 5.02×10^16 mol/L

at pH=10
x = 5.02×10^–6 *10^(28 - 2∙*10) = 5.02×10^–6 *10^(8) = 5.02×10^2 mol/L