At 22 ∘C, an excess amount of a generic metal hydroxide, M(OH)2, is mixed with pure water. The resulting equilibrium solution has a pH of 10.22. What is the 𝐾sp of the salt at 22 ∘C? 𝐾sp=

Answer 1

M(OH)₂(s) <==> M²⁺(aq) + 2OH-(aq)

Ksp = [M²⁺][OH-]²

From the pH, we can find the [OH-]:

pH + pOH = 14

pOH = 14 - 10.22 = 3.78

[OH-] = 1x10^-3.78

[OH-] = 1.66x10^-4 M

If [OH-] = 1.66x10^-4 M, then [M²⁺] must be 1/2 that or 8.30x10^-5 M

Ksp = (8.30x10^-5)(1.66x10^-4)²

Ksp = 2.3x10^-12