At 22 °C, an excess amount of a generic metal hydroxide, M(OH)2, is mixed with pure water.

Question

The resulting equilibrium solution has a pH of 10.26.  What is the 𝐾sp of the salt at 22 °C?

Ksp= ?

Answer 1

M(OH)2(s) <==> M²⁺(aq) + 2OH^-(aq)

From pH = 10.26 we find pOH = 14 - 10.26 = 3.74 (the Kw ~10-14 @ 22ºC so we will go with that)

From pOH = 3.74, we find [OH^-] = 1x10^-3.74 = 1.82x10^-4 M

Ksp = [M²⁺][OH^-]²

Since [OH^-] = 1.82x10^-4 M, the [M²⁺] would be 1/2 that = 9.1x10^-5 M

Ksp = [9.1x10^-5][1.8x10^-4]²

Ksp ≈ 2.9x10^-12

Answer 2

Did your teacher mention that K_w is not equal to 10^-14 at 25^oC? Because the pOH is not going to be equal to 14-10.26.