Question

The generic metal hydroxide M(OH)2 has Ksp = 7.25×10−12. (NOTE: In this particular problem, because of the magnitude of the Ksp and the stoichiometry of the compound, the contribution of OH− from water can be ignored. However, this may not always be the case.)

Answer 1

This is an incomplete question, here is a complete question.

The generic metal hydroxide M(OH)₂ has Ksp = 7.25 × 10⁻¹². (NOTE: In this particular problem, because of the magnitude of the Ksp and the stoichiometry of the compound, the contribution of OH⁻ from water can be ignored. However, this may not always be the case.)

What is the solubility of M(OH)₂ in pure water?  Express your answer with the appropriate units.

Answer : The solubility of M(OH)₂ in pure water is, $1.22\times 10^{-4}M$

Explanation :

The equilibrium chemical reaction will be:

$M(OH)_2(s)\righleftharpoons M^{2+}(aq)+2OH^-(aq)$

The solubility constant expression for this reaction is:

$K_{sp}=[M^{2+}][OH^-]^2$

Let the solubility be, 'x'

$M(OH)_2(s)\rightleftharpoons M^{2+}(aq)+2OH^-(aq)$

     x                 x                 2x

Now put all the given values in this expression, we get:

$7.25\times 10^{-12}=(x)\times (2x)^2$

$7.25\times 10^{-12}=4x^3$

$x=1.22\times 10^{-4}M$

Thus, the solubility of M(OH)₂ in pure water is, $1.22\times 10^{-4}M$