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* 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* 10^(-12)=(x)* (2x)^2`

`7.25* 10^(-12)=4x^3`

`x=1.22* 10^(-4)M`

Thus, the solubility of M(OH)₂ in pure water is,
`1.22* 10^(-4)M`