Question

The solubility in water of a sparingly soluble salt ${AB}_2$ is $1.0\times {10}^{-5}mol{L}^{-1}$. Its solubility product will be:

• A. $4\times {10}^{-15}$
• B. $4\times {10}^{-10}$
• C. $1\times {10}^{-15}$
• D. $1\times {10}^{-10}$

Answer 1

The correct option is A $4\times {10}^{-15}$

Solubility process is:

$\underset{1}{{AB}_2}\to \underset{S}{A^{2+}}+\underset{2S}{2B^{-}}$

Using the relation of solubility and solubility product as:

$K_{sp}=\left[A^{2+}\right][B^{-}{]}^2$

$K_{sp}=S.(2S{)}^2$

Substituting $S={10}^{-5}M$

$K_{sp}={10}^{-5}\times (2\times {10}^{-5}{)}^2$

$K_{sp}=4\times {10}^{-15}$

Therefore option A is correct.