Question

The solubility in water of a sparingly soluble salt $A{B}_2$ is $1.0\times {10}^{-5}mol{L}^{-1}$. Its solubility product number 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 $4\times {10}^{-15}$

The solubility of $A{B}_2$ is $x$.

The dissociation equilibrium is as given below.

$\underset{1.0\times {10}^{-5}}{A{B}_2}\rightleftharpoons \underset{1.0\times {10}^{-5}}{A^{2+}}+\underset{2.0\times {10}^{-5}}{2B^{-}}$.

The expression for the solubility product is $K_{sp}=\left[A^{2+}\right][2B^{-}{]}^2$.

Substitute values in the above expression.

$K_{sp}=(1.0\times {10}^{-5})(2.0\times {10}^{-5}{)}^2=4\times {10}^{-15}$.

Hence, the correct option is $\text{A}$