Question

$A{X}_2$ is a sparingly soluble salt. The solubility product of $A{X}_2$ is $3.2\times {10}^{-11}.$ Find the value of solubility $\left(s\right)$ in $mol{L}^{-1}$ ?

• A. $2\times {10}^{-8}$
• B. $2\times {10}^{-4}$
• C. $8\times {10}^{-12}$
• D. $2\times {10}^{-11}$

Answer 1

The correct option is B $2\times {10}^{-4}$

Given $K_{sp}=3.2\times {10}^{-11}$
Lets consider solid $A{X}_2$ is in contact with its saturated aqueous solution.The equilibrium is established between undissolved $A{X}_2$ and its ions. The equilibrium reaction is given by : :
$A{X}_2\left(s\right)\rightleftharpoons A^{2+}\left(aq\right)+2X^{-}\left(aq\right)t=0c00t=t_{eq}c-ss2s$
The solubility product is given by:
$K_{sp}=\left[A^{2+}\right][X^{-}{]}^2$
$K_{sp}=s\times (2s{)}^2=4s^3$
$K_{sp}=4s^3=3.2\times {10}^{-11}$
$s^3=8\times {10}^{-12}s=2\times {10}^{-4}$
The value of solubility $\left(s\right)$ in $mol{L}^{-1}$