Question

Calculate the solubility of a sparingly soluble compund $A_2{X}_3$ at ${25}^{\circ }C$.
The solubility product $K_{sp}$ of $A_2{X}_3$ is $1.08\times {10}^{-23}$ at the given temperature.

• A. $1\times {10}^{-7}mol{L}^{-1}$
• B. $1.08\times {10}^{-7}mol{L}^{-1}$
• C. $1\times {10}^{-5}mol{L}^{-1}$
• D. $1.08\times {10}^{-3}mol{L}^{-1}$

Answer 1

The correct option is C $1\times {10}^{-5}mol{L}^{-1}$

The equilibrium reaction is:
$\begin{array}{ccccc}{A}_2{X}_3\left(s\right)& \rightleftharpoons & 2A^{3+}\left(aq\right)& +& 3X^{2-}\left(aq\right)\\ c& & 0& & 0\\ c-s& & 2s& & 3s\end{array}$
Here $s$ is the solubility of $A_2{X}_3$
Since,
$K_{sp}=\left[A^{3+}{]}^2\right[X^{2-}{]}^3$
$\Rightarrow K_{sp}=(2s{)}^2\times (3s{)}^3$
$\Rightarrow K_{sp}=4s^2\times 27s^3$
$\Rightarrow 108s^5=K_{sp}$
$\Rightarrow s={\left(\frac{K_{sp}}{108}\right)}^{\frac{1}{5}}$

Putting the value of $K_{sp}$
$\Rightarrow s={\left(\frac{1.08\times {10}^{-23}}{108}\right)}^{\frac{1}{5}}=({10}^{-25}{)}^{\frac{1}{5}}{K}_{sp}=1\times {10}^{-5}mol{L}^{-1}$