Question

Calculate the solubility of $A_2{X}_3$ in pure water, assuming that neither kind of ion reacts with water. The solubility product of $A_2{X}_3,K_{sp}=1.1\times {10}^{-23}.$

• A. $2.5\times {10}^{-5}mol{L}^{-1}$
• B. $2.0\times {10}^{-5}mol{L}^{-1}$
• C. $1.0\times {10}^{-5}mol{L}^{-1}$
• D. $1.5\times {10}^{-5}mol{L}^{-1}$

Answer 1

The correct option is C $1.0\times {10}^{-5}mol{L}^{-1}$

The equilibrium established will be,
$\begin{array}{ccccc}{A}_2{X}_3& \rightleftharpoons & 2A^{3+}\left(aq\right)& +& 3X^{2-}\left(aq\right)\\ 1& & 0& & 0\\ 1-s& & 2s& & 3s\end{array}$

Using the formula,
$K_{sp}=\left[A^{3+}{]}^2\right[X^{2-}{]}^3$
$\Rightarrow K_{sp}=(2s{)}^2\times (3s{)}^3$
$\Rightarrow K_{sp}=4s^2\times 27s^3$
$\Rightarrow 108s^5=K_{sp}$
$\Rightarrow s={\left(\frac{K_{sp}}{108}\right)}^{1/5}$

putting the value of $K_{sp}$
$\Rightarrow s={\left(\frac{1.1\times {10}^{-23}}{108}\right)}^{1/5}=1\times {10}^{-5}mol{L}^{-1}$