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 $A_2{X}_3,K_{sp}=1.1\times {10}^{-23}$

Answer 1

$A_2{X}_3⟶{2A}^{3+}+3X^{-2}$

$K_{sp}={\left[A^{3+}\right]}^2{\left[X^{2-}\right]}^3$

Let solubility of $A_2{X}_3$ be $S$

$\therefore$ $K_{sp}={\left[2S\right]}^2{\left[3S\right]}^3={108S}^5$

$\therefore$ $1.1\times {10}^{-23}={108S}^5$

$\therefore$ $S^5=\frac{1.1\times {10}^{-23}}{108}=1\times {10}^{-25}$

$\therefore$ $S=1.0\times {10}^{-5}mol/L$