Question

Calculate the solubility of A₂X₃ in pure water, assuming that neither kind of ion reacts with water. The solubility product of A₂X₃, K_sp = 1.1 × 10^-23

Answer 1

Given Data:

${\mathrm{K}}_{\mathrm{sp}}=1.1\mathrm{x}{10}^{-23}$

GIven Compound = ${\mathrm{A}}_2{\mathrm{X}}_3$

Applying Solubility Formula:-

$$\begin{gathered} {\mathrm{A}}_2{\mathrm{X}}_3\to 2{\mathrm{A}}^{3+}+3{\mathrm{X}}^{2-} \\ {\mathrm{K}}_{\mathrm{SP}}={\left[{\mathrm{A}}^{3+}\right]}^2{\left[{\mathrm{X}}^{2-}\right]}^3 \\ \mathrm{Let}\mathrm{solubility}\mathrm{of}{\mathrm{A}}_2{\mathrm{X}}_3\mathrm{be}\mathrm{S} \\ \therefore {\mathrm{K}}_{\mathrm{sp}}={\left[2\mathrm{S}\right]}^{2+}{\left[3\mathrm{S}\right]}^3=108{\mathrm{S}}^5 \\ \Rightarrow 1.1\mathrm{x}{10}^{-23}=108{\mathrm{S}}^5 \\ \therefore {\mathrm{S}}^5=1\mathrm{x}{10}^{-25} \\ \therefore \mathrm{S}=1.0\mathrm{x}{10}^{-5}\mathrm{mol}/\mathrm{L} \end{gathered}$$

Therefore, the solubility of A₂X₃ in pure water, assuming that neither kind of ion reacts with water is $1.0\mathrm{x}{10}^{-5}\mathrm{mol}/\mathrm{L}$.