Question

Calculate the solubility of solid Ca₃(PO₄)₂ in a 0.20 M Na₃PO₄ solution

Short answer

The solubility of solid Ca₃(PO₄)₂ in a 0.20 MNa₃PO₄ solution is 2.35 x 10^-11mol/L

Step-by-step solution

Definition of solubility

The maximum amount of a substance that will dissolve in a given amount of solvent at a specified temperature is defined as solubility.

Calculation of solubility

The dissolution equilibrium of Ca₃(PO₄)₂

_{$$\begin{gathered} \mathit{C}{\mathit{a}}_{\mathbf{3}}\mathbf{(}\mathbf{P}{\mathbf{O}}_{\mathbf{4}}{\mathbf{)}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{3}\mathit{C}{\mathit{a}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{P}{\mathit{O}}_{\mathbf{4}}^{\mathbf{3}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)} \\ {\mathit{K}}_{\mathbf{s}\mathbf{p}}\mathbf{=}{\left[\mathit{C}{\mathit{a}}^{2+}\right]}^{\mathbf{3}}{\left[\mathit{P}{\mathit{O}}_4^3\right]}_{\mathbf{2}} \end{gathered}$$}

Let ‘s’ mol/L represent the molar solubility of Ca₃(PO₄)₂

.

Based on the dissolution stoichiometry,

$$\begin{gathered} \left[C{a}^{2+}\right]=\text{'}3s\text{'} mol/L \\ \left[P{O}_4^{3-}\right]=\text{'}2s\text{'}mol/L \end{gathered}$$

There is an increase in the concentration of the $P{O}_4^{3-}$ ion in the presence of 0.20 M Na₃PO₄solution.

$$\begin{gathered} \left[P{O}_4^{3-}\right]=2s+0.20 mol/L \\ {K}_{sp}={\left[C{a}^{2+}\right]}^3{\left[P{O}_4^{3-}\right]}_2 \\ s=2.35\times {10}^{-11}mol/L \end{gathered}$$