Question

Assuming that the solubility of ${\mathbf{Ca}}_{\mathbf{2}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$is $\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{mol}\mathbf{/}\mathbf{L}$at${\mathbf{25}}^{\mathbf{o}}\mathbf{C}$ , calculate the ${\mathbf{K}}_{\mathbf{sp}}$ for this salt. Ignore any potential reactions of the ions with water.

Short answer

The ${\mathrm{K}}_{\mathrm{sp}}$ of ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$ in the solubility $1.6\times {10}^{-7}\mathrm{mol}/\mathrm{L}$ is $1.13\times {10}^{-32}$.

Step-by-step solution

Definition of solubility product constant

The equilibrium constant for the dissolution of a solid substance into an aqueous solution is defined as solubility product constant.

Calculations

The dissolution equilibrium of ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$ is

$\begin{array}{l}{\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2\left(\mathrm{s}\right)\rightleftharpoons 3{\mathrm{Ca}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{PO}}_4^{3-}\left(\mathrm{aq}\right)\\ {\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Ca}}^{2+}\right]}^3{\left[{\mathrm{PO}}_4^{3-}\right]}_2\end{array}$

Let $\text{‘s’mol/L}$ represent the molar solubility of ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$.

$\begin{array}{l}\left[{\mathrm{Ca}}^{2+}\right]=\text{'}3\mathrm{s}\text{'}\mathrm{mol}/\mathrm{L}\\ \left[{\mathrm{PO}}_4^{3-}\right]=\text{'}2\mathrm{s}\text{'}\mathrm{mol}/\mathrm{L}\end{array}$

The${\mathrm{K}}_{\mathrm{sp}}$ of ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$ is

$\begin{array}{l}{\mathrm{K}}_{\mathrm{sp}}={\left[3\mathrm{s}\right]}^3{\left[\mathrm{s}\right]}^2\\ {\mathrm{K}}_{\mathrm{sp}}=108{\mathrm{s}}^5\\ {\mathrm{K}}_{\mathrm{sp}}=1.13\times {10}^{-32}\end{array}$