A solution contains $\mathbf{0}\mathbf{.}\mathbf{0500}{\mathbf{MCa}}^{\mathbf{2}\mathbf{+}}$and $\mathbf{0}\mathbf{.}\mathbf{0300}{\mathbf{MAg}}^{\mathbf{+}}$. Can 99%of ${\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}$be precipitated by sulphate without precipitating${\mathbf{Ag}}^{\mathbf{+}}$? What will be the concentration of${\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}$when${\mathbf{Ag}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$begins to precipitate?

For a sparingly soluble salt, the solubility product is the mathematical product of the concentration of its dissolved constituent ions raised up to the power of their stoichiometric coefficient. It is the equilibrium constant for the dissociation reaction for a sparingly soluble salt.

$$\begin{gathered} M_y{X}_{z\left(s\right)}\rightleftharpoons y{M}^z+\left(aq\right)+z{X}^{y-}\left(aq\right) \\ {K}_{sp}={\left[M^{z+}\right]}^y{\left[X^{y-}\right]}^z \end{gathered}$$

The solubility product constant $\left({\mathrm{K}}_{\mathrm{sp}}\right)$value for calcium sulphate is $2.4\times {10}^{-5}$and for ${\mathrm{Ag}}_2{\mathrm{SO}}_4$is $1.5\times {10}^{-5}$. Eliminating 99% of ${\mathrm{Ca}}^{2+}$reduces concentration of ${\mathrm{Ca}}^{2+}$to 0.0005M .

Thus we need to find the concentration of $S{O}_4^{-2}$that will be precipitated with ions.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{+2}\right]\left[{\mathrm{SO}}_4^{-2}\right] \\ \left[{\mathrm{SO}}_4^{-2}\right]=\frac{2.4\times {10}^{-5}}{0.0005} \\ =0.048\mathrm{M} \end{gathered}$$

Let us calculate the reaction quotient for ${\mathrm{AgSO}}_4$.

$$\begin{gathered} Q={\left[A{g}^{+}\right]}^2\left[S{O}_4^{2-}\right] \\ ={\left(0.0300\right)}^2\left(0.048\right) \\ =4.3\times {10}^{-5} \end{gathered}$$

The reaction quotient is greater than the solubility product. Thus,will form precipitate. So, the separation is not possible.

Concentration of ${\mathrm{Ca}}^{2+}$when ${\mathrm{AgSO}}_4$initiates to precipitate

When${\mathrm{Ag}}^{+}$ precipitate early,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Ag}}^{+}\right]}^2\left[{\mathrm{SO}}_4^{-2}\right] \\ \left[{\mathrm{SO}}_4^{-2}\right]=\frac{1.5\times {10}^{-5}}{{\left(0.030\right)}^2}=1.67\times {10}^{-2}\mathrm{M} \\ {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{+2}\right]\left[{\mathrm{SO}}_4^{-2}\right] \\ \left[{\mathrm{Ca}}^{+2}\right]=\frac{2.4\times {10}^{-5}}{1.67\times {10}^{-2}}=0.0014\mathrm{M} \end{gathered}$$

The concentration ${\mathrm{Ca}}^{+2}$of ions is 0.0014 M.