Calculate the molar solubility of ${\mathbf{Hg}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{}\mathbf{}(K_{\mathrm{sp}}=1.75\times {10}^{-13})$ in $\mathbf{0}\mathbf{.}\mathbf{13}\mathbf{}\mathbf{M}\mathbf{}\mathbf{Hg}{\left({\mathrm{NO}}_3\right)}_{\mathbf{2}}$.

A solid's solubility (commonly referred to as molar solubility) is measured as the concentration of the "dissolved solute" in a saturated solution. The solubility of a sparingly soluble salt can be estimated from the solubility product. The solubility product is the product of the concentrations of the constituent ions present in the solution raised to their stoichiometric coefficients.

$\begin{array}{c}\mathrm{c}\left({\mathrm{Hg}}_2{\left({\mathrm{NO}}_3\right)}_2\right)=0.13\mathrm{M}\\ \mathrm{Ksp}=1.75\times {10}^{-13}.\end{array}$

We can write the dissolution equation fordata-custom-editor="chemistry" ${\mathrm{Hg}}_2{\mathrm{C}}_2{\mathrm{O}}_4$:

role="math" localid="1663257760525" $$\begin{gathered} {\mathrm{Hg}}_2{\mathrm{C}}_2{\mathrm{O}}_4\left(s\right)\rightleftharpoons {\mathrm{Hg}}_2^{2+}\left(\mathrm{aq}\right)+{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\left(\text{aq}\right) \\ \mathrm{ksp}=\left[{\mathrm{Hg}}_{\mathit{2}}^{\mathit{2}\mathit{+}}\right]\left[{\mathrm{C}}_{\mathit{2}}{\mathrm{O}}_{\mathit{4}}^{\mathit{2}\mathit{-}}\right]=1.75\times {10}^{-13} \end{gathered}$$

Let us prepare an ICE table to estimate the equilibrium concentrations of the species in solution. The initial concentration of ions will not be zero because $0.13 \mathrm{M} \mathrm{of} {\mathrm{Hg}}_2{\left({\mathrm{NO}}_3\right)}_2$.

$\left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]=S$

Because of the presence of $0.13\mathrm{M} {\mathrm{Hg}}_2{\left({\mathrm{NO}}_3\right)}_2$the concentration of ${\mathrm{Hg}}_2^{2+}$will be $0.13 \mathrm{M}$, and we can use that to calculate solubility:

$\begin{array}{l}{\mathrm{K}}_{\mathrm{sp}}= \left[\mathrm{H}{{\mathrm{g}}_2}^{2+}\right]\left[{\mathrm{C}}_2{\mathrm{O}}_4^{-2}\right]\\ 1.75\times {10}^{-13}=\left(0.13+\mathrm{S}\right)\mathrm{S}\end{array}$

$\begin{array}{l}\mathrm{since},\mathrm{S}<<0.13 \mathrm{M},\mathrm{wecanwrite}:\\ 1.75\times {10}^{-13}=0.13\times \mathrm{S}\\ \mathrm{S}=\frac{1.75\times {10}^{-13}}{0.13} \mathrm{M}\\ \mathrm{S}=1.34\times {10}^{-12} \mathrm{M}\end{array}$

Therefore, the required molar solubility is $1.34\times {10}^{-12}\text{M}$.