Question

Calcium oxalate (${\mathbf{CaC}}_2{\mathbf{O}}_4$) is relatively insoluble in water $\left({\mathbf{K}}_{sp}=\mathbf{2}\mathbf{x}{\mathbf{10}}^{-9}\right)$. However, calcium oxalate is moresoluble in acidic solution. How much more soluble is calcium oxalate in$\mathbf{0}\mathbf{.}\mathbf{10}{\mathbf{MH}}^{\mathbf{+}}$ than in pure water? In pure water, ignore the basic properties of ${\mathbf{C}}_2{\mathbf{O}}_4^{2-}$.

Short answer

Calcium oxalate in ${\text{0.10MH}}^{\text{+}}$is $\text{46.9}$times more soluble than in pure water.

Step-by-step solution

Introduction to the Concept

The product of the concentration of ions dissolved in water,${\mathrm{K}}_{\mathrm{sp}}$ is the solubility product constant of a solution raised to the power of its appropriate stoichiometry coefficient.

Determination of solubility of calcium oxalate in pure water 

From the given,

${\text{K}}_{sp}$of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}{\text{=2x10}}^{\text{-9}}$

${\mathrm{CaC}}_2{\mathrm{O}}_4\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Ca}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\left(\mathrm{aq}\right)$

Initial	-	-
Eqm	s	s

$\begin{array}{c}{\text{K}}_{\text{SP}}\text{=}\left[{\text{Ca}}^{\text{2+}}\right]\left[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\right]\\ 2x{10}^{-9}=s^2\end{array}$

Hence,$\mathrm{s}=4.47\mathrm{x}{10}^{-5}\text{M}$

Determination of solubility of calcium oxalate in 0.10M H+

${\mathrm{CaC}}_2{\mathrm{O}}_4\left(\mathrm{s}\right)\iff {\mathrm{a}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\left(\mathrm{aq}\right)-----\left(1\right)$

${\text{K}}_{\text{SP}}{\text{=2x10}}^{\text{-9}}$

${\mathrm{C}}_2{\mathrm{O}}_4^{2-}\left(\mathrm{aq}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\iff {\mathrm{HC}}_2{\mathrm{O}}_4^{-}\left(\mathrm{aq}\right)-----\left(2\right)$

$\text{K\hspace{0.33em}=\hspace{0.33em}}\frac{\text{1}}{{\text{K}}_{\text{a2}}}{\text{\hspace{0.33em}=\hspace{0.33em}1.6×10}}^{\text{4}}$

${\mathrm{HC}}_2{\mathrm{O}}_4^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\iff {\mathrm{HC}}_2{\mathrm{O}}_4\left(\mathrm{aq}\right)-----\left(3\right)$

$\text{K\hspace{0.33em}=\hspace{0.33em}}\frac{\text{1}}{{\text{K}}_{\text{a1}}}\text{\hspace{0.33em}=\hspace{0.33em}15}$

By adding equation (1), (2) and (3).

${\mathrm{CaC}}_2{\mathrm{O}}_4\left(\mathrm{s}\right)+2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\iff {\mathrm{Ca}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\left(\mathrm{aq}\right)$

${\text{K}}_{\text{overall}}{\text{=\hspace{0.33em}4.8\hspace{0.33em}×\hspace{0.33em}10}}^{\text{-4}}$

Initial	0.10 M	-	-
Eqm	0.10 - s	s	s

$\begin{array}{c}4.8\times {10}^{-4}=\frac{s^2}{{\left(0.10-2s\right)}^2}\\ 0.022=\frac{s}{{\left(0.10-2s\right)}^2}\end{array}$

Hence, $\mathrm{s}=0.0021\text{M}$.

$\begin{array}{l}\frac{\mathrm{Solubility}\mathrm{in}0.10{\mathrm{MH}}^{+}}{\mathrm{Solubility}\mathrm{in}\text{purewater}}=\frac{0.0021\text{M}}{4.47\times {10}^{-5}\text{M}}\\ \mathrm{Solubility}\mathrm{in}\text{purewater\hspace{0.33em}}=46.9\end{array}$

Therefore, calcium oxalate is$\text{46.9}$ times more soluble than pure water.