The values of ${\mathrm{K}}_{\mathrm{sp}}$ of ${\mathrm{CaCO}}_3\text{ and }{\mathrm{CaC}}_2{\mathrm{O}}_4$ are $4.7\times {10}^{-9}$ and $1.3\times {10}^{-9}$respectively at ${25}^{\mathrm{o}}\mathrm{C}$. If the mixture of these two is washed with water, what is the concentration of ${\mathrm{Ca}}^{2+}$ ions in water?

$$\begin{gathered} {\mathrm{CaCO}}_3⟶{\mathrm{Ca}}^{2+}+{\mathrm{CO}}_3^{2-} \\ {\mathrm{CaC}}_2{\mathrm{O}}_4⟶{\mathrm{Ca}}^{\stackrel{x}{2+}}+{\mathrm{C}}_2{\mathrm{O}}_4^{\stackrel{x}{2-}} \end{gathered}$$

$$\begin{gathered} \text{ Now, }\left[{\mathrm{Ca}}^{2+}\right]=x+y \\ \text{ and }x(x+y)=4.7\times {10}^{-9 } .............\left(i\right) \\ y(x+y)=1.3\times {10}^{-9} ..............\left(ii\right) \\ \text{ Dividing equation (i) and (ii) we get } \\ \frac{x}{y}=3.6 \\ \therefore x=3.6y \\ \text{ Putting this value in equation (ii), we get } \\ y(3.6y+y)=1.3\times {10}^{-9} \\ \text{ On solving, we get }y=1.6\times {10}^{-5} \\ \text{ and }x=3.6\times 1.6\times {10}^{-5}=5.76\times {10}^{-5} \\ \therefore \left[{\mathrm{Ca}}^{2+}\right]=(x+y)=\left(1.6\times {10}^{-5}\right)+\left(5.76\times {10}^{-5}\right) \\ \therefore \left[{\mathrm{Ca}}^{2+}\right]=7.46\times {10}^{-5}\mathrm{M} \end{gathered}$$