The values of $K_{s p}$ of $\mathrm{CaCO}_{3}$ and $\mathrm{CaC}_{2} \mathrm{O}_{4}$ are $4.7 imes

Question

The values of $K_{s p}$ of $\mathrm{CaCO}_{3}$ and $\mathrm{CaC}_{2} \mathrm{O}_{4}$ are $4.7 	imes 10^{-9}$ and $1.3 	imes 10^{-9}$ respectively at $25^{\circ} \mathrm{C}$. If the mixture of these two is washed with water, what is the concentration of $\mathrm{Ca}^{2+}$ ions in water?\

Professor PC Thomas Classes · Answer

(d) : $\mathrm{CaCO}_{3} ightarrow \mathrm{Ca}^{2+}+\mathrm{CO}_{3}^{2-}$ $$\begin{equation*} \mathrm{CaC}_{2} \mathrm{O}_{4} ightarrow \underset{y}{\mathrm{Ca}^{2+}}+\underset{y}{x}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} 	ag{ii} \end{equation*}$$ Now, $\left[\mathrm{Ca}^{2+}ight]=x+y$\ and $x(x+y)=4.7 	imes 10^{-9} \quad$...(i) $\quad y(x+y)=1.3 	imes 10^{-9}$\ Dividing equation (i) and (ii) we get\ $\frac{x}{y}=3.6 \quad 	herefore x=3.6 y$\ Putting this value in equation (ii), we get $$ y(3.6 y+y)=1.3 	imes 10^{-9} $$ On solving, we get $y=1.68 	imes 10^{-5}$\ and $x=3.6 	imes 1.68 	imes 10^{-5}=6.05 	imes 10^{-5}$\ $	herefore \quad\left[\mathrm{Ca}^{2+}ight]=(x+y)=\left(1.68 	imes 10^{-5}ight)+\left(6.05 	imes 10^{-5}ight)$\ $	herefore \quad\left[\mathrm{Ca}^{2+}ight]=7.73 	imes 10^{-5} \mathrm{M}$\