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The values of $K_{sp}$ of $CaCO_3$ and $CaC_2O_4$ are $4.7 \times 10^{-9}$ and $1.3 \times 10^{-9}$ respectively at $25\,^oC.$ If the mixture of these two is washed with water, what is the concentration of $Ca^{2+}$ ions in water ..............$ \times 10^{-5}\, M$?
$5.831$
$6.856$
$3.606$
$7.746$
Solution
$\mathrm{CaCO}_{3} \rightarrow \mathrm{Ca}^{2+}+\mathrm{CO}_{3}^{2-}$
$\quad \quad \quad \quad \quad \quad x \quad \quad x$
$\mathrm{CaC}_{2} \mathrm{O}_{4} \rightarrow \mathrm{Ca}^{2+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$
$\quad \quad \quad \quad \quad \quad y \quad \quad y$
Now, $\left[\mathrm{Ca}^{2+}\right]=x+y$
and $x(x+y)=4.7 \times 10^{-9}$
$y(x+y)=1.3 \times 10^{-9}$
Dividing equation $(i)$ and $(ii)$ we get $\frac{x}{y}=3.6$
$\therefore \quad x=3.6 y$
Putting this value in equation $(ii)$, we get
$y(3.6 y+y)=1.3 \times 10^{-9}$
On solving, we get $y=1.68 \times 10^{-5}$ and $x=3.6 \times 1.68 \times 10^{-5}=6.048 \times 10^{-5}$
$\therefore \quad\left[\mathrm{Ca}^{2+}\right]=(x+y)=\left(1.68 \times 10^{-5}\right)+\left(6.048 \times 10^{-5}\right)$
$\therefore \quad\left[\mathrm{Ca}^{2+}\right]=7.728 \times 10^{-5}\, \mathrm{M}$
