Q65 / 180 · NEET 2013 (Karnataka) · Chemistry
The value of $\mathrm{K}_{s p}$ of $\mathrm{CaCO}_3$ and $\mathrm{CaC}_2 \mathrm{O}_4$ are $4.7 \times 10^{-9}$ and $1.3 \times 10^{-9}$ 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?
- A$5.831 \times 10^{-5} \mathrm{M}$
- B$6.856 \times 10^{-5} \mathrm{M}$
- C$3.606 \times 10^{-5} \mathrm{M}$
- Correct answerD$7.746 \times 10^{-5} \mathrm{M}$
Solution
$\begin{aligned}
\mathrm{CaCO}_3 & \rightarrow \mathrm{Ca}^{2+}+\mathrm{CO}_3{ }^{2-} \\
\mathrm{CaC}_2 \mathrm{O}_4 & \rightarrow \mathrm{Ca}^{2+}+\mathrm{C}_2 \mathrm{O}_4^{2-} \\
y & \\
{\left[\mathrm{Ca}^{2+}\right] } & =x+y \\
x(x+y) & =4.7 \times 10^{-9} \\
y(x+y) & =1.3 \times 10^{-9}
\end{aligned}$
$\begin{aligned}
& \frac{x}{y}=3.6 \\
& \therefore \quad x=3.6 y \\
& y(3.6 y+y)=1.3 \times 10^{-9} \\
& y=1.6 \times 10^{-5} \\
& {\left[\mathrm{Ca}^{2+}\right] }=(x+y)=\left(1.6 \times 10^{-5}\right) \\
&+\left(5.76 \times 10^{-5}\right) \\
&=7.46 \times 10^{-5} \mathrm{M}
\end{aligned}$
Hence, the correct answer is option: 'D'.