Q2 / 160 · AP EAMCET 2020 (17 Sep Shift 2) · Chemistry
When equal volumes of \(\mathrm{Ca}^{2+}\) and \(\mathrm{F}^{-}\)solutions are mixed, in which of the solutions precipitation will not occur? \(\left(K_{\mathrm{sp}}\right.\) of \(\left.\mathrm{CaF}_2=1.6 \times 10^{-10}\right)\)
- Correct answerA\(10^{2-} \mathrm{MCa}^{2+}+10^{5-} \mathrm{MF}^{-}\)
- B\(10^{3-} \mathrm{MCa}^{2+}+10^{3-} \mathrm{MF}^{-}\)
- C\(10^{4-} \mathrm{M} \mathrm{Ca}^{2+}+10^{2-} \mathrm{M} \mathrm{F}^{-}\)
- D\(10^{2-} \mathrm{MCa}^{2+}+10^{3-} \mathrm{M} \mathrm{F}^{-}\)
Solution
At a given temperature, when ionic product \(\left(K_{\text {ip }}\right)\) over comes solubility product \(\left(K_{\text {sp }}\right)\), precipitation of the sparingly soluble electrolyte (here, it is \(\mathrm{CaF}_2\) ) takes place.
\(\begin{aligned}
K_{\text {ip }} & =\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{F}^{-}\right]^2=\left(10^{-2}\right)\left(10^{-5}\right)^2=10^{-12} < K_{\text {sp }} \\
& =1.6 \times 10^{-10}
\end{aligned}\)
So, at this condition \(\left(10^{-2} \mathrm{M} \mathrm{Ca}^{2+}+10^{-5} \mathrm{M} \mathrm{F}^{-}\right)\), precipitation of \(\mathrm{CaF}_2\) will not occur.
Hence, the correct answer is option: 'A'.