Question

If $K_{sp}$ of $CaF_2$ at $25^°C$ is $1.7 \times 10^{-10}$, the combination amongst the following which gives a precipitate of $CaF_2$ is

• A. $1 \times 10^{-2} M \,Ca^{2+}$ and $1 \times 10^{-3} MF^-$
• B. $1 \times 10^{-4} M \,Ca^{2+}$ and $1 \times 10^{-4} MF^-$
• C. $1 \times 10^{-2} M \,Ca^{2+}$ and $1 \times 10^{-5} MF^-$
• D. $1 \times 10^{-3} M \,Ca^{2+}$ and $1 \times 10^{-5} MF^-$

Answer 1

Answer: (A)

When ionic product i.e. the product of the concentration of ions in the solution exceeds the value of solubility product, formation of precitpiate occurs.
$CaF_2 \rightleftharpoons Ca^{2+} + 2F^-$
Ionic product $= [Ca^{2+}] [F^-]^2$
when, $[Ca^{2+}] = 1 \times 10 ^{-2} \,M$
$[F^-]^2 = (1 \times 10^{-3})^2\,M$
$=1 \times 10^{-6}\,M$
$\therefore \left[Ca^{2+}\right] \left[F^{-}\right]^{2}= \left(1\times10^{-2}\right) \left(1\times 10^{-6}\right) = 1\times 10^{-8}$
In this case,
Ionic product $\left(1\times 10^{8}\right) >$ solubility product $\left(1.7\times 10^{-10}\right)$
$\therefore $ Hence (a) is correct option.