Question

If $K_{sp}$ of $Ca{F}_2$ at $25^{°}C$ is $1.7\times 10^{-10}$, the combination amongst the following which gives a precipitate of $Ca{F}_2$ is

• A. $1\times 10^{-2}MC{a}^{2+}$ and $1\times 10^{-3}M{F}^{-}$
• B. $1\times 10^{-4}MC{a}^{2+}$ and $1\times 10^{-4}M{F}^{-}$
• C. $1\times 10^{-2}MC{a}^{2+}$ and $1\times 10^{-5}M{F}^{-}$
• D. $1\times 10^{-3}MC{a}^{2+}$ and $1\times 10^{-5}M{F}^{-}$

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.
$Ca{F}_2\rightleftharpoons C{a}^{2+}+2F^{-}$
Ionic product $=[C{a}^{2+}][F^{-}{]}^2$
when, $[C{a}^{2+}]=1\times 10^{-2}M$
$[F^{-}{]}^2=(1\times 10^{-3}{)}^{2}M$
$=1\times 10^{-6}M$
$\therefore \left[C{a}^{2+}\right]{\left[F^{-}\right]}^2=\left(1\times 10^{-2}\right)\left(1\times 10^{-6}\right)=1\times 10^{-8}$
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.