Question

How many times solubility of $Ca{F}_2$ is decreased in $4\times {10}^{-3}$ M $KF$(aq.) solution as compare to pure water at ${25}^o$C?

Given: $K_{sp}\left(Ca{F}_2\right)=3.2\times {10}^{-11}$

• A. $50$
• B. $100$
• C. $500$
• D. $1000$

Answer 1

The correct option is B $100$

Let solubility of $Ca{F}_2$ be $s$ moles/lit.

$Ca{F}_2\rightleftharpoons \underset{s}{C{a}^{2+}}+\underset{2s}{2F^{-}}$

$K_{sp}$ of $Ca{F}_2=\left[C{a}^{2+}\right][F^{-}{]}^2=s\times (2s{)}^2=4s^3$

$\therefore 4s^3=3.2\times {10}^{-11}$

$\Rightarrow s=2\times {10}^{-4}$

In presence of $4\times {10}^{-3}M$ $KF\left(aq\right)$ solubility is $s^{\prime }$

$\Rightarrow s^{\prime }\times (4\times {10}^{-3}+2s^{\prime }{)}^2=3.2\times {10}^{-11}$

$\Rightarrow s^{\prime }\times 16\times {10}^{-6}=3.2\times {10}^{-11}$ $[4\times {10}^{-3}+2s^{\prime }\approx 4\times {10}^{-3}]$

$\Rightarrow s^{\prime }=2\times {10}^{-6}$

Solubility decreases= $\frac{s}{s^{\prime }}=\frac{2\times {10}^{-4}}{2\times {10}^{-6}}=100$ times as compared to pure water.