Question

The solubility product of $Mg{F}_2$ is 7.4 x 10${}^{-11}$. Calculate the solubility of $Mg{F}_2$ in 0.1 M NaF solution.

• A. $7.4\times {10}^{-9}$
• B. $3.7\times {10}^{-9}$
• C. $3.7\times {10}^{-11}$
• D. $7.4\times {10}^{-11}$

Answer 1

Answer: (A)

$7.4\times {10}^{-9}$

Answer:-

Given:-

$K_{SP}=7.4\times {10}^{-11}$

$\left[NaF\right]=0.1M$

Solution:-

$Mg{F}_2\rightleftharpoons {Mg}^{2+}+2F^{-}$

$NaF⟶{Na}^{+}+F^{-}$

$\left[F^{-}\right]=\left[{Na}^{+}\right]=0.1M$

Initially -

$\left[{Mg}^{2+}\right]=0$

$\left[2F^{-}\right]=0.1$

At equillibrium -

$\left[{Mg}^{2+}\right]=0+x$

$\left[2F^{-}\right]=0.1+2x$

According to the definition of the solubility product constant,

$K_{SP}=\left[{Mg}^{2+}\right]{\left[F^{-}\right]}^2$

$7.4\times {10}^{-11}=x{(0.1+2x)}^2$

As $K_{SP}$ is small, therefore

$0.1+2x\approx 0.1$

$\therefore 7.4\times {10}^{-11}=x\times {\left(0.1\right)}^2$

$\Rightarrow x=7.4\times {10}^{-9}$