Question

Solid $Ba{F}_2$ is added to a solution that has 0.3 mole of a sodium oxalate solution (1 litre) until equilibrium is reached. The $K_{sp}$ of $Ba{F}_2$ and $Ba{C}_2{O}_4\left(s\right)$ is ${10}^{-6}$ and $2\times {10}^{-7}$ respectively. Assume that the addition of $Ba{F}_2$ does not cause any change in the volume and causes no hydrolysis of any of the cations or anions. If the concentration of $B{a}^{2+}$ ions in the resulting solution at equilibrium is represented as $y\times {10}^{-x}$, then $(x+y)$ is:

Answer 1

$Ba{F}_2\left(s\right)toB{a}^{2+}(aq.)+2F^{-}\left(aq\right)$ Let $S$ be the solubility of $Ba{F}_2$ , but $B{a}^{2+}$ reacts with $C_2{O}_4^{2-}$ present in this solution & almost completely converts into $Ba{C}_2{O}_4$ $B{a}^{2+}+C_2{O}_4^{2-}toBa{C}_2{O}_4\left(s\right)$
Let y mole per litre of $B{a}^{2+}$ be left after reaching equilibrium
So, $\underset{y}{B{a}^{2+}}+\underset{0.3-s}{C_2{O}_4^{2-}}\to Ba{C}_2{O}_4\left(s\right)$ $K=2\times {10}^{-7}$ So, we get the following two equations,
$y(0.3-s)=2\times {10}^{-7}$ & $y(2s{)}^2={10}^{-6}=(K_{sp}{)}_{Ba{F}_2}$
Solving these two equations, we get, $s=0.25M$
So , $\left[C_2{O}_4^{2-}\right)=0.3-0.25=0.05M$ $\left[F^{-}\right]=2s=2\times 0.25=0.5M$ $\left[B{a}^{2+}\right]=y=4\times {10}^{-6}M$