Solid  ${\text{BaF}}_{\text{2}}$  is added to a solution containing 0.1 mole of sodium (1 litre) until equilibrium is reached. If the K_sp of ${\text{BaF}}_{\text{2}}$   and  $Ba{C}_2{O}_4\left(s\right)$  is ${10}^{-6}\&{10}^{-7}$  respectively. Assume addition of   ${\text{BaF}}_{\text{2}}$ does not cause any change in volume and no hydrolysis of any of the cations or anions. (Given : $\sqrt{116}=10.77$ ) If concentration of  $B{a}^{2+}$  ions in resulting solution at equilibrium is represented as ${\text{P×10}}^{\text{-5}}$ , then value if P is

$Ba{F}_2\left(s\right)\to B{a}^{2+}\left(aq\right)+2F^{-}\left(aq\right)$  Lets ‘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 convert into  $Ba{C}_2{O}_4$

$B{a}^{2+}+C_2{O}_4^{2-}\to Ba{C}_2{O}_4\left(s\right)$   Let y mole per lt. of  $B{a}^{2+}$ is left after reaching equilibrium.

So,  $\underset{y}{B{a}^{2+}}+\underset{0.1-s}{C_2{O}_4^{2-}}\to Ba{C}_2{O}_4\left(s\right)Keq={10}^{-7}$  So, we get the following two equations,

$y(0.1-s)={10}^{-7}\&y{\left(2s\right)}^2={10}^{-6}=\left(K_{sp}\right)Ba{F}_2$  Solving there two equations,
We get,  $S=0.096M$   $\left[C_2{O}_4^{2-}\right)=0.1-0.096=4\times {10}^{-3}M$

$$\begin{gathered} \left[F^{-}\right]=2s=2\times 0.096=0.192m \\ \left[B{a}^{2+}\right]=y=2.7\times {10}^{-5}M \end{gathered}$$