Question
Solid BaF2 is added to a solution containing 0.1 mole of Na2C2O4 solution (1 L) until equilibrium is reached. If the Ksp of BaF2 and BaC2O4 is 10−6 mol3 L–3 and 10–10 mol2 L–2, respectively, then find the equilibrium concentration of Ba2+ in the solution. Assume that addition of BaF2 does not cause any change in volume.
Options
Solution
$BaF_{2}(s) + \underset{(0.1 - x)\text{ M}}{C_{2}O_{4}^{2 -}}(aq) \Rightarrow BaC_{2}O_{4}(s) + \underset{2x\text{ M}}{2F^{-}}(aq)$
$K_{eq} = \frac{\left\lbrack F^{-} \right\rbrack}{\left\lbrack C_{2}O_{4}^{2 -} \right\rbrack} = \frac{K_{sp}\left( BrF_{2} \right)}{K_{sp}\left( BrC_{2}O_{4} \right)} = \frac{10^{- 6}}{10^{- 10}} = 10^{4} $$${\therefore x \approx 0.1 \Rightarrow \left\lbrack F^{-} \right\rbrack = 0.2\text{ M} }{\therefore\left\lbrack Ba^{2 +} \right\rbrack = \frac{10^{- 6}}{(0.2)^{2}} = 2.5 \times 10^{- 5}\text{ M}}$$
Create a free account to view solution
View Solution Free