Solid $\mathrm{BaF}_2$ is added to a solution containing 0.1 mole of $\mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4$ solution $(1 \mathrm{~L})$ until equilibrium is reached. If the $K_{\mathrm{sp}}$ of $\mathrm{BaF}_2$ and $\mathrm{BaC}_2 \mathrm{O}_4$ is $10^{-6}$ $\mathrm{mol}^3 \mathrm{~L}^{-3}$ and $10^{-10} \mathrm{~mol}^2 \mathrm{~L}^{-2},$ respectively, then find the equilibrium concentration of $\mathrm{Ba}^{2+}$ in the solution. Assume that addition of $\mathrm{BaF}_2$ does not cause any change in volume.

$2.5 \times 10^{-6} \mathrm{M}$

JEE CHEMISTRY MCQ: Equilibrium

1 Medium

1 of 1

Analyzing +4 | -1

Solid

\mathrm{BaF}_2

is added to a solution containing 0.1 mole of

\mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4

solution

(1 \mathrm{~L})

until equilibrium is reached. If the

K_{\mathrm{sp}}

\mathrm{BaF}_2

and

\mathrm{BaC}_2 \mathrm{O}_4

10^{-6}

\mathrm{mol}^3 \mathrm{~L}^{-3}

and

10^{-10} \mathrm{~mol}^2 \mathrm{~L}^{-2},

respectively, then find the equilibrium concentration of

\mathrm{Ba}^{2+}

in the solution. Assume that addition of

\mathrm{BaF}_2

does not cause any change in volume.

View answer and explanation

Correct option: C

2.5 \times 10^{-5} \mathrm{M}

Option C is correct because the equilibrium concentration of

\mathrm{Ba}^{2+}

is calculated using the solubility product of

\mathrm{BaF}_2

and the fluoride ion concentration derived from the reaction with oxalate.