JEE Main 2013 — Ionic Equilibrium Question with Solution

Given ${\mathrm{Na}}_2{\mathrm{CO}}_3=1.0\times 10^{-4}\mathrm{M}$ $\begin{aligned} & \therefore \quad\left[\mathrm{CO}_3^{--}\right]=1.0 \times 10^{-4} \mathrm{M} \\ & \text { i.e. } \quad \mathrm{s}=1.0 \times 10^{-4} \mathrm{M} \end{aligned}$ At equilibrium $\begin{aligned} & {\left[\mathrm{Ba}^{++}\right]\left[\mathrm{CO}_3{ }^{--}\right]=\mathrm{K}_{\mathrm{sp}} \text { of } \mathrm{BaCO}_3} \\ & {\left[\mathrm{Ba}^{++}\right]=\frac{\mathrm{K}_{\mathrm{sp}}}{\left[\mathrm{CO}_3{ }^{--}\right]}=\frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}}} \end{aligned}$ $=5.1\times 10^{-5}\mathrm{M}$