Solid Ba (NO_3)_2 is gradually dissolved in a 1.0 × 10^ -4 ~M ~Na_2...

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$$\begin{aligned} & {\left[\mathrm{Ba}^{++}\right]\left[\mathrm{CO}_3{ }^{--}\right]=\mathrm{K}_{\mathrm{sp}} \text{\text{ of } } \mathrm{BaCO}_3} \\ & {\left[\mathrm{Ba}^{++}\right]=\frac{\mathrm{K}_{\mathrm{sp}}}{\left[\mathrm{CO}_3{ }^{--}\right]}=\frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}}} \end{aligned}$$=5.1 \times 10^{-5} \mathrm{M}$