Given Na2CO3=1.0×10−4M $\begin{aligned}
& \therefore \quad\left[\mathrm{CO}_3^{--}\right]=1.0 \times 10^{-4} \mathrm{M} \
& \text { i.e. } \quad \mathrm{s}=1.0 \times 10^{-4} \mathrm{M}
\end{aligned}Atequilibrium\begin{aligned}
& {\left[\mathrm{Ba}^{++}\right]\left[\mathrm{CO}3{ }^{--}\right]=\mathrm{K}{\mathrm{sp}} \text { of } \mathrm{BaCO}3} \
& {\left[\mathrm{Ba}^{++}\right]=\frac{\mathrm{K}{\mathrm{sp}}}{\left[\mathrm{CO}_3{ }^{--}\right]}=\frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}}}
\end{aligned}=5.1 \times 10^{-5} \mathrm{M}$