Solid $$Ba(NO_3)_2$$ is gradually dissolved in a $$1.0 \times 10^{-4}$$ M $$Na_2CO_3$$ solution. At what concentration of $$Ba^{2+}$$ will a precipitate begin to form ? ($$K_{sp}$$ for $$BaCO_3 = 5.1 \times 10^{-9}$$).

In the heterogeneous equilibrium for the precipitation of barium carbonate, the solubility-product expression is

$$BaCO_3 (s) \rightleftharpoons Ba^{2+} + CO_3^{2-}$$
$$K_{sp} \;=\;[Ba^{2+}]\,[CO_3^{2-}]$$

A 1.0 × 10−4 M solution of $$Na_2CO_3$$ is essentially fully dissociated because sodium salts are highly soluble. Hence, before any $$Ba^{2+}$$ is added, the carbonate-ion concentration is

$$[CO_3^{2-}] = 1.0 \times 10^{-4}\,\text{M}$$

Precipitation of $$BaCO_3$$ will start the moment the ionic product equals $$K_{sp}$$. Thus, setting $$[Ba^{2+}]$$ at the threshold value $$[Ba^{2+}]_{crit}$$,

$$[Ba^{2+}]_{crit}\,[CO_3^{2-}] = K_{sp}$$

$$[Ba^{2+}]_{crit} = \frac{K_{sp}}{[CO_3^{2-}]} = \frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}} = 5.1 \times 10^{-5}\,\text{M}$$

Therefore, a precipitate of $$BaCO_3$$ will begin to form when the barium-ion concentration reaches

Option B which is: $$5.1 \times 10^{-5}\ \text{M}$$