The solubility of is . Calculate the solubility product of from this information and from the fact t — Ionic Equilibrium Chemistry Question
Question
The solubility of $\text{CaCO}_3$ is $7\text{ mg/litre}$. Calculate the solubility product of $\text{BaCO}_3$ from this information and from the fact that when $\text{Na}_2\text{CO}_3$ is added slowly to a solution containing equimolar concentration of $\text{Ca}^{2+}$ and $\text{Ba}^{2+}$, no precipitate of $\text{CaCO}_3$ is formed until 90% of $\text{Ba}^{2+}$ has been precipitated as $\text{BaCO}_3$.
💡 Solution & Explanation
Molar solubility of $\text{CaCO}_3 = \frac{7 \times 10^{-3}\text{ g/L}}{100\text{ g/mol}} = 7 \times 10^{-5}\text{ M}$. $K_{sp}(\text{CaCO}_3) = (7 \times 10^{-5})^2 = 4.9 \times 10^{-9}$. Let initial equimolar concentration be $C$. When 90% of $\text{Ba}^{2+}$ is precipitated, $[\text{Ba}^{2+}] = 0.1C$. At this moment, $\text{CaCO}_3$ just begins precipitating, so $[\text{Ca}^{2+}] = C$. Both share the same $[\text{CO}_3^{2-}]$, thus $\frac{K_{sp}(\text{CaCO}_3)}{[\text{Ca}^{2+}]} = \frac{K_{sp}(\text{BaCO}_3)}{[\text{Ba}^{2+}]}$. Substituting gives $\frac{4.9 \times 10^{-9}}{C} = \frac{K_{sp}(\text{BaCO}_3)}{0.1C}$. Therefore, $K_{sp}(\text{BaCO}_3) = 0.1 \times 4.9 \times 10^{-9} = 4.9 \times 10^{-10}$.