Question

The solubility of $CaC{O}_3$ is 7 mg/L. Calculate the $K_{sp}$ of $BaC{O}_3$ when $N{a}_{2}C{O}_3$ is added slowly a solution containing equimolar concentration of $C{a}^{2+}$ and $B{a}^{2+}$ and no precipitate is formed until 90% of $B{a}^{2+}$ has been precipitated as $BaC{O}_3$.

• A. $1.1\times {10}^{-10}$
• B. $5\times {10}^{-10}$
• C. $1.1\times {10}^{-8}$
• D. $5\times {10}^{-6}$

Answer 1

The correct option is B $5\times {10}^{-10}$

$MwofCaC{O}_3=100gmo{l}^{-1}$.

$\left[CaC{O}_3\right]=\frac{7\times {10}^{-3}}{100\times 1}$

$Q_{sp}$ for $CaC{O}_3={\left(\frac{7\times {10}^{-3}}{100}\right)}^2=49\times {10}^{-10}$

When only $B{a}^{2+}$ is 90% precipitated then only $CaC{O}_3$ starts precipitation. Then, let solution contains x M of $C{a}^{2+}$ and $B{a}^{2+}$ each.

$\left[C{a}^{2+}\right]\left[C{O}_3^{2-}\right]=49\times {10}^{-10}$

$\left[C{O}_3^{2-}\right]=\frac{49\times {10}^{-10}}{x}$

$\left[B{a}^{2+}\right]=\frac{x\times 10}{100}$

(90% $B{a}^{2+}$ precipitated 10% $B{a}^{2+}$ left)

$K_{sp}$ of $BaC{O}_3=\left[B{a}^{2+}\right]\left[C{O}_3^{2-}\right]=\frac{x\times 10}{100}\times \frac{49\times {10}^{-10}}{x}=4.9\times {10}^{-10}\approx 5\times {10}^{-10}$