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. $55\times {10}^{-10}$
• C. $2.5\times {10}^{-10}$
• D. $4.9\times {10}^{-10}$

Answer 1

The correct option is D $4.9\times {10}^{-10}$

Molecular weight of $CaC{O}_3=100gmo{l}^{-1}$

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

$Q_{sp}$ 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. So, let solution contains $xM$ 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 $BaC{O}_3=\left[B{a}^{2+}\right]\left[C{O}_3^{2-}\right]$

$=\frac{x\times 10}{100}\times \frac{49\times {10}^{-10}}{x}=4.9\times {10}^{-10}$