Question

The solubility of $Ca{CO}_3$ is $7$ mg/litre. Calculate the solubility product of $Ba{CO}_3$ from this information and from the fact that when ${Na}_2{CO}_3$ is added slowly to a sloution containing equimolar concentration of ${Ca}^{2+}$ and ${Ba}^{2+}$, no precipitate is formed until $90$% of ${Ba}^{2+}$ has been precipitated as $Ba{CO}_3$. The solubility product of $Ba{CO}_3$ is

• A. $3\times {10}^{-10}$
• B. $4\times {10}^{-10}$
• C. $5\times {10}^{-10}$
• D. $6\times {10}^{-10}$

Answer 1

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

$K_{sp}$ of $Ca{CO}_3={\left(\frac{7\times {10}^{-3}}{100}\right)}^2=49\times {10}^{-10}$

When only ${Ba}^{+}$ is $90$% precipitated then only $Ca{CO}_3$ starts precipitation then if solution contains $a$ mol ${litre}^{-1}$ of ${Ca}^{2+}$ amd ${Ba}^{2+}$ each

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

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

Now for $Ba{CO}_3$: $K_{sp}=\left[{Ba}^{2+}\right]\left[C{O}_3^{2-}\right]=\frac{a\times 10}{100}\times \frac{49\times {10}^{-10}}{a}$

$=4.9\times {10}^{-10}$ ${mol}^2{litre}^{-2}$