Question

The solubility of $CaC{O}_3$ is $7mg/litre$. Calculate the solubility product of $BaC{O}_3$ from this information and from the fact that when $N{a}_{2}C{O}_3$ is added slowly to a colution containing equimolar concentration of $C{a}^{+2}andB{a}^{+2}$, no precipitate of $CaC{O}_3$ is formed until $90\%$ of $B{a}^{+2}$ has been precipitated as $BaC{O}_3$ :

• A. $4.9\times {10}^{-10}$
• B. $9.8\times {10}^{-10}$
• C. $4.9\times {10}^{-9}$
• D. $1.96\times {10}^{-9}$

Answer 1

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

${\text{Concentration of CaCO}}_3=\left[\begin{array}{c}\frac{7\times {10}^{-3}}{100}\end{array}\right]\text{mole/lit}=7\times {10}^{-6}\text{mole/lit}$

$Ksp$ of $CaC{O}_3$= $49\times {10}^{-10}\text{mole/lit}$

When only $\left[{\text{Ba}}^{2+}\right]$ is $90\%$ precipitated then only ${CaCO}_3$ starts precipitation

$\left[C{a}^{+2}\right]\left[C{O}_3^{-2}\right]=49\times {10}^{-10}\left[C{O}_3^{-2}\right]=\left[\begin{array}{c}\frac{49\times {10}^{-10}}{\text{a}}\end{array}\right]$

Now, for $BaC{O}_3{\text{K}}_{\text{sp}}={\text{[Ba}}^{+2}\left]{\text{[CO}}_3^{-2}\right]=\frac{a\times 10}{100}\times \frac{49\times {10}^{-10}}{\text{a}}=4.9\times {10}^{-10}$