Question

$Ca{CO}_3$ and $Ba{CO}_3$ have solubility product values $1\times {10}^{-8}$ and $5\times {10}^{-9}$ respectively. If water is shaken up with both solids till equilibrium is reached, the concentration of ${CO}_3^{2-}$ ions is:

• A. $1.5\times {10}^{-8}$
• B. $1.225\times {10}^{-4}$
• C. $2.25\times {10}^{-9}$
• D. none of these

Answer 1

The correct option is B $1.225\times {10}^{-4}$

$\begin{array}{c}\text{Since, Ksp of}{CaCO}_3\text{and}BaC{O}_3\text{are}\\ \text{very close.}so\text{, concentration of any}\\ \text{species cannot be neglected.}\\ \text{Let the solubility of}{CaCO}_3\text{and}{BaCO}_3\\ \text{are}x\text{and}y\text{M}\end{array}$

$\therefore {CaCO}_3\rightleftharpoons C{a}^{2+}\left(x\right)+C{O}_3^{2-}\left(x\right)$

${BaCO}_3\rightleftharpoons B{a}^{2+}\left(y\right)+C{O}_3^{2-}\left(y\right)$

$\text{Total}\left[C{O}_3^{2-}\right]=(x+y)$

$\begin{array}{cc}\therefore k_{sp}\text{of}CaC{O}_3& =\left[C{a}^{2+}\right]\left[C{O}_3^{2-}\right]\\ & =x(x+y)={10}^{-8}-\left(i\right)\end{array}$

$\begin{array}{cc}{k}_{sp}\text{of}BaC{O}_3& =\left[B{a}^{2+}\right]\left[C{O}_3^{2-}\right]\\ & =y(x+y)=5\times {10}^{-9}\end{array}$

$\begin{array}{c}\frac{y(y+x)}{x(x+y)}=\frac{5\times {10}^{-9}}{{10}^{-8}}\\ \therefore y=0.5x\\ \text{substitute the value of}y\text{in}\left(i\right)\\ \mathrm{and}\left(ii\right):x=0.8\times {10}^{-4}\end{array}$

$\begin{array}{c}\therefore y=0.5x=0.4\times {10}^{-4}\\ \left[{C{O}_3}^{2-}\right]=(x+y)=1.2\times {10}^{-4}\end{array}$