Question

One litre of saturated solution of $CaC{O}_3$ is evaporated to dryness due to which 7.0 mg of residue is left. The solubility product for $CaC{O}_3$ is:

• A. $4.9\times {10}^{-5}$
• B. $4.9\times {10}^9$
• C. $4.9\times {10}^{-9}$
• D. $4.9\times {10}^{-7}$

Answer 1

The correct option is C $4.9\times {10}^{-9}$

Moles of $CaC{O}_3$ in the residue =$\frac{\text{residue mass}}{\text{molar mass of calcium carbonate}}=\frac{7\times {10}^{-3}}{100}=7\times {10}^{-5}$
Concentration of $CaC{O}_3$ in 1 litre solution =$7\times {10}^{-5}$
$CaC{O}_3\left(s\right)\rightleftharpoons C{a}^2\left(aq\right)+C{O}_3^{2-}\left(aq\right)1001-sss$

$K_{sp}=\left[C{a}^{2+}\right]\left[C{O}_3^{2-}\right]$
$=7\times {10}^{-5}\times 7\times {10}^{-5}=4.9\times {10}^{-9}$