Answer
The molar solubility of $ZnCO_3$ in this solution is equal to $2.8 \times 10^{-10}M$.
Work Step by Step
$0.050M [K_2CO_3] = 0.050M [C{O_3}^{2-}]$
1. Write the $K_{sp}$ expression:
$ ZnCO_3(s) \lt -- \gt 1C{O_3}^{2-}(aq) + 1Zn^{2+}(aq)$
$1.4 \times 10^{-11} = [C{O_3}^{2-}]^ 1[Zn^{2+}]^ 1$
$1.4 \times 10^{-11} = (0.05 + S)^ 1( 1S)^ 1$
2. Find the molar solubility.
Since 'S' has a very small value, we can approximate: $[C{O_3}^{2-}] = 0.05$
$1.4 \times 10^{-11}= 0.05 \times ( 1S)^ 1$
$ \frac{1.4 \times 10^{-11}}{0.05} = ( 1S)^ 1$
$2.8 \times 10^{-10} = S$