## Chemistry: The Molecular Science (5th Edition)

The molar solubility of $ZnCO_3$ in this solution is equal to $2.8 \times 10^{-10}M$.
$0.050M [K_2CO_3] = 0.050M [C{O_3}^{2-}]$ 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: $[C{O_3}^{2-}] = 0.05$ $1.4 \times 10^{-11}= 0.05 \times ( 1S)^ 1$ $\frac{1.4 \times 10^{-11}}{0.05} = ( 1S)^ 1$ $2.8 \times 10^{-10} = S$