Question

A swimming pool was sufficiently alkaline so that CO${}_2$ absorbed from the air produced in the pool a solution which $2\times {10}^{-4}M$ in $C{O}_3^{2-}M$. If the pool water was originally $4\times {10}^{-3}M$ in $M{g}^{2+},6\times {10}^{-4}M$ in $C{a}^{2+}$ and $8\times {10}^{-7}M$ in $F{e}^{2+}$, then a precipitate should form of:
$(K_{sp}$ for $CaC{O}_3,MgC{O}_3,FeC{O}_3$ are $2.8\times {10}^{-9},3.5\times {10}^{-8},3,2\times {10}^{-11}$ respectively$)$

• A. only $MgC{O}_3$
• B. only $CaC{O}_3$
• C. only $FeC{O}_3$
• D. only $MgC{O}_3$ and $FeC{O}_3$
• E. $MgC{O}_3,CaC{O}_3$ and $FeC{O}_3$

Answer 1

The correct option is D only $MgC{O}_3$ and $FeC{O}_3$

From given values of concentations, the ionic product is calculated and compared with solubility product values.
For $CaC{O}_3,$, the ionic product is $\left[C{a}^{2+}\right]\left[C{O}_3^{2-}\right]=6\times {10}^{-4}\times 2\times {10}^{-4}=1.2\times {10}^{-9}$.
The ionic product is less than the solubility product which is $2.8\times {10}^{-9}$. Hence, $CaC{O}_3,$ will not precipitate.
For $MgC{O}_3$, the ionic product is $\left[M{g}^{2+}\right]\left[C{O}_3^{2-}\right]=4\times {10}^{-3}\times 2\times {10}^{-4}=8\times {10}^{-7}$.
The ionic product is more than the solubility product which is $3.5\times {10}^{-8}$. Hence, $MgC{O}_3$ will precipitate.
For $FeC{O}_3$, the ionic product is $\left[F{e}^{2+}\right]\left[C{O}_3^{2-}\right]=8\times {10}^{-7}\times 2\times {10}^{-4}=1.6\times {10}^{-12}$.
The ionic product is more than the solubility product which is $3.2\times {10}^{-11}$. Hence, $FeC{O}_3$ will precipitate.