Question

What is the maximum concentration of an equimolar solution of ferrous sulphate$\left(FeS{O}_4\right)$ and sodium sulphide $\left(N{a}_{2}S\right)$ so that when mixed in equal volumes, there is not precipitation of iron sulphide $\left(FeS\right)$?
(For FeS, $K_{sp}=5\times {10}^{-18}$)

• A. $15.5\times {10}^{-6}M$
• B. $4.47\times {10}^{-9}M$
• C. $7.8\times {10}^{-10}M$
• D. $2.21\times {10}^{-9}M$

Answer 1

The correct option is B $4.47\times {10}^{-9}M$

The dissociation reactions are given below :
$FeS{O}_4\to F{e}^{2+}+S{O}_4^{2-}$ and $N{a}_{2}S\to 2N{a}^{+}+S^{2-}$

Let in the solution the concentration of each $FeS{O}_4$​ and $N{a}_{2}S$ is ${}^{\prime }{s}^{\prime }M.$
Since equal volume of $FeS{O}_4$​ and $N{a}_{2}S$ are mixed.
$\left[FeS{O}_4\right]=\left[N{a}_{2}S\right]=\frac{s}{2}M.$
i.e. $\left[F^{2+}\right]=\left[S^{2-}\right]=\frac{s}{2}M.$
After mixing , $FeS$ is formed.

Now, $FeS\rightleftharpoons \left[F{e}^{2+}\right]\left[S^{2-}\right]$
$K_{sp}=\left[F{e}^{2+}\right]\left[S^{2-}\right]=5\times {10}^{-18}\frac{s^2}{4}=5\times {10}^{-18}{s}^2=20\times {10}^{-18}s=4.47\times {10}^{-9}M$

Hence the highest molarity for the solution is ${}^{\prime }{s}^{\prime }$ for which there will be precipitation.
$\Rightarrow s=4.47\times {10}^{-9}M$