Question

What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide?
(For iron sulphide, $K_{sp}=6.3\times {10}^{-18}$).

Answer 1

Let the maximum concentration of each solution be x mol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., x/2 .

$\therefore \left[FeS{O}_4\right]=\left[N{a}_{2}S\right]=\frac{x}{2}M$
Then, $\left[F{e}^{2+}\right]=\left[FeS{O}_4\right]=\frac{x}{2}M$
Also, $\left[S^{2-}\right]=\left[N{a}_{s}S\right]=\frac{x}{2}MFe(S{)}_{\left(x\right)}\leftrightarrow F{e}_{\left(aq\right)}^{2+}+S_{\left(aq\right)}^{2-}{K}_{sp}=[F{e}^{2+}\left]\right[S^{2-}]6.3\times {10}^{-18}=\left(\frac{x}{2}\right)\left(\frac{x}{2}\right)\frac{x^2}{4}=6.3\times {10}^{-18}\Rightarrow x=5.02\times {10}^{-9}$
If the concentrations of both solutions are equal to or less than $5.02\times {10}^{–9}M$, then there will be no precipitation of iron sulphide.

$\Rightarrow$ $x=5.02\times {10}^{-9}$