Question

What is the maximum concentration of an equimolar solution of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is not precipitation of iron sulphide?
(For $FeS,K_{sp}=6.3\times {10}^{-18}$)

Answer 1

${FeSO}_4\rightleftharpoons {Fe}^{2+}+{SO}_4^{2-}$ and ${Na}_{2}S\rightleftharpoons 2{Na}^{+}+S^{2-}$

Now, $FeS\rightleftharpoons \left[{Fe}^{2+}\right]\left[S^{2-}\right]$

$K_{sp}=\left[{Fe}^{2+}\right]\left[S^{2-}\right]=6.3\times {10}^{-18}$

Let say minimum concentration of ${Fe}^{2+}$ is $S$ for which there will be precipitation.

$\Rightarrow S=2.51\times {10}^{-9}$

So in the solution the concentration of ${FeSO}_4$ and ${Na}_{2}S$ is $2.51\times {10}^{-9}M$.

Since equal volume of ${FeSO}_4$ and ${Na}_{2}S$ are mixed. The maximum concentration of both solutions $=2\times 2.51\times {10}^{-9}M=5.02\times {10}^{-9}M$.