Question

What is the maximum concentration of equimolar solutions of $FeS{O}_4$ and $N{a}_{2}S$ so that when mixed in equal volumes, there is no precipitation of FeS?

[$K_{sp}$ of FeS $=6.25\times {10}^{-18}$]

• A. $5\times {10}^{-9}$ M
• B. $5\times {10}^{-8}$ M
• C. $5\times {10}^{-18}$ M
• D. $5\times {10}^{-19}$ M

Answer 1

Answer: (A)

$5\times {10}^{-9}$ M

Let us assume that the maximum concentration of each solution is $xmol{L}^{-1}$

Therefore,

$\left[FeS{O}_4\right]=\left[N{a}_{2}S\right]=\frac{x}{2}$

Therefore,

$\left[F{e}^{2+}\right]=\left[FeS{O}_4\right]=\frac{x}{2}$

we also known from the reaction that ,

$\left[S^{2-}\right]=\left[N{a}_{2}S\right]=\frac{x}{2}$

Hence,

$FeS\to F{e}^{2+}+S^{2-}$

$K_p=\left[F{e}^{2+}\right]\left[S^{2-}\right]$

Therefore,

$6.3\times {10}^{-18}=\frac{x}{2}\times \frac{x}{2}=\frac{x^2}{4}$

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