Question

For the reaction:
$2{Fe}^{3+}\left(aq\right)+{{(Hg}_2)}^{2+}\left(aq\right)\rightleftharpoons 2{Fe}^{2+}\left(aq\right)+{2Hg}^{2+}\left(aq\right)$
$K_C=9.14\times {10}^{-6}$ at 25C. If the initial concentration of the ions are ${Fe}^{3+}=0.5M$, ${Hg}_2^{2+}=0.5M$, ${Fe}^{2+}=0.03M$ and ${Hg}^{2+}=0.03M$ , what will be the concentration of ions at equilibrium?

• A. ${Fe}^{3+}=0.4973M$,${Hg}_2^{2+}=0.4987M$, ${Fe}^{2+}=0.0327M$, ${Hg}^{2+}=0.0327M$
• B. ${Fe}^{3+}=0.0327M$,${Hg}_2^{2+}=0.0327M$, ${Fe}^{2+}=0.4973M$, ${Hg}^{2+}=0.4973M$
• C. ${Fe}^{3+}=0.3973M$,${Hg}_2^{2+}=0.3987M$, ${Fe}^{2+}=0.1327M$, ${Hg}^{2+}=0.1327M$
• D. None of these

Answer 1

The correct option is A ${Fe}^{3+}=0.4973M$,${Hg}_2^{2+}=0.4987M$, ${Fe}^{2+}=0.0327M$, ${Hg}^{2+}=0.0327M$

The initial concentrations of $F{e}^{3+},H{g}_2^{2+},F{e}^{2+}andH{g}^{2+}$ are $0.5M,0.5M,0.03Mand0.03M$ respectively.
The equilibrium concentrations of $F{e}^{3+},H{g}_2^{2+},F{e}^{2+}andH{g}^{2+}$ are $0.5-2xM,0.5-xM,0.03+2xMand0.03+2xM$ respectively.
The equilibrium constant expression is
$K_c=\frac{\left[F{e}^{2+}{]}^2\right[H{g}^{2+}{]}^2}{\left[F{e}^{3+}{]}^2\right[H{g}_2^{2+}]}$
Substitute values in the above expression.
$9.14\times {10}^{-6}=\frac{(0.03+2x{)}^2(0.03+2x{)}^2}{(0.5-2x{)}^2(0.5-x)}$
$x=0.0013$
Hence, the equilibrium concentrations are as shown below.
$\left[F{e}^{3+}\right]=0.5-2x=0.5-2\left(0.0013\right)=0.4973$ M.
$\left[H{g}_2^{2+}\right]=0.5-x=0.5-0.0013=0.4987$ M.
$\left[F{e}^{2+}\right]=0.03+2x=0.03+2\left(0.0013\right)=0.0327$ M.
$\left[H{g}^{2+}\right]=0.03+2x=0.03+2\left(0.0013\right)=0.0327$ M.