Question

For the reaction,
$2{Fe}^{3+}\left(aq\right)+{\left({Hg}_2\right)}^{2+}\left(aq\right)\rightleftharpoons 2{Fe}^{2+}\left(aq\right)+2{Hg}^{2+}\left(aq\right)$

$K_c=9.14\times {10}^{-6}$ at $25$. If the initial concentration of the ions are $\left[{Fe}^{3+}\right]=0.5M,\left[{Hg}_2^{2+}\right]=0.5M,\left[{Fe}^{2+}\right]=0.03M$ and $\left[{Hg}^{2+}\right]=0.03M$. What will be the concentrations of ions at equilibrium?

• A. $\left[F{e}^{3+}\right]=0.2986M$, $\left[H{g}_2^{2+}\right]=0.2988M$, $\left[F{e}^{2+}\right]=0.0268M$, $\left[H{g}^{2+}\right]=0.0327M$
• B. $\left[F{e}^{3+}\right]=0.2513M$, $\left[H{g}_2^{2+}\right]=0.2506M$, $\left[F{e}^{2+}\right]=0.0131M$, $\left[H{g}^{2+}\right]=0.0131M$
• C. $\left[F{e}^{3+}\right]=0.4973M$, $\left[H{g}_2^{2+}\right]=0.4987M$, $\left[F{e}^{2+}\right]=0.0327M$, $\left[H{g}^{2+}\right]=0.0327M$
• D. $\left[F{e}^{3+}\right]=0.5027M$, $\left[H{g}_2^{2+}\right]=0.50135M$,$\left[F{e}^{2+}\right]=0.0273M$, $\left[H{g}^{2+}\right]=0.0273M$

Answer 1

Answer: (C)

$\left[F{e}^{3+}\right]=0.4973M$, $\left[H{g}_2^{2+}\right]=0.4987M$, $\left[F{e}^{2+}\right]=0.0327M$, $\left[H{g}^{2+}\right]=0.0327M$

$2{Fe}^{3+}\left(aq\right)+{\left({Hg}_2\right)}^{2+}\left(aq\right)\rightleftharpoons 2{Fe}^{2+}\left(aq\right)+2{Hg}^{2+}\left(aq\right)$

Concentration before reaction respeactively : $0.50.50.030.03$

Concentration after reaction at equilibrium: $(0.5-a)(0.5-a/2)(0.03+a)(0.03+a)$

$K_c=9.14\times {10}^{-6}=\frac{{\left[{Fe}^{2+}\right]}^2{\left[{Hg}^{2+}\right]}^2}{{\left[{Fe}^{3+}\right]}^2\left[{\left({Hg}_2\right)}^{2+}\right]}$

$\therefore 9.14\times {10}^{-6}=\frac{{(0.03+a)}^2{(0.03+a)}^2}{{(0.5-a)}^2(0.5-a/2)}$

$\therefore a=0.0027$

$\therefore \left[{Fe}^{3+}\right]=0.5-0.0027=0.4973M$

$\left[{Hg}_2^{2+}\right]=0.5-\frac{0.0027}{2}=0.4987M$

$\left[{Fe}^{2+}\right]=0.03+0.0027=0.0327M$

$\left[{Hg}^{2+}\right]=0.03+0.0027=0.0327M$