Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\begin{array}{cccccccc}& {\text{Ag}}^{+}\left(\text{aq}\right)& +& {\text{Fe}}^{2+}\left(\text{aq}\right)& \rightleftharpoons & {\text{Fe}}^{3+}\left(\text{aq}\right)& +& \text{Ag}\left(\text{s}\right)\\ \text{Initial}& 75\text{ mM}& & 545\text{ mM}& & 0& & 0\\ \text{Reaction}& -46\text{ mM}& & -46\text{ mM}& & +46\text{ mM}& & +46\text{ mM}\\ & \text{---------------}& & \text{---------------}& & \text{---------------}& & \text{---------------}\\ \text{Final}& 29\text{ mM}& & 499\text{ mM}& & 46\text{ mM}& & 46\text{ mM}\end{array}$

m.eqs of${\text{KMnO}}_4=30\times 0\mathrm{·}0832\times 5=12\mathrm{·}48\text{ m.eqs}$
(Z factor =5)

m.eqs of${\text{Fe}}^{2+}$ = 12·48 Meqs = 12·48 mM
(Z factor= 1)

Total mM of${\text{Fe}}^{2+}$left$=12\mathrm{·}48\times 40=499\text{ mM}$

At Equilibrium concentrations

$\left[{\text{Ag}}^{+}\right]=\frac{29\text{ mM}}{1000\text{ mL}}=0\mathrm{·}029\text{ M}$

$\left[{\text{Fe}}^{2+}\right]=\frac{499\text{ mM}}{1000\text{ mL}}=0\mathrm{·}499\text{ M}$

$\left[{\text{Fe}}^{3+}\right]=\frac{46\text{ mM}}{1000\text{ mL}}=0\mathrm{·}046\text{ M}$

${\text{K}}_{\text{C}}=\frac{\left[{\text{Fe}}^{3+}\right]}{\left[{\text{Ag}}^{+}\right]\left[{\text{Fe}}^{2+}\right]}=\frac{0·046}{0·499\times 0·029}=3.18{\text{M}}^{-1}$