Question

To $500mL$ of $0.150MAgN{O}_3$ solution were added $500mL$ of $1.09M{Fe}^{2+}$ solution and the reaction is allowed to reach an equilibrium at $25$.
${Ag}^{+}\left(aq\right)+{Fe}^{2+}\left(aq\right)\rightleftharpoons {Fe}^{3+}\left(aq\right)+Ag\left(s\right)$
For $25mL$ of the solution, $30mL$ of $0.0832MKMn{O}_4$ were required for oxidation. The equilibrium constant for the reaction at $25$${}^{o}C$ is_________.

Answer 1

${Ag}^{+}\left(aq\right)+{Fe}^{2+}\left(aq\right)\rightleftharpoons {Fe}^{3+}\left(aq\right)+Ag\left(s\right)$
Millimole before reaction $500\times 0.150500\times 1.09$
or $=75=54500$
Millimole after reaction $(75-x)(545-x)xx$
$\because mm=$Meq. (both ${Ag}^{+}/Ag$ and ${Fe}^{2+}/{Fe}^{3+}$ have valency factor unity)
$\because K_c=\frac{\left[{Fe}^{3+}\right]}{\left[{Ag}^{+}\right]\left[{Fe}^{2+}\right]}$
$\because$ concentration $=\frac{Millimole}{Totalvolume}$
$\left[{Ag}^{+}\right]=\frac{75-x}{1000}$
$\left[{Fe}^{2+}\right]=\frac{545-x}{100}$
$\left[{Fe}^{3+}\right]=\frac{x}{1000}$
$\therefore K_c=\frac{\frac{x}{1000}}{\left(\frac{75-x}{1000}\right)\left(\frac{545-x}{1000}\right)}$ ...(i)
Now, $25mL$ of mixture requires $30mL$ of $0.0832M$ or $0.0832\times 5NKMn{O}_4$.
$\because {Fe}^{2+}$ is oxidised by $KMn{O}_4$
$\therefore$ Milliequivalent of ${Fe}^{2+}$ left at equilibrium in $1000mL$
= Milliequivalent of $KMn{O}_4$ for $1000mL$
$=\frac{30\times 0.0832\times 5\times 1000}{25}=499.2$
$\therefore 545-x=499.2$
$\therefore x=545-499.2=45.8$
Thus, by equation (i)
$K_c=\frac{\frac{45.8}{1000}}{\left(\frac{75-45.8}{1000}\right)\times \left(\frac{545-45.8}{1000}\right)}=\frac{45.8\times 1000}{29.2\times 499.2}$
$K_c=3.1420$