Question

To 500 mL of 0.150 M ${AgNO}_3$ solution were added 500 mL of 1.09 M ${Fe}^{+2}$ solution and the reaction is allowed to reach an equilibrium at 25C.

${Ag}^{\oplus }\left(aq\right)+{Fe}^{+2}\left(aq\right)\rightleftharpoons {Fe}^{+3}\left(aq\right)+Ag\left(s\right)$

For 25 mL of the solution, 30 mL of 0.0832 M ${KMnO}_4$ was required for oxidation. Calculate the equilibrium constant for the reaction 25C.

• A. $K_c$= 0.19
• B. $K_c$= 739
• C. $K_c$= 31.420
• D. $K_c$= 3.1420

Answer 1

Answer: (D)

$K_c$= 3.1420

${Ag}^{\oplus }\left(aq\right)+{Fe}^{+2}\left(aq\right),\rightleftharpoons {Fe}^{+3}\left(aq\right)+Ag\left(s\right)$
500 x 0.150 500 x 0.109 0 0
= 75 = 545 0 0
$(75-x)$ $(545-x)$ (Millimoles before reaction)
$x$ $x$
(Millimoles after reaction)
$\because$ mM=mEq(both${Ag}^{\oplus }/Agand{Fe}^{+2}/{Fe}^{+3}$ have valency factor unity)
$\because$ $K_c$= $\frac{{[Fe}^{+3}]}{{[Ag}^{\oplus }\left]\right[{Fe}^{+2}]}$
$\therefore$
$K_c$= $\frac{\frac{x}{1000}}{\frac{75-x}{1000}\left(\frac{545-x}{1000}\right)}$
$\because$ concentration=$\frac{Millimole}{Totalvolume}$
${[Ag}^{\oplus }]$=$\frac{75-x}{1000}$
${[Fe}^{+2}]$=$\frac{545-x}{1000}$
${[Fe}^{+3}]$=$\frac{x}{1000}$
Now 25 mL of mixture requires 30 mL of 0.0832 M or 0.0832 x 5 N ${KMnO}_4$
$\because {Fe}^{+2}$ is oxidised by ${KMnO}_4$
Milliequivalent of ${Fe}^{+2}$ left at equilibrium in1000mL
$=$ Milliequivalent of ${KMnO}_4$ for 1000 mL
$=\frac{30\times 0.0832\times 5\times 1000}{25}$
$=$ 499.2
$\therefore$ $545-x$ = 499.2
$\therefore$ $x$ = 545 -499.2 = 45.8
$K_c=\frac{\frac{45.8}{1000}}{\left(\frac{75-45.8}{1000}\right)\left(\frac{545-45.8}{1000}\right)}$
$=\frac{45.8\times 1000}{29.2\times 499.2}$
$K_c$=3.1420