At $25^{\circ} \mathrm{C}$, the resistance of a cell filled with $0.01 \mathrm{M} \mathrm{KCl}$ solution is…

At $25^{\circ} \mathrm{C}$, the resistance of a cell filled with $0.01 \mathrm{M} \mathrm{KCl}$ solution is 525 ohms. The resistance of the same cell filled with $0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}$ is 2030 ohms. By calculating the degree of dissociation, calculate the equilibrium constant of $\mathrm{NH}_{4} \mathrm{OH}$.
Given : $\lambda^{\infty}\left(\mathbf{K}^{+}ight)=73.52, \lambda^{\infty}\left(\mathbf{C l}^{-}ight)=76.34, \lambda^{\infty}\left(\mathbf{N H}_{4}^{+}ight)=73.4$ and
$\lambda^{\infty}\left(\mathrm{OH}^{-}ight)=197.6 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
  1. $2.05 \times 10^{-4} \mathrm{M}$
  2. $2.15 \times 10^{-5} \mathrm{M}$
  3. $1.05 \times 10^{-7} \mathrm{M}$
  4. $2.05 \times 10^{-5} \mathrm{M}$

Solution

Molar conductivity at infintite dilution of $\mathrm{KCl}$ is
$\Lambda_{\mathrm{m}}^{\infty}(\mathrm{KC})=\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{K}^{+}ight)+\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{C}^{-}ight)$
$=(73.52+76.34) \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}=149.86 \Omega^{-} \mathrm{cm}^{2} \mathrm{~mol}^{-1}$
Conductivity of $0.01 \mathrm{M} \mathrm{KCl}$ is
$\mathbf{K}_{1}=\Lambda_{\mathrm{m}}^{\infty} \mathrm{c}=\left(149.86 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}ight)$
$\left(0.01 \times 10^{-3} \mathrm{~mol} \mathrm{~cm}^{-3}ight)=1.4986 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}$
Cell constant is
$K=k_{1} R=\left(1.4986 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}ight)(525 \Omega)=0.7868 \mathrm{~cm}^{-1}$
Conductivity of $0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}$ is
$K_{2}=\frac{K}{R}=\frac{0.7868 \mathrm{~cm}^{-1}}{2030 \Omega}=3.876 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}$ Molar conductivity of $0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}$ is
$\Lambda_{\mathrm{o}}\left(\mathrm{NH}_{4} \mathrm{OH}ight)=\frac{k_{2}}{\mathrm{c}}=\frac{3.876 \times 10^{-1} \mathrm{\Omega}^{-1} \mathrm{~cm}^{-1}}{0.1 \times 10^{-3} \mathrm{~mol} \mathrm{~cm}^{-3}}$
$=3.876 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
Molar conducitivity of $\mathrm{NH}_{4} \mathrm{OH}$ at infinite dilution is
$\Lambda_{\mathrm{m}}^{\infty}\left(\mathrm{NH}_{4} \mathrm{OH}ight)=\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{NH}_{4}^{+}ight)+\lambda_{\mathrm{m}}^{\infty}(\mathrm{OH})$
$=(73.4+197.6) \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
$=271.0 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
Degree of dissociation of $\mathrm{NH}_{4} \mathrm{OH}$ is
$\alpha=\frac{\Lambda_{\mathrm{o}}}{\Lambda_{\mathrm{m}}^{\mathrm{o}}}=\frac{3.876 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}{271.0 \mathrm{\Omega}^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}=0.0143$
Equilibrium constant of $\mathrm{NH}_{4} \mathrm{OH}$ is $\mathrm{K}=\frac{\left[\mathrm{NH}_{4}^{+}ight]\left[\mathrm{OH}^{-}ight]}{\left[\mathrm{NH}_{4} \mathrm{OH}ight]}=\frac{(\mathrm{c} \alpha)(\mathrm{co})}{\mathrm{c}(1-\alpha)} \simeq \mathrm{ca}^{2} \quad=(0.1 \mathrm{M})(0.0143)^{2}=2.05 \times 10^{-5} \mathrm{M}$ ^

Asked in: JEE-TOPICTESTS-CHEMISTRY

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