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

At $25^{\circ} \mathrm{C}$, the resistance of a cell filled with $0.01 \mathrm{~M} \mathrm{~KCl}$ solution is $525 \mathrm{~ohms}$. The resistance of the same cell filled with $0.1 \mathrm{~M} \mathrm{~NH}_{4} \mathrm{OH}$ is $2030 \mathrm{~ohms}$. By calculating the degree of dissociation, calculate the equilibrium constant of $\mathrm{NH}_{4} \mathrm{OH}$. Molar conductivity at infinite dilution for K+ and Cl- is 73.52 and 76.34 respectively.
  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 infinite dilution of $\mathrm{KCl}$ is
$\Lambda_{\mathrm{m}}^{\infty}(\mathrm{KCl})=\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{K}^{+}ight)+\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{Cl}^{-}ight)$
$=(73.52+76.34) \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
$=149.86 ~\Omega^{-} \mathrm{cm}^{2} \mathrm{~mol}^{-1}$
Conductivity of $0.01 \mathrm{~M} \mathrm{~KCl}$ is
$\mathrm{K}_{1}=\Lambda_{\mathrm{m}}^{\infty} \mathrm{c}=\left(149.86 ~\Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}ight)$
$\left(0.01 \times 10^{-3} \mathrm{~mol} \mathrm{~cm}^{-3}ight)=1.4986 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}$
Cell constant is
$\mathrm{K}=k_{1} \mathrm{R}=\left(1.4986 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}ight)(525 ~\Omega)=0.7868 \mathrm{~cm}^{-1}$
Conductivity of $0.1 \mathrm{~M} \mathrm{~NH}_{4} \mathrm{OH}$ is
$\Lambda_{\mathrm{c}}\left(\mathrm{NH}_{4} \mathrm{OH}ight)=\frac{k_{2}}{\mathrm{c}}=\frac{3.876 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}}{0.1 \times 10^{-3} \mathrm{~mol} \mathrm{~cm}^{-3}}$
Molar conductivity of $0.1 \mathrm{~M} \mathrm{~NH}_{4} \mathrm{OH}$ is
$\Lambda_{\mathrm{c}}\left(\mathrm{NH}_{4} \mathrm{OH}ight)=\frac{k_{2}}{\mathrm{c}}=\frac{3.876 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}}{0.1 \times 10^{-3} \mathrm{~mol} \mathrm{~cm}^{-3}}$
$=3.876 ~\Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
Molar conductivity of $\mathrm{NH}_{4} \mathrm{OH}$ at infinite dilution is
$\Lambda_{\mathrm{m}}^{\infty}\left(\mathrm{NH}_{4} \mathrm{OH}ight)=\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{NH}_{4}^{+}ight)+\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{OH}^{-}ight)$
$=(73.4+197.6) \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
$=271.0 ~\Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
Degree of dissociation of $\mathrm{NH}_{4} \mathrm{OH}$ is
$\alpha=\frac{\Lambda_{\mathrm{c}}}{\Lambda_{\mathrm{m}}^{\infty}}=\frac{3.876 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}{271.0 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}=0.0143$
Equilibrium constant of $\mathrm{NH}_{4} \mathrm{OH}$ is
$\mathrm{K}=\frac{\left[\mathrm{NH}_{4}^{+}ight]\left[\mathrm{OH}^{-}ight]}{\left[\mathrm{NH}_{4} \mathrm{OH}ight]}=\frac{(\mathrm{c} \alpha)(\mathrm{c} \alpha)}{\mathrm{c}(1-\alpha)} \simeq \mathrm{c} \alpha^{2}$
$=(0.1 \mathrm{M})(0.0143)^{2}=2.05 \times 10^{-5} \mathrm{~M}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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