Electrochemistry Result Question 5
5. $\Lambda_{\mathrm{m}}^{\circ}$ for $\mathrm{NaCl}, \mathrm{HCl}$ and $\mathrm{Na} A$ are 126.4, 425.9 and $100.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, respectively. If the conductivity of $0.001 \mathrm{M} $ $\mathrm{H} A$ is $5 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$, degree of dissociation of $\mathrm{H} A$ is
(2019 Main, 12 Jan II)
(a) $0.25$
(b) $0.50$
(c) $0.75$
(d) $0.125$
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Answer:
Correct Answer: 5. ( d )
Solution:
- According to Kohlrausch’s law, the molar conductivity of $\mathrm{H} A$ at infinite dilution is given as,
$ \begin{aligned} & \Lambda_{\mathrm{m}}^{\circ}(\mathrm{H} A)=\left[\Lambda_{\mathrm{m}}^{\circ}\left(\mathrm{H}^{+}\right)+\Lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)\right]+\left[\Lambda_m^{\circ}\left(\mathrm{Na}^{+}\right)+\Lambda_m^{\circ}\left(A^{-}\right)\right] -\left[\Lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)+\Lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)\right] \\ & =425.9+100.5-126.4=400 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ \end{aligned} $
Also, molar conductivity at given concentration is given as,
$ \Lambda_{\mathrm{m}}=\frac{1000 \times \kappa}{M} $
Given, $\kappa=$ conductivity $\Rightarrow 5 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$
$ \begin{aligned} M & =\text { Molarity } \Rightarrow 0.001 \mathrm{M} \\ \therefore \quad \Lambda_{\mathrm{m}} & =\frac{1000 \times 5 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}}{10^{-3} \mathrm{M}}=50 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \end{aligned} $
Therefore, degree of dissociation ( $\alpha$ ), of $\mathrm{H} A$ is,
$ \alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}=\frac{50 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}}{400 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}}=0.125 $
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