Electrochemistry 2 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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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 $\Rightarrow 5 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$
$$ \begin{aligned} & M=\text { Molarity } \Rightarrow 0.001 \mathrm{M} \ \therefore \quad & \Lambda_{\mathrm{m}}=\frac{1000 \times 5 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}}{10^{-3} \mathrm{M}}=50 \mathrm{Scm}^{2} \mathrm{~mol}^{-1} \end{aligned} $$
Therefore, degree of dissociation ( $\alpha$ ), of $\mathrm{H} A$ is,
$$ \alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}=\frac{50 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}{400 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}=0.125 $$
