Electrochemistry 2 Question 53
52. The conductance of a $0.0015 \mathrm{M}$ aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinised Pt electrodes. The distance between the electrodes is $120 \mathrm{~cm}$ with an area of cross section of $1 \mathrm{~cm}$ 2 . The conductance of this solution was found to be $5 \times 10^{-7}$ $\mathrm{S}$. The $\mathrm{pH}$ of the solution is 4 . The value of limiting molar conductivity $\left(\Lambda_{\mathrm{m}}^{\circ}\right)$ of this weak monobasic acid in aqueous solution is $Z \times 10^{2} \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1}$. The value of $Z$ is
(2017 Adv.)
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Solution:
- $\mathrm{pH}=C \alpha=10^{-4}$
$$ \Rightarrow \quad \alpha=\frac{10^{-4}}{0.0015} $$
Also, conductance $(G)=\kappa\left(\frac{A}{l}\right)$
$$ \begin{array}{rlrl} \Rightarrow & \kappa & =G\left(\frac{l}{A}\right)=5 \times 10^{-7} \times \frac{120}{1}=6 \times 10^{-5} \ \Rightarrow & & \Lambda^{c} & =\frac{\kappa \times 1000}{C} \ & & =\frac{6 \times 10^{-5} \times 1000}{0.0015} \ \Rightarrow & & \Lambda^{\infty} & =\frac{\Lambda^{c}}{\alpha}=\frac{6 \times 10^{-5} \times 1000}{0.0015} \times \frac{0.0015}{10^{-4}} \ & & =600=6 \times 10^{2} \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1} \end{array} $$
