Ncert Exercises & Solutions

3.11 Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If for acetic acid is 390.5 S cm2 mol–1, what is its dissociation constant?

Hint: First calculate the value of α then dissociate constant
Step 1:

First calculate the value of molar conductivity as follows:

The given values are-

$\mathrm{k}=7.896 \times 10^{-5} \mathrm{~S} \mathrm{~c m}^{-1} \\ Conc.=0.00241 \mathrm{~mol} \mathrm{~L^{-1}}\\$

$\wedge_{\mathrm{m}}=\frac{\mathrm{k}\times1000}{\mathrm{c}}$

$=\frac{7.896 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}}{0.00241 \mathrm{~mol} \mathrm{~L}^{-1}} \times \frac{1000 \mathrm{~cm}^{3}}{\mathrm{~L}}\\ \Lambda_{\mathrm{m}} =32.76 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\\ \Lambda_{\mathrm{m}}^{0}=390.5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

Step 2:

Calculate the value of α as follows:

$\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{0}}=\frac{32.76 \mathrm{~S} \mathrm{} \mathrm{cm}^{2} \mathrm{~mol}^{-1}}{390.5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}\\$
= 0.084

Step 3:

Calculate the value of dissociate constant as follows:

$\mathrm{K}_{\mathrm{a}}=\frac{c \mathrm{a}^{2}}{(1-\alpha)}\\ =\frac{\left(0.00241 \mathrm{~mol} \mathrm{~L}^{-1}\right)(0.084)^{2}}{(1-0.084)}\\ =1.86 \times 10^{-5}$

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