The molar conductance of M/32 solution of a weak monobasic acid is $8.0 ext{ ohm}^{-1} ext{ cm}^2$ and at

Question

The molar conductance of M/32 solution of a weak monobasic acid is $8.0 	ext{ ohm}^{-1} 	ext{ cm}^2$ and at infinite dilution is $400 	ext{ ohm}^{-1} 	ext{ cm}^2$. The dissociation constant of this acid is:

Accepted Answer

1.  **Calculate Degree of Dissociation ($\alpha$):**
    For a weak electrolyte, the degree of dissociation $\alpha$ is the ratio of molar conductivity at a specific concentration ($\Lambda_m$) to the limiting molar conductivity ($\Lambda_m^\circ$).
    $$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$$
    Given: $\Lambda_m = 8.0 	ext{ S cm}^2 	ext{ mol}^{-1}$ and $\Lambda_m^\circ = 400 	ext{ S cm}^2 	ext{ mol}^{-1}$.
    $$\alpha = \frac{8.0}{400} = 0.02$$

2.  **Calculate Dissociation Constant ($K_a$):**
    The dissociation constant for a weak acid is given by:
    $$K_a = \frac{c\alpha^2}{1 - \alpha}$$
    Given: Concentration $c = 	ext{M}/32 = 1/32 	ext{ mol L}^{-1} \approx 0.03125 	ext{ M}$.
    Since $\alpha (0.02)$ is very small compared to 1, we can approximate $1 - \alpha \approx 1$.
    $$K_a \approx c\alpha^2$$
    $$K_a = \frac{1}{32} 	imes (0.02)^2$$
    $$K_a = \frac{1}{32} 	imes 4 	imes 10^{-4}$$
    $$K_a = \frac{1}{8} 	imes 10^{-4} = 0.125 	imes 10^{-4}$$
    $$K_a = 1.25 	imes 10^{-5}$$

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