The specific conductance of 0.0025 M acetic | vedclass

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(D) The molar conductivity $\wedge_m$ is calculated as: $\wedge_m = \frac{k 	imes 1000}{C}$ Given $k = 5 	imes 10^{-5} \ S \ cm^{-1}$ and $C = 0.002

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(D) The molar conductivity $\wedge_m$ is calculated as: $\wedge_m = \frac{k 	imes 1000}{C}$ Given $k = 5 	imes 10^{-5} \ S \ cm^{-1}$ and $C = 0.0025 \ M$. $\wedge_m = \frac{5 	imes 10^{-5} 	imes 1000}{0.0025} = \frac{0.05}{0.0025} = 20 \ S \ cm^2 \ mol^{-1}$. The degree of dissociation $\alpha$ is given by $\alpha = \frac{\wedge_m}{\wedge_m^\circ} = \frac{20}{400} = 0.05$. The dissociation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{1 - \alpha}$. Since $\alpha$ is very small,$1 - \alpha \approx 1$,so $K_a \approx C \alpha^2 = 0.0025 	imes (0.05)^2 = 0.0025 	imes 0.0025 = 6.25 	imes 10^{-6} = 62.5 	imes 10^{-7}$. Rounding to the nearest integer,we get $63 	imes 10^{-7}$.

The specific conductance of $0.0025 \ M$ acetic acid is $5 \times 10^{-5} \ S \ cm^{-1}$ at a certain temperature. The dissociation constant of acetic acid is $...... \times 10^{-7}$. (Nearest integer) Consider limiting molar conductivity of $CH_3COOH$ as $400 \ S \ cm^2 \ mol^{-1}$.

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