The conductivity of a weak acid HA of concent | vedclass

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(B) First,calculate the molar conductivity $(\Lambda_{m})$ using the formula: $\Lambda_{m} = \frac{1000 	imes \kappa}{C}$$\Lambda_{m} = \frac{1000

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) First,calculate the molar conductivity $(\Lambda_{m})$ using the formula: $\Lambda_{m} = \frac{1000 	imes \kappa}{C}$$\Lambda_{m} = \frac{1000 	imes 2.0 	imes 10^{-5}}{0.001} = 20 \, S \, cm^{2} \, mol^{-1}$Next,calculate the degree of dissociation $(\alpha)$:$\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}} = \frac{20}{190} = \frac{2}{19}$For a weak acid $HA$,the ionization constant $K_{a}$ is given by:$K_{a} = \frac{C \alpha^{2}}{1 - \alpha}$Since $\alpha$ is very small,$1 - \alpha \approx 1$,but we can calculate it precisely:$K_{a} = \frac{0.001 	imes (2/19)^{2}}{1 - (2/19)} = \frac{0.001 	imes (4/361)}{17/19} = \frac{0.001 	imes 4}{361} 	imes \frac{19}{17} = \frac{0.004}{19 	imes 17} = \frac{0.004}{323} \approx 1.238 	imes 10^{-5} = 12.38 	imes 10^{-6}$Rounding to the nearest integer,we get $12 	imes 10^{-6}$.

The conductivity of a weak acid $HA$ of concentration $0.001 \, mol \, L^{-1}$ is $2.0 \times 10^{-5} \, S \, cm^{-1} .$ If $\Lambda_{m}^{\circ} (HA)=190 \, S \, cm^{2} \, mol^{-1}$,the ionization constant $(K_{a})$ of $HA$ is equal to $....\, \times 10^{-6} .$ (Round off to the Nearest Integer)

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