The dissociation constant of acetic acid is 1 | vedclass

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(B) For a weak electrolyte,the degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$.Given $K_a = 1.6 	imes 10^{-5}$ and $C = 0

Vedclass · Accepted Answer

$1.50 	imes 10^{-6} \, S \, cm^{-1}$

Vedclass · Answer

(B) For a weak electrolyte,the degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$.Given $K_a = 1.6 	imes 10^{-5}$ and $C = 0.01 \, M$,we have $\alpha = \sqrt{\frac{1.6 	imes 10^{-5}}{0.01}} = \sqrt{1.6 	imes 10^{-3}} \approx 0.04$. Now,the molar conductivity $\Lambda_M = \alpha 	imes \Lambda_M^o = 0.04 	imes 370.6 	imes 10^{-4} \, S \, m^2 \, mol^{-1} = 1.4824 	imes 10^{-5} \, S \, m^2 \, mol^{-1}$.Since $1 \, S \, m^2 \, mol^{-1} = 10^4 \, S \, cm^2 \, mol^{-1}$,$\Lambda_M = 1.4824 	imes 10^{-5} 	imes 10^4 = 0.14824 \, S \, cm^2 \, mol^{-1}$.The specific conductance $\kappa$ is given by $\kappa = \Lambda_M 	imes C$ (in $mol/cm^3$).Since $C = 0.01 \, mol/L = 10^{-5} \, mol/cm^3$,we have $\kappa = 0.14824 	imes 10^{-5} \approx 1.5 	imes 10^{-6} \, S \, cm^{-1}$.

The dissociation constant of acetic acid is $1.6 \times 10^{-5}$ and $\Lambda _{CH_3COOH}^o = 370.6 \times 10^{-4} \, S \, m^2 \, mol^{-1}$. The specific conductance of $0.01 \, M$ acetic acid solution will be approximately:

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