The dissociation constant of acetic acid is 1 | vedclass

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Question

(A) Given: $K_a = 1.75 	imes 10^{-5}$,$C = 0.01 \, M$,$\Lambda _m^o = 370.6 	imes 10^{-4} \, S \, m^2 \, mol^{-1} = 370.6 \, S \, cm^2 \, mol^{-1}$.

Vedclass · Accepted Answer

$1.55 	imes 10^{-4} \, S \, cm^{-1}$

Vedclass · Answer

(A) Given: $K_a = 1.75 	imes 10^{-5}$,$C = 0.01 \, M$,$\Lambda _m^o = 370.6 	imes 10^{-4} \, S \, m^2 \, mol^{-1} = 370.6 \, S \, cm^2 \, mol^{-1}$.Degree of dissociation $\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.75 	imes 10^{-5}}{0.01}} = \sqrt{1.75 	imes 10^{-3}} \approx 0.0418$.Molar conductivity $\Lambda _m = \alpha 	imes \Lambda _m^o = 0.0418 	imes 370.6 \approx 15.5 \, S \, cm^2 \, mol^{-1}$.Specific conductance $\kappa = \frac{\Lambda _m 	imes C}{1000} = \frac{15.5 	imes 0.01}{1000} = 1.55 	imes 10^{-4} \, S \, cm^{-1}$.

The dissociation constant of acetic acid is $1.75 \times 10^{-5}$ 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:

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