The degree of ionization of 0.10 M lactic ac | vedclass

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(B) Given: Concentration $(C) = 0.10 \ M$,Degree of ionization $(\alpha) = 4.0 \ \% = 0.04$. For the dissociation of lactic acid: $CH_3CH(OH)COOH \rig

Vedclass · Accepted Answer

$1.66 	imes 10^{-4}$

Vedclass · Answer

(B) Given: Concentration $(C) = 0.10 \ M$,Degree of ionization $(\alpha) = 4.0 \ \% = 0.04$. For the dissociation of lactic acid: $CH_3CH(OH)COOH ightleftharpoons CH_3CH(OH)COO^{-} + H^{+}$. The equilibrium constant $K_a$ is given by the expression: $K_a = \frac{C\alpha^2}{1-\alpha}$. Substituting the values: $K_a = \frac{0.1 	imes (0.04)^2}{1 - 0.04}$. $K_a = \frac{0.1 	imes 0.0016}{0.96} = \frac{0.00016}{0.96} = \frac{1.6 	imes 10^{-4}}{0.96} \approx 1.66 	imes 10^{-4}$.

The degree of ionization of $0.10 \ M$ lactic acid is $4.0 \ \%$. The value of $K_a$ is

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