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 $(CH_3CH(OH)COOH)$:$CH_3CH(

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$1.66 	imes 10^{-4}$

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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 $(CH_3CH(OH)COOH)$:$CH_3CH(OH)COOH ightleftharpoons CH_3CH(OH)COO^- + H^+$At equilibrium, the concentrations are:$[CH_3CH(OH)COOH] = C(1 - \alpha)$$[CH_3CH(OH)COO^-] = C\alpha$$[H^+] = C\alpha$The dissociation constant $K_a$ is given by:$K_a = \frac{[CH_3CH(OH)COO^-][H^+]}{[CH_3CH(OH)COOH]} = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha}$Substituting the values:$K_a = \frac{0.1 	imes (0.04)^2}{1 - 0.04} = \frac{0.1 	imes 0.0016}{0.96} = \frac{0.00016}{0.96} = \frac{1.6 	imes 10^{-4}}{0.96} \approx 1.66 	imes 10^{-4}$

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

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