At 298 K,0.1 M solution of acetic acid is 1 | vedclass

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(B) The dissociation of acetic acid is given by: $CH_3COOH_{(aq)} \rightleftharpoons CH_3COO^-_{(aq)} + H^+_{(aq)}$Given concentration $C = 0.1 \ M$ a

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$1.8 	imes 10^{-5}$

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(B) The dissociation of acetic acid is given by: $CH_3COOH_{(aq)} ightleftharpoons CH_3COO^-_{(aq)} + H^+_{(aq)}$Given concentration $C = 0.1 \ M$ and degree of ionization $\alpha = 1.34 \ \% = 0.0134$.The dissociation constant $K_a$ is given by the formula: $K_a = \frac{C\alpha^2}{1-\alpha}$.Since $\alpha$ is very small,we can approximate $1-\alpha \approx 1$.Thus,$K_a \approx C\alpha^2$.Substituting the values: $K_a = 0.1 	imes (0.0134)^2$.$K_a = 0.1 	imes 0.00017956 = 1.7956 	imes 10^{-5} \approx 1.8 	imes 10^{-5}$.

At $298 \ K$,$0.1 \ M$ solution of acetic acid is $1.34 \ \%$ ionized. What is the dissociation constant of acetic acid?

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