What is the percent dissociation of acetic ac | vedclass

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(D) For the dissociation of acetic acid: $CH_3COOH_{(aq)} \rightleftharpoons CH_3COO^-_{(aq)} + H^+_{(aq)}$Given concentration $C = 0.01 \ M$ and diss

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$1.15$

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(D) For the dissociation of acetic acid: $CH_3COOH_{(aq)} ightleftharpoons CH_3COO^-_{(aq)} + H^+_{(aq)}$Given concentration $C = 0.01 \ M$ and dissociation constant $K_a = 1.34 	imes 10^{-6}$.The expression for dissociation constant is $K_a = \frac{C \alpha^2}{1-\alpha}$.Since $K_a$ is very small,$1-\alpha \approx 1$,so $K_a \approx C \alpha^2$.Thus,$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.34 	imes 10^{-6}}{0.01}} = \sqrt{1.34 	imes 10^{-4}} = 1.157 	imes 10^{-2}$.Percent dissociation $\% \alpha = \alpha 	imes 100 = 1.157 	imes 10^{-2} 	imes 100 = 1.157 \% \approx 1.15 \%$.

What is the percent dissociation of acetic acid in its $0.01 \ M$ solution if the dissociation constant of the acid is $1.34 \times 10^{-6}$ (in $\%$)?

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