0.1 , M acetic acid is 1% ionized. If its ion | vedclass

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(A) For a weak acid,the dissociation constant is given by $K_a = C \alpha^2$. Initially,$C_1 = 0.1 \, M$ and $\alpha_1 = \frac{1}{100} = 0.01$. $K_a =

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$0.001$

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(A) For a weak acid,the dissociation constant is given by $K_a = C \alpha^2$. Initially,$C_1 = 0.1 \, M$ and $\alpha_1 = \frac{1}{100} = 0.01$. $K_a = 0.1 	imes (0.01)^2 = 0.1 	imes 10^{-4} = 10^{-5}$. Now,for the new concentration $C_2$ where $\alpha_2 = \frac{10}{100} = 0.1$: $K_a = C_2 	imes (\alpha_2)^2$ $10^{-5} = C_2 	imes (0.1)^2$ $10^{-5} = C_2 	imes 10^{-2}$ $C_2 = \frac{10^{-5}}{10^{-2}} = 10^{-3} \, M = 0.001 \, M$.

$0.1 \, M$ acetic acid is $1\%$ ionized. If its ionization increases to $10\%$,what will be its new concentration (in $, M$)?

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