If a 0.1 N solution of CH_3COOH is 1.3% ioni | vedclass

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(B) The concentration of the solution is $C = 0.1 \ N$. Since $CH_3COOH$ is a monobasic acid,its normality is equal to its molarity,so $C = 0.1 \ M$.G

Vedclass · Accepted Answer

$2.89$

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(B) The concentration of the solution is $C = 0.1 \ N$. Since $CH_3COOH$ is a monobasic acid,its normality is equal to its molarity,so $C = 0.1 \ M$.Given the degree of ionization $\alpha = 1.3\% = 0.013$.The concentration of hydrogen ions $[H^+]$ is given by $[H^+] = C 	imes \alpha$.$[H^+] = 0.1 	imes 0.013 = 0.0013 \ M = 1.3 	imes 10^{-3} \ M$.The $pH$ of the solution is calculated as $pH = -\log[H^+]$.$pH = -\log(1.3 	imes 10^{-3}) = -(\log 1.3 + \log 10^{-3})$.$pH = -(0.11 - 3) = -(-2.89) = 2.89$.

If a $0.1 \ N$ solution of $CH_3COOH$ is $1.3\%$ ionized,what will be the $pH$ of the solution? (Given: $\log 1.3 = 0.11$)

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