During the titration of N/10 acetic acid with | vedclass

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Question

(A) The reaction is: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$. Since this is a titration of a weak acid with a strong base,the $pH$ is calculate

Vedclass · Accepted Answer

$5 + \log(1/3), 5, 5 + \log(3)$

Vedclass · Answer

(A) The reaction is: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$. Since this is a titration of a weak acid with a strong base,the $pH$ is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log(\frac{[Salt]}{[Acid]})$. Given $K_a = 10^{-5}$,so $pK_a = -\log(10^{-5}) = 5$. At $25\%$ titration: $[Salt] = 25$,$[Acid] = 75$. $pH = 5 + \log(25/75) = 5 + \log(1/3)$. At $50\%$ titration: $[Salt] = 50$,$[Acid] = 50$. $pH = 5 + \log(50/50) = 5 + 0 = 5$. At $75\%$ titration: $[Salt] = 75$,$[Acid] = 25$. $pH = 5 + \log(75/25) = 5 + \log(3)$.

During the titration of $N/10$ acetic acid with $N/10 \, NaOH$,what is the $pH$ of the solution when $25\%$,$50\%$,and $75\%$ of the titration is complete? $[K_a = 10^{-5}]$

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