When NaOH is added to a CH_3COOH solution,60% | vedclass

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Question

(A) The reaction is $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$. Since $60\%$ of the acid is neutralized,the concentration of salt $[Salt] = 0.6$ a

Vedclass · Accepted Answer

Greater than $4.7$ but less than $5.0$

Vedclass · Answer

(A) The reaction is $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$. Since $60\%$ of the acid is neutralized,the concentration of salt $[Salt] = 0.6$ and the remaining acid $[Acid] = 0.4$. Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$ $pH = 4.7 + \log \frac{0.6}{0.4}$ $pH = 4.7 + \log(1.5)$ Since $\log(1.5) \approx 0.176$,$pH = 4.7 + 0.176 = 4.876$. This value is greater than $4.7$ and less than $5.0$.

When $NaOH$ is added to a $CH_3COOH$ solution,$60\%$ of the acid is neutralized. If the $pK_a$ is $4.7$,the $pH$ of the resulting solution is:

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