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(B) $1$. Calculate the initial moles of acetic acid $(CH_3COOH)$: Since $20 	ext{ mL}$ of acetic acid is neutralized by $28.4 	ext{ mL}$ of $0.1 	e

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$4.75$

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(B) $1$. Calculate the initial moles of acetic acid $(CH_3COOH)$: Since $20 	ext{ mL}$ of acetic acid is neutralized by $28.4 	ext{ mL}$ of $0.1 	ext{ M } NaOH$,the moles of acid = moles of base = $0.1 	ext{ M} 	imes 28.4 	ext{ mL} = 2.84 	ext{ mmol}$.$2$. Calculate the composition of solution $(X)$: Solution $(X)$ is formed by mixing $2.84 	ext{ mmol}$ of $CH_3COOH$ and $1.42 	ext{ mmol}$ of $NaOH$ $(0.1 	ext{ M} 	imes 14.2 	ext{ mL} = 1.42 	ext{ mmol})$.$3$. Reaction: $CH_3COOH + NaOH ightarrow CH_3COONa + H_2O$.$4$. After reaction: Remaining $CH_3COOH = 2.84 - 1.42 = 1.42 	ext{ mmol}$. Formed $CH_3COONa$ (salt) = $1.42 	ext{ mmol}$.$5$. Apply Henderson-Hasselbalch equation: $pH = pK_a + \log \left( \frac{[Salt]}{[Acid]} ight)$.$6$. Since $[Salt] = [Acid] = 1.42 	ext{ mmol}$,the ratio is $1$. Thus,$pH = 4.75 + \log(1) = 4.75 + 0 = 4.75$.

$20$ mL of a solution of acetic acid required $28.4$ mL of $0.1$ $M$ NaOH for its neutralization. $A$ solution $(X)$ was prepared by mixing $20$ mL of the above acetic acid and $14.2$ mL of $0.1$ $M$ NaOH solution. What is the pH of the solution $(X)$? ($pK_a$ value of acetic acid is $4.75$).

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