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(D) The number of millimoles $(mmol)$ of acetic acid $(CH_3COOH)$ is $100 \, mL 	imes 0.1 \, mol \, L^{-1} = 10 \, mmol$.The number of millimoles $(m

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$4.76$

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(D) The number of millimoles $(mmol)$ of acetic acid $(CH_3COOH)$ is $100 \, mL 	imes 0.1 \, mol \, L^{-1} = 10 \, mmol$.The number of millimoles $(mmol)$ of sodium acetate $(CH_3COONa)$ is $50 \, mL 	imes 0.2 \, mol \, L^{-1} = 10 \, mmol$.According to the Henderson-Hasselbalch equation for an acidic buffer:$pH = pK_a + \log \frac{[Salt]}{[Acid]}$Since the volume is the same for both components in the mixture,the ratio of concentrations is equal to the ratio of millimoles:$pH = pK_a + \log \frac{mmol \, of \, CH_3COO^-}{mmol \, of \, CH_3COOH}$Substituting the values:$pH = 4.76 + \log \frac{10}{10}$$pH = 4.76 + \log 1$Since $\log 1 = 0$,we get:$pH = 4.76$

An aqueous buffer is prepared by adding $100 \, mL$ of $0.1 \, mol \, L^{-1}$ acetic acid to $50 \, mL$ of $0.2 \, mol \, L^{-1}$ of sodium acetate. If $pK_a$ of acetic acid is $4.76$,the $pH$ of the buffer is

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