50 , mL of 0.1 , M , CH_3COOH is being titrat | vedclass

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(C) Initial moles of $CH_3COOH = 0.1 \, M 	imes 50 \, mL = 5 \, mmol$.Moles of $NaOH$ added = $0.1 \, M 	imes 25 \, mL = 2.5 \, mmol$.The reaction i

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$476$

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(C) Initial moles of $CH_3COOH = 0.1 \, M 	imes 50 \, mL = 5 \, mmol$.Moles of $NaOH$ added = $0.1 \, M 	imes 25 \, mL = 2.5 \, mmol$.The reaction is: $CH_3COOH + NaOH \longrightarrow CH_3COONa + H_2O$.After the reaction,$2.5 \, mmol$ of $CH_3COOH$ remains and $2.5 \, mmol$ of $CH_3COONa$ is formed.Since we have a weak acid and its conjugate base,a buffer solution is formed.Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \left(\frac{[CH_3COO^-]}{[CH_3COOH]}ight)$.Since the volume is the same for both,the ratio of concentrations is equal to the ratio of moles: $pH = 4.76 + \log \left(\frac{2.5}{2.5}ight) = 4.76 + \log(1) = 4.76$.Thus,$pH = 476 	imes 10^{-2}$.

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