A buffer solution is prepared by mixing 10 m | vedclass

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Question

(D) The number of millimoles of acetic acid $(CH_3COOH)$ $= 10 \ mL 	imes 1.0 \ M = 10 \ mmol$. The number of millimoles of sodium acetate $(CH_3COON

Vedclass · Accepted Answer

$4.76$

Vedclass · Answer

(D) The number of millimoles of acetic acid $(CH_3COOH)$ $= 10 \ mL 	imes 1.0 \ M = 10 \ mmol$. The number of millimoles of sodium acetate $(CH_3COONa)$ $= 20 \ mL 	imes 0.5 \ M = 10 \ mmol$. According to the Henderson-Hasselbalch equation: $pH = pK_a + \log \left( \frac{[salt]}{[acid]} ight)$. Since both components are in the same total volume $(100 \ mL)$,the ratio of their concentrations is equal to the ratio of their millimoles: $pH = 4.76 + \log \left( \frac{10 \ mmol}{10 \ mmol} ight)$. $pH = 4.76 + \log(1)$. Since $\log(1) = 0$,we get $pH = 4.76$. Therefore,option $(D)$ is correct.

$A$ buffer solution is prepared by mixing $10 \ mL$ of $1.0 \ M$ acetic acid and $20 \ mL$ of $0.5 \ M$ sodium acetate and then diluted to $100 \ mL$ with distilled water. The $pH$ of the buffer solution is ($pK_a$ of acetic acid is $4.76$).

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