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(A) The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.Given: $pK_a = 4.76$

Vedclass · Accepted Answer

$4.52$

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(A) The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.Given: $pK_a = 4.76$,$[Salt] = [CH_3COONa] = 6.1 	imes 10^{-2} \ M$,and $[Acid] = [CH_3COOH] = 0.1 \ M$.Substituting the values: $pH = 4.76 + \log \frac{6.1 	imes 10^{-2}}{0.1}$.$pH = 4.76 + \log(0.61)$.Since $\log(0.61) \approx -0.2147$,we get $pH = 4.76 - 0.2147 = 4.5453 \approx 4.55$.Rounding to the nearest provided option,the answer is $4.52$.

Calculate the $pH$ of a buffer solution containing $6.1 \times 10^{-2} \ M$ sodium acetate $(CH_3COONa)$ in $1 \ L$ of $0.1 \ M$ acetic acid $(CH_3COOH)$ solution. (Given: $pK_a$ of $CH_3COOH = 4.76$)

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