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(A) The solution is a buffer solution consisting of a weak acid $(CH_3COOH)$ and its conjugate base $(CH_3COO^-)$. The Henderson-Hasselbalch equation

Vedclass · Accepted Answer

$4.76$

Vedclass · Answer

(A) The solution is a buffer solution consisting of a weak acid $(CH_3COOH)$ and its conjugate base $(CH_3COO^-)$. The Henderson-Hasselbalch equation is: $pH = pK_{a} + \log \frac{[	ext{Salt}]}{[	ext{Acid}]}$ Calculate the concentration of acid: $[	ext{Acid}] = \frac{10 \ mL 	imes 1.0 \ M}{100 \ mL} = 0.1 \ M$ Calculate the concentration of salt: $[	ext{Salt}] = \frac{20 \ mL 	imes 0.5 \ M}{100 \ mL} = 0.1 \ M$ Substitute the values into the equation: $pH = 4.76 + \log \frac{0.1}{0.1}$ Since $\log(1) = 0$,we get $pH = 4.76 + 0 = 4.76$.

$A$ solution is prepared by mixing $10 \ mL$ of $1.0 \ M$ acetic acid and $20 \ mL$ of $0.5 \ M$ sodium acetate and diluted to $100 \ mL$. If the $pK_{a}$ of acetic acid is $4.76$,then the $pH$ of the solution is

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