A solution contains 0.1 mol of acetic acid a | vedclass

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(A) The given solution is a buffer solution consisting of a weak acid $(CH_3COOH)$ and its conjugate base $(CH_3COONa)$.According to the Henderson-Has

Vedclass · Accepted Answer

$1.82 	imes 10^{-5}$

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(A) The given solution is a buffer solution consisting of a weak acid $(CH_3COOH)$ and its conjugate base $(CH_3COONa)$.According to the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.Given: $pH = 4.74$,$[Salt] = 0.1 \ mol / 0.5 \ L = 0.2 \ M$,and $[Acid] = 0.1 \ mol / 0.5 \ L = 0.2 \ M$.Substituting the values: $4.74 = pK_a + \log \frac{0.2}{0.2}$.Since $\log(1) = 0$,we get $pK_a = 4.74$.We know that $pK_a = -\log(K_a)$,so $K_a = 10^{-4.74}$.$K_a \approx 1.82 	imes 10^{-5}$.

$A$ solution contains $0.1 \ mol$ of acetic acid and $0.1 \ mol$ of sodium acetate in $500 \ mL$ of solution. If the $pH$ of the solution is $4.74$,find the ionization constant $(K_a)$ of acetic acid.

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