20 text{ mL} of 0.1 text{ M} acetic acid is m | vedclass

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(B) The mixture of acetic acid $(CH_3COOH)$ and potassium acetate $(CH_3COOK)$ forms an acidic buffer. Using the Henderson-Hasselbalch equation: $pH =

Vedclass · Accepted Answer

$0.04$

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(B) The mixture of acetic acid $(CH_3COOH)$ and potassium acetate $(CH_3COOK)$ forms an acidic buffer. Using the Henderson-Hasselbalch equation: $pH = pK_a + \log\left(\frac{[	ext{Salt}]}{[	ext{Acid}]}ight)$. First,calculate $pK_a$: $pK_a = -\log(1.8 	imes 10^{-5}) \approx 4.74$. Let the concentration of potassium acetate be $M$. The number of moles of acid $= 0.1 	imes 20 = 2 	ext{ mmol}$. The number of moles of salt $= M 	imes 50 = 50M 	ext{ mmol}$. Substituting the values into the equation: $4.8 = 4.74 + \log\left(\frac{50M}{2}ight)$. $0.06 = \log(25M)$. Taking antilog on both sides: $10^{0.06} = 25M$. Since $10^{0.06} \approx 1.15$,we have $25M = 1.15$. $M = \frac{1.15}{25} = 0.046 	ext{ M}$. Rounding to the nearest provided option,$M \approx 0.04 	ext{ M}$.

$20 \text{ mL}$ of $0.1 \text{ M}$ acetic acid is mixed with $50 \text{ mL}$ of potassium acetate. $K_a$ of acetic acid $= 1.8 \times 10^{-5}$ at $27^\circ \text{C}$. Calculate the concentration of potassium acetate if the $pH$ of the mixture is $4.8$. (in $text{ M}$)

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