50 mL of 2 N acetic acid mixed with 10 mL | vedclass

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(A) The mixture forms an acidic buffer solution.Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.Given: $pK_a$ of ac

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The mixture forms an acidic buffer solution.Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.Given: $pK_a$ of acetic acid is approximately $4.74$ (or $5$ for approximation).Number of milliequivalents of acid $= 50 \ mL 	imes 2 \ N = 100 \ meq$.Number of milliequivalents of salt $= 10 \ mL 	imes 1 \ N = 10 \ meq$.$pH = 4.74 + \log \frac{10}{100} = 4.74 + \log(0.1) = 4.74 - 1 = 3.74$.The approximate value is $4$.

$50 \ mL$ of $2 \ N$ acetic acid mixed with $10 \ mL$ of $1 \ N$ sodium acetate solution will have an approximate $pH$ of:

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