If 50 , mL of 2 , N acetic acid is mixed with | vedclass

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(A) The solution is a buffer solution consisting of a weak acid and its salt.Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}

Vedclass · Accepted Answer

$4$

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(A) The solution is a buffer solution consisting of a weak acid and its salt.Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$Given: $K_a = 10^{-5}$,so $pK_a = -\log(10^{-5}) = 5$.Number of equivalents of salt = $1 \, N 	imes 10 \, mL = 10 \, meq$.Number of equivalents of acid = $2 \, N 	imes 50 \, mL = 100 \, meq$.Substituting the values: $pH = 5 + \log \frac{10}{100} = 5 + \log(0.1) = 5 + (-1) = 4$.

If $50 \, mL$ of $2 \, N$ acetic acid is mixed with $10 \, mL$ of $1 \, N$ sodium acetate solution,what will be the approximate $pH$ of the resulting solution? (Given: $K_a = 10^{-5}$)

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