pH of a solution of 10 mL 1 N sodium acetat | vedclass

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(A) The solution is a buffer containing a weak acid (acetic acid) and its salt (sodium acetate).Using the Henderson-Hasselbalch equation: $pH = pK_a +

Vedclass · Accepted Answer

$4$

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(A) The solution is a buffer containing a weak acid (acetic acid) and its salt (sodium acetate).Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$Number of milliequivalents of salt = $10 \ mL 	imes 1 \ N = 10 \ meq$.Number of milliequivalents of acid = $50 \ mL 	imes 2 \ N = 100 \ meq$.Total volume = $10 \ mL + 50 \ mL = 60 \ mL$.Since the volume cancels out in the ratio,we use milliequivalents: $\frac{[Salt]}{[Acid]} = \frac{10}{100} = 0.1$.$pK_a = -\log(1.8 	imes 10^{-5}) = 5 - \log(1.8) \approx 5 - 0.255 = 4.745$.$pH = 4.745 + \log(0.1) = 4.745 - 1 = 3.745$.The value is approximately $4$.

$pH$ of a solution of $10 \ mL$ $1 \ N$ sodium acetate and $50 \ mL$ $2 \ N$ acetic acid $(K_a = 1.8 \times 10^{-5})$ is approximately

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