2 g acetic acid and 3 g sodium acetate are | vedclass

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(C) The solution is a buffer solution containing a weak acid $(CH_3COOH)$ and its salt with a strong base $(CH_3COONa)$.$pH$ is calculated using the H

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$4.78$

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(C) The solution is a buffer solution containing a weak acid $(CH_3COOH)$ and its salt with a strong base $(CH_3COONa)$.$pH$ is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log(\frac{[Salt]}{[Acid]})$.$1$. Calculate moles of acetic acid $(CH_3COOH)$: Molar mass = $60 \ g/mol$. Moles = $2 \ g / 60 \ g/mol = 0.0333 \ mol$.$2$. Calculate moles of sodium acetate $(CH_3COONa)$: Molar mass = $82 \ g/mol$. Moles = $3 \ g / 82 \ g/mol = 0.0366 \ mol$.$3$. Since the volume is the same $(100 \ mL)$,the ratio of concentrations is equal to the ratio of moles: $\frac{[Salt]}{[Acid]} = \frac{0.0366}{0.0333} \approx 1.1$.$4$. Calculate $pK_a$: $pK_a = -\log(1.8 	imes 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$.$5$. Calculate $pH$: $pH = 4.745 + \log(1.1) = 4.745 + 0.041 = 4.786$.Thus,the $pH$ is approximately $4.78$.

$2 \ g$ acetic acid and $3 \ g$ sodium acetate are present in $100 \ mL$ aqueous solution. What will be the $pH$ of the solution if the ionisation constant of acetic acid is $1.8 \times 10^{-5}$?

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