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(D) The solution is a buffer solution consisting of a weak acid (acetic acid) and its salt (sodium acetate).The Henderson-Hasselbalch equation is: $pH

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$4.76 < pH < 5.0$

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(D) The solution is a buffer solution consisting of a weak acid (acetic acid) and its salt (sodium acetate).The Henderson-Hasselbalch equation is: $pH = pK_a + \log \frac{[salt]}{[acid]}$First,calculate the molar concentrations (or simply the ratio of moles since the volume is the same for both):Moles of acetic acid $(CH_3COOH)$ = $\frac{5 \ g}{60 \ g/mol} = 0.0833 \ mol$Moles of sodium acetate $(CH_3COONa)$ = $\frac{7.5 \ g}{82 \ g/mol} = 0.0915 \ mol$Using the ratio of moles in the Henderson-Hasselbalch equation:$pH = 4.76 + \log \left( \frac{0.0915}{0.0833} \right) = 4.76 + \log(1.098) \approx 4.76 + 0.04 = 4.80$Since $4.80$ falls in the range $4.76 < pH < 5.0$,the correct option is $D$.

What would be the $pH$ of a solution obtained by mixing $5 \ g$ of acetic acid and $7.5 \ g$ of sodium acetate and making the volume equal to $500 \ mL$? $(K_a = 1.75 \times 10^{-5}, pK_a = 4.76)$

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