What would be the exact pH of a solution obtained by mixing of acetic acid and of sodium acetate and — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Moles of $CH_3COOH = \frac{5}{60} = 0.083 \text{ mol}$. Moles of $CH_3COONa = \frac{7.5}{82} = 0.091 \text{ mol}$. The mixture forms an acidic buffer. By Henderson-Hasselbalch equation: $pH = pK_a + \log\frac{[Salt]}{[Acid]} = 4.76 + \log\left(\frac{0.091/V}{0.083/V}\right)$. Since $\frac{0.091}{0.083} > 1$, the log term is positive, making $pH > 4.76$. However, the ratio is less than $10$, so the increase is less than $1$ unit, meaning $4.76 < pH < 5.0$.