A 0.1 M acetic acid solution is titrated against 0.1 M – NaOH solution. What would be the difference — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

At the 1/4 stage, $\frac{1}{4}$ of the acid is converted to salt, leaving $\frac{3}{4}$ as acid, so $[\text{Salt}]/[\text{Acid}] = 1/3$. $\text{pH}_1 = pK_a + \log(1/3)$. At the 3/4 stage, $\frac{3}{4}$ is converted to salt, leaving $\frac{1}{4}$ as acid, so $[\text{Salt}]/[\text{Acid}] = 3/1$. $\text{pH}_2 = pK_a + \log(3)$. The difference is $\Delta\text{pH} = \text{pH}_2 - \text{pH}_1 = \log(3) - \log(1/3) = \log(3) - (-\log 3) = 2\log 3$.