An acid type indicator, HIn differs in colour from its conjugate base (In). The human eye is sensiti — Ionic Equilibrium Chemistry Question
Question
An acid type indicator, HIn differs in colour from its conjugate base (In$^-$). The human eye is sensitive to colour differences only when the ratio $[\text{In}^-]/[\text{HIn}]$ is greater than 10 or smaller than 0.1. What should be the minimum change in the pH of the solution to observe a complete colour change ($K_a = 1.0 \times 10^{-5}$)?
💡 Solution & Explanation
The pH for the first distinct color occurs at $[\text{In}^-]/[\text{HIn}] = 0.1$, where $\text{pH}_1 = pK_a + \log(0.1) = pK_a - 1$. The pH for the second distinct color occurs at $[\text{In}^-]/[\text{HIn}] = 10$, where $\text{pH}_2 = pK_a + \log(10) = pK_a + 1$. The minimum pH change required to transition fully between the two visible boundary states is $\Delta\text{pH} = \text{pH}_2 - \text{pH}_1 = (pK_a + 1) - (pK_a - 1) = 2.0$.