An acid–base indicator has . The acid form of the indicator is red and the basic form is blue. The r — Ionic Equilibrium Chemistry Question
Question
An acid–base indicator has $K_a = 3.0 \times 10^{-5}$. The acid form of the indicator is red and the basic form is blue. The $[\text{H}^+]$ required to change the indicator from 75% blue to 75% red is
💡 Solution & Explanation
For the indicator $\text{HIn} \rightleftharpoons \text{H}^+ + \text{In}^-$, $K_a = \frac{[\text{H}^+][\text{In}^-]}{[\text{HIn}]}$. At 75% blue (basic form $\text{In}^-$), $[\text{In}^-]/[\text{HIn}] = 75/25 = 3$, so $[\text{H}^+]_1 = \frac{K_a}{3} = \frac{3.0 \times 10^{-5}}{3} = 1.0 \times 10^{-5}\text{ M}$. At 75% red (acidic form HIn), $[\text{HIn}]/[\text{In}^-] = 75/25 = 3$, so $[\text{H}^+]_2 = K_a \times 3 = 3.0 \times 10^{-5} \times 3 = 9.0 \times 10^{-5}\text{ M}$. The change in hydrogen ion concentration required is $\Delta[\text{H}^+] = (9.0 - 1.0) \times 10^{-5}\text{ M} = 8.0 \times 10^{-5}\text{ M}$. Therefore, correct answer is A.