For the indicator thymol blue, the value of pH is 2.0 when half of the indicator is present in the u — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

When half of the indicator is unionized, $[\text{In}^-] = [\text{HIn}]$, which means $\text{pH} = pK_a$. Thus, $pK_a = 2.0$, so $K_a = 10^{-2}$. Using the equilibrium equation $K_a = \frac{[\text{H}^+][\text{In}^-]}{[\text{HIn}]}$, we rearrange to find the ratio: $\frac{[\text{In}^-]}{[\text{HIn}]} = \frac{K_a}{[\text{H}^+]} = \frac{10^{-2}}{4.0 \times 10^{-3}} = \frac{10}{4} = 2.5$. The fraction of unionized indicator is $\frac{[\text{HIn}]}{[\text{HIn}] + [\text{In}^-]} = \frac{1}{1 + 2.5} = \frac{1}{3.5} \approx 28.57\%$.