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
Question
For the indicator thymol blue, the value of pH is 2.0 when half of the indicator is present in the unionized form. The percentage of the indicator in the unionized form in a solution of $4.0 \times 10^{-3}$ M hydrogen ion concentration is
💡 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 $\frac{[\text{HIn}]}{[\text{HIn}] + [\text{In}^-]} = \frac{1}{1 + 2.5} = \frac{1}{3.5} \approx 28.57\%$.