The indicator constant for an acidic indicator, HIn, is M. This indicator appears only in the colour — Ionic Equilibrium Chemistry Question
Question
The indicator constant for an acidic indicator, HIn, is $5 \times 10^{-6}$ M. This indicator appears only in the colour of acidic form when $\frac{[\text{In}^-]}{[\text{HIn}]} \le \frac{1}{20}$ and it appears only in the colour of basic form when $\frac{[\text{HIn}]}{[\text{In}^-]} \le 40$. The pH range of indicator is ($\log 2 = 0.3$)
💡 Solution & Explanation
$pK_a = -\log(5 \times 10^{-6}) = 6 - \log 5 = 6 - 0.7 = 5.3$. For the acidic color bound: $\frac{[\text{In}^-]}{[\text{HIn}]} = \frac{1}{20}$, so $\text{pH}_{acid} = pK_a + \log(1/20) = 5.3 - 1.3 = 4.0$. For the basic color bound: $\frac{[\text{HIn}]}{[\text{In}^-]} = 40 \implies \frac{[\text{In}^-]}{[\text{HIn}]} = \frac{1}{40}$, so $\text{pH}_{base} = pK_a + \log(40) = 5.3 + 1.6 = 6.9$. Therefore, the transition pH range is 4.0 – 6.9.