A certain acid-type indicator, , has an indicator constant . A single drop of this indicator is plac — Ionic Equilibrium Chemistry Question
Question
A certain acid-type indicator, $HIn \rightleftharpoons H^+ + In^-$, has an indicator constant $K_{In} = 5.0 \times 10^{-5}$. A single drop of this indicator is placed into a large volume of an aqueous solution whose pH is firmly maintained at 4.30. Calculate the exact ratio of the concentration of the unionized acid form to the ionized basic form, $\frac{[HIn]}{[In^-]}$, in this solution. (Given $\log 5 = 0.70$)
💡 Solution & Explanation
The generic dissociation constant expression is $K_{In} = \frac{[H^+][In^-]}{[HIn]}$. Rearranging provides the ratio $\frac{[HIn]}{[In^-]} = \frac{[H^+]}{K_{In}}$. The fixed pH of the solution is 4.30, meaning $[H^+] = 10^{-4.30} \text{ M}$. We are given $\log 5 = 0.70$, so $5 = 10^{0.70}$, meaning the constant $5.0 \times 10^{-5} = 10^{0.70} \times 10^{-5} = 10^{-4.30}$. Therefore, the ambient $[H^+]$ perfectly equals $K_{In}$. The ratio $\frac{[HIn]}{[In^-]} = \frac{10^{-4.30}}{10^{-4.30}} = 1$.