Question

A certain indicator HIn has a pK_a of 3.00 and a color change becomes visible when 7.00% of the indicator has been converted to In^-. At what pH is this color change visible?

Short answer

At thepH of 1.88, the color change is visible.

Step-by-step solution

Given Information

It is given that:

${\mathrm{pK}}_{\mathrm{a}}=1.88$

The color change is visible when 7 % of the indicator converts to ln^-.

Conversion of pKa to Ka

On converting,

$\begin{array}{l}{\mathrm{pK}}_{\mathrm{a}}=3\\ -{\mathrm{logK}}_{\mathrm{a}}=3\\ {\mathrm{logK}}_{\mathrm{a}}=-3\\ {\mathrm{K}}_{\mathrm{a}}={10}^{-3}\\ {\mathrm{K}}_{\mathrm{a}}=1\times {10}^{-3}\end{array}$

Reaction showing the HIn indicator dissociation

The reaction which shows the dissociation of the HIn indicator can be denoted as:

$\mathrm{HIn}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{In}}^{-}\left(\mathrm{aq}\right)$

The equilibrium constant for this reaction can be expressed as:

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{In}}^{-}\right]}{\left[\mathrm{HIn}\right]}$

The value of $\left[\mathrm{HIn}\right]\&\left[{\mathrm{In}}^{-}\right]$ are 0.93 & 0.07. On substituting the values,

$\begin{array}{l}1\times {10}^{-3}=\frac{\left[{\mathrm{H}}^{+}\right](0.07)}{(0.93)}\\ \left[{\mathrm{H}}^{+}\right]=\frac{1\times {10}^{-3}\times 0.93}{0.07}\\ \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]\\ =-\log (1.33\times {10}^{-2})\\ \mathrm{pH}=1.88\end{array}$