# pKa value of an indicator

Can we deduce the actual value of $$\mathrm{p}K_\mathrm{a}$$ of an indicator if the $$\mathrm{pH}$$ range has been given?

I know that $$\mathrm{pH}$$ range is physically the range of values over which an indicator changes its colour from its acid form to its base form, but can we judge the $$\mathrm{p}K_\mathrm{a}$$ value from the data?

For example: if $$\mathrm{pH}$$ range of $$3.4$$ to $$4.6$$ has been provided, then I know for certain that the $$\mathrm{p}K_\mathrm{a}$$ value of the indicator lies in between, but can we deduce the actual value?

Visually, we can only estimate the $$\mathrm{p}K_\mathrm{a}$$ value to be near the middle of the indicator range, shifted to more intense colour.

The $$\mathrm{p}K_\mathrm{a}$$ of an indicator could be determined more precisely by combining $$\mathrm{pH}$$ meter with photometry, where we would get the well known " round step function" of absorbance versus $$\mathrm{pH}$$.

E.g phenolphthalein has indicating range $$\mathrm{pH}=8.2 - 10.0$$ for reaction $$\ce{HInd(clear) <=> H+ + Ind^-(violet)}$$

(I will not fight over the colour name. It may be magenta, carmine, purpur, dark pink - you know the 16 colours of Windows jokes. )

Let suppose the phenolphthalein $$\mathrm{p}K_\mathrm{a}$$ is in the middle of the range, i.e. $$\mathrm{p}K_\mathrm{a}=9.1$$.
Let suppose the ratio of concentrations of clear and violet forms at $$\mathrm{pH}=8.2$$ is $$\ce{[Ind-]/[HInd]}=X$$.

The equation for indicates are the same as for any other weak acid:

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log \frac {\ce{[Ind-]}}{\ce{[HInd]}}$$

If we consider the mentioned border $$\mathrm{pH}$$ values, then \begin{align} 8.2 &=9.1+\log X \\ 10.0 &=9.1+\log \frac 1X \\ X&=0.126 \\ \end{align}

The problem is the relative eye sensibility to small additions of one form of an indicator to the other form.
We notice much easier a small amount of violet substance in clear substance, than a small amount of clear substance in violet substance.

\begin{align} \mathrm{pH_{low}}=\mathrm{p}K_\mathrm{a} + \log \frac {\ce{[Ind^-_{noticeble}]}}{\ce{[HInd_{abundance}]}} \\ \mathrm{pH_{high}}=\mathrm{p}K_\mathrm{a} + \log \frac {\ce{[Ind^-_{abundance}]}}{\ce{[HInd_{noticeble}]}} \\ \end{align}

As the consequence, the $$\mathrm{pH}=10.0$$ is significantly closer to $$\mathrm{p}K_\mathrm{a}$$ than $$\mathrm{pH}=8.2$$, therefore

$$\mathrm{p}K_\mathrm{a} \gt \frac {8.2+10.0}{2}$$

Let suppose for now the real $$\mathrm{p}K_\mathrm{a}=9.7$$.

Then $$\ce{[Ind-]/[HInd]}$$ is

• 2 for pH 10.0, i.e full violet weakened by 1/3

• 0.03 for pH 8.2, i.e 3% of the full violet.

The similar visual shifts happen to other indicators as well.

• Could you please elaborate this , "Let suppose pKa=9.1 and the border values are reciprocal values [Ind−]/[HInd]"? May 4 '19 at 4:26
• "still think insisting on writing pKa as \mathrm{p}K_\mathrm{a} is kind of perversion." Please avoid leaving irrelevant complaints in the answers and keep it for Meta. Yes, MathJax is cumbersome, but deal with it. Set a hotkey or a snippet for \mathrm{ if it's such big issue for you. Also note that in "pH" both letters are upright (I'm not sure why you reverted my edit). May 4 '19 at 4:43
• @andselisk I have not reverted your edit, I was not aware of it. You may have edit it in parallel. I did ask myself 25 min ago as well, why is not pH written upright ? The issue is not with writing MathJax but with staying MathJax :-) I am sorry, I could not resist. May 4 '19 at 4:54
• In fact I knew it I may have been corrupted by MathJax formatting terrorism :-) May 4 '19 at 4:58
• I was writing my graduation thesis in era, where it was still usual to write it on a typewriter, and my one was one of the first, printed on 9pin Epson printer. LaTeX was for gloves in that time. May 4 '19 at 5:01