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.

  • $\begingroup$ Could you please elaborate this , "Let suppose pKa=9.1 and the border values are reciprocal values [Ind−]/[HInd]"? $\endgroup$
    – Arshiya
    May 4 '19 at 4:26
  • $\begingroup$ "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). $\endgroup$
    – andselisk
    May 4 '19 at 4:43
  • 1
    $\begingroup$ @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. $\endgroup$
    – Poutnik
    May 4 '19 at 4:54
  • 1
    $\begingroup$ In fact I knew it I may have been corrupted by MathJax formatting terrorism :-) $\endgroup$
    – Poutnik
    May 4 '19 at 4:58
  • 1
    $\begingroup$ 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. $\endgroup$
    – Poutnik
    May 4 '19 at 5:01

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