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 $\mathrm{pH}$.




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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.


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