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Graph 1

The $\mathrm{p}K_\mathrm{a}$ of indicators can be determined via plotting a graph of $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right)$ vs $\mathrm{pH}$ with the help of a spectrophotometer, from the equation $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = \mathrm{pH} - \mathrm{p}K_\mathrm{a}$.

From this, the graph should have a gradient of 1, and $x$-intercept = -($y$-intercept) = $\mathrm{p}K_\mathrm{a}$. However, my graph has a slope of ~0.8 which means my $x$-intercept ≠ -($y$-intercept) so which value do I use as my calculated $\mathrm{p}K_\mathrm{a}$?

From the papers I've read, most if not all of them use the intercept with the $\mathrm{pH}$ axis instead of the $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right)$ axis in this situation, why is this?

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  • $\begingroup$ What exactly did you measure with spectrophotometer, and can you post a picture of the plot? $\endgroup$
    – S R Maiti
    Feb 27 at 16:57
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    $\begingroup$ The spectrophotometer was used to measure the absorption at 592nm of bromophenol blue at different pH's $\endgroup$
    – Owen
    Feb 27 at 17:47
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    $\begingroup$ So how do you transform the measured absorption at $\pu{592 nm}$ into two values, namely $\ce{ [HI]}$ and $\ce{[I-]}$ ? $\endgroup$
    – Maurice
    Feb 27 at 17:51
  • $\begingroup$ Your term of $\log \left(\frac{\ce{I-}}{\ce{HI}}\right)$ is misleading. $\endgroup$ Feb 27 at 17:51
  • $\begingroup$ I wasn't sure about your techniques. But according to the graphs and the equation, when $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = 0$, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} $. $\endgroup$ Feb 27 at 18:12
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OP's question: From the papers I've read, most if not all of them use the intercept with the $\mathrm{pH}$ axis instead of the $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right)$ axis in this situation, why is this?

Answer for this question is easy. According to the graphs plotted using the equation $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = \mathrm{pH} - \mathrm{p}K_\mathrm{a}$ (Henderson-Hasslebalch equation), when $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = 0$ (meaning $[\ce{In-}] = [\ce{HIn}]$), then $\mathrm{pH} = \mathrm{p}K_\mathrm{a} $. Therefore, the $x$-value where the straight-line crosses the $x$-axis is the best value for $\mathrm{p}K_\mathrm{a} $.

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    $\begingroup$ It is not exactly a straight line (experimental data, no information on error bars). Nice explanation why to focus on the x-intercept. $\endgroup$ Feb 28 at 17:06

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