# Slope ≠ 1 in Henderson-Hasselbalch graph for determination of pKa

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?

• What exactly did you measure with spectrophotometer, and can you post a picture of the plot? Feb 27 at 16:57
• The spectrophotometer was used to measure the absorption at 592nm of bromophenol blue at different pH's
– Owen
Feb 27 at 17:47
• So how do you transform the measured absorption at $\pu{592 nm}$ into two values, namely $\ce{ [HI]}$ and $\ce{[I-]}$ ? Feb 27 at 17:51
• Your term of $\log \left(\frac{\ce{I-}}{\ce{HI}}\right)$ is misleading. Feb 27 at 17:51
• I wasn't sure about your techniques. But according to the graphs and the equation, when $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = 0$, $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$. Feb 27 at 18:12

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 $$\mathrm{p}K_\mathrm{a}$$.