5
$\begingroup$

I have done an experiment where I dissolved four different substituted phenols in acidic, basic, and buffer solutions, and recorded the UV-Vis spectra for each. I'm supposed to pick a wavelength where the absorbances for the protonated, normal and dissociated forms are very different. I then use these equations given to me to calculate the $\mathrm{p}K_\mathrm{a}$ of that phenol:

$$x = \frac{A_\text{buffer} - A_\text{acid}}{A_\text{base} - A_\text{acid}}\tag{1}$$

where $x$ is the mole fraction; $A$ is the absorbance in corresponding medium, and

$$\mathrm{p}K_\mathrm{a} = \mathrm{pH} - \log\left(\frac{1-x}{x}\right)\tag{2}$$

I've calculated the $\mathrm{p}K_\mathrm{a}$ values of each phenol, but none of them are close to literature values. I'm not sure what I've done wrong. Can anyone help explain these equations and any ways I could have messed up?

$\endgroup$
5
$\begingroup$

To quote the key points of an easy to read publication by Salgado and Vargas-Hernández (doi 10.4236/ajac.2014.517135, open access):

All starts with the dissociation equilibrium between the acid $\ce{HA}$ and its anion $\ce{A-}$ for which you write

$$\ce{HA + H2O <=> H3O+ + A-}$$

By consequence of this, the recorded total absorbance $A_\mathrm{t}$ for system, assuming only the acid and its anion absorb light in the region inspected, equates to

$$A_\mathrm{t} = d(c_{\ce{HA}} \cdot \varepsilon_\ce{HA} + c_\ce{A-} \cdot \varepsilon_\ce{A-}) = d \cdot c_\mathrm{t} \cdot \varepsilon_\mathrm{t}$$

with $c_\mathrm{t}$ as total concentration; $\varepsilon$, the absorption coefficient; and $d$, the optical path length.

For one, you record a spectrum at low pH to convert all of your sample into the acid form ($c_\mathrm{t} = c_\mathrm{HA}$, and $c_\ce{A^-} = 0$) where all absorption is due to the acid. Equally, you record a spectrum at a pH this high that all or your sample is deprotonated (then $c_\mathrm{t} = c_\mathrm{A^-}$). You need then spectra between the two extrema, with some of $\ce{HA}$ in presence of some of $\ce{A-}$, because the proportion of the concentrations of acid to its anion may be expressed by

$$\frac{c_\ce{A-}}{c_\ce{HA}} = \frac{[\ce{A-}]}{[\ce{HA}]} = \frac{A_\mathrm{t} - A_\ce{HA}}{A_\ce{A-} - A_\mathrm{t}}$$

This is why you search for the isobestic point, where the absorption spectra recorded do not vary while varying the pH value of your analyte, and where $\varepsilon_\ce{HA} = \varepsilon_\ce{A^-}$.

enter image description here

(credit: doi 10.4236/ajac.2014.517135)

Because the equilibrium between the acid and its anion may be described by

$$\log{K_\mathrm{a}} = \log[\ce{H3O+}] + \log{\frac{[\ce{A^-}]} {[\ce{HA}]}} $$

and the definition of $pK_\mathrm{a} = - \log{K_\mathrm{a}}$ you may substitute absorptions by concentrations to yield

$$\log{\left(\frac{A_\mathrm{t} - A_\ce{HA}} {A_\ce{A-} - A_\mathrm{t}}\right)} = \mathrm{pH} - pK_\mathrm{a} $$

Hence, plotting $\log{\left(\frac{A_\mathrm{t} - A_\ce{HA}} {A_\ce{A-} - A_\mathrm{t}}\right)}$ in function of $\mathrm{pH}$ yields a slope with an intercept of $y = - pK_\mathrm{a}$.

enter image description here

(credit: loc. cit.)


Addition: With WebPlotDigitizer it was possible to convert above mentioned figure 8 into numbers, separate for the basic (b), the acidic (a) and the intermediate / about neuter run at pH = 7.7 (n). (The result, a .csv file, is deposit here.) For a wavelength of about $550\,\mathrm{nm}$ the readout of Abs equals to 1.9935 (b), 0.6502 (n) and 0.2241 (a), and the computation

$$ pK_\mathrm{a} = 7.7 - \log_{10}{\frac {0.4534 - 0.2984} {1.2121 - 0.4534}} $$ $$ pK_\mathrm{a} = 7.7 - \log_{10}{\frac {0.1550} {0.7587}} $$ $$ pK_\mathrm{a} (550\,\mathrm{nm}) = 8.39 $$

By similar token, $pK_\mathrm{a} (604\,\mathrm{nm}) = 8.20$ (with Abs of 1.9935 (b), 0.6502 (n), and 0.2241 (a)); or $pK_\mathrm{a} (432\,\mathrm{nm}) = 8.51$ (with Abs of 0.3411 (b), 0.8198 (n), and 0.8940 (a)).

To put this into perspectivive, the arithmetical mean of the four methods compared in Salgado's paper equals to $pK_\mathrm{a} = 8.277$. Possibly, the difference between their result and the «recuperation» here were smaller if either a) WebPlotDigitizer were used with a lesser data increment, b) using an interpolation / smoothing function among the data extracted by the program, c) figure 7 were reconstructed from their figure 4.

$\endgroup$
  • $\begingroup$ Thank you for the clarification. I'm still wondering about the second equation I mentioned in my question. Is pH just the pH of my buffer solution? The article you included has a similar equation where it is the pH of the intermediate solution, but that equation was used over all wavelengths. I'm required to look only at the absorbances at a specific wavelength though, so I'm a little bit confused still. $\endgroup$ – cjperkie Nov 14 at 19:14
  • 2
    $\begingroup$ As far as I remember, you measure the pure acid and the pure base and then the buffered solutions. Those buffered solutions have a defined pH value and these are the ones you take. $\endgroup$ – pH13 - Yet another Philipp Nov 17 at 7:53
  • 2
    $\begingroup$ We do this practical with our students. Either they use the nominal value of the pH of the buffered stock solutions or they use a pH-meter to determine the values of the pH. With rare exceptions both methods yields statistically compatible results. $\endgroup$ – PAEP Nov 17 at 12:26
  • 1
    $\begingroup$ @cjperkie I equally think that you use the pH value of your buffer solution including your phenoles. Of course, be sure to provide enough buffer capacity, otherwise the addition of the phenols may alter significantly the pH value of this solution in common. (Which may be checked by a pH electrode.) $\endgroup$ – Buttonwood Nov 18 at 10:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.