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I was wondering which $\mathrm{p}K_\mathrm{a}$ to use when calculating the ratio of $\ce{HZ}$ to $\ce{Z-}$ of amino acids, the Henderson–Hasselbalch formula used:

$$\mathrm{pH} = \mathrm{p}K_\text{a} +\log\frac{[\ce{Z-}]}{[\ce{ZH}]}$$

Because most amino acids have two $\mathrm{p}K_\mathrm{a}$'s, one for the amino group and one for the carboxyl group (and sometimes a third $\mathrm{p}K_\mathrm{a}$ for the side chain). For example, leucine:

$$\begin{array}{cc} \hline \text{Residue} & \mathrm{p}K_\mathrm{a} \\ \hline \ce{NH3} & 9.74 \\ \ce{COOH} & 2.33 \\ \hline \end{array}$$

Let's say I want to know the ratio of $\ce{HZ}$ to $\ce{Z-}$ for a $\mathrm{pH}$ of $1.00$ and $11.00$, which $\mathrm{p}K_\mathrm{a}$ should be used in the Henderson–Hasselbalch equation and why?

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    $\begingroup$ Both? I don't see why you can't use both at the same time. When $\mathrm{pH} = \mathrm{p}K_{\mathrm{a}}$, the corresponding dissociation is at 50%. When they're not that close, then one form will greatly predominate over the other... $\endgroup$ – Zhe Oct 26 '16 at 17:14
  • $\begingroup$ how would you calculate the Hz/Z- ratio based on the two Pka's given at Ph = 1 e.g.? Offcourse it's possible to use both the Pka's in two different calculations but how would you combine these to get the Hz/Z- ratio? @Zhe $\endgroup$ – KingBoomie Oct 26 '16 at 17:37
  • $\begingroup$ Check out this page $\endgroup$ – Karsten Theis Mar 17 at 20:41
  • $\begingroup$ By a rule of thumb, pK farther than 3 units from pH can be omitted, as the component is ( at 99.9% level ) exclusively present in just of the A/B forms. $\endgroup$ – Poutnik Sep 15 at 10:57
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There are two approaches.

  1. The two groups are treated as independent. You would calculate the ratio of protonated and deprotonated group one using the pKa for that group. Then, you would repeat that for the second group. Independent treatment would be a good approach, for example, to estimate the average charge of the hundreds of side chains in a typical protein that have a pKa near physiological pH.
  2. The two pKa values are obtained experimentally from a titration.$$\ce{AH2 <=> AH- + H+ <=> A^2- + 2 H+}$$ Here, the $\ce{AH-}$ species could be ambiguous if the pKa values are close to each other (group 1 deprotonated and group 2 protonated or vice versa). Or, for a case like hydronium, water and hydroxide where the same atom can accept two protons, $\ce{AH-}$ would represent a single defined species. You can get the ratios of the concentrations from the pH and the respective pKa ($\ce{AH2}$ vs $\ce{AH-}$ via $K_1$, and $\ce{AH2}$ vs. $\ce{AH-}$ via $K_2$), and turn that into fractions of the total for each species. You can use that for a weighted average to calculate the average charge, for example.

Let's say I want to know the ratio of HZ to Z− for a pH of 1.00 and 11.00, which pKa should be used in the Henderson–Hasselbalch equation and why?

This depends what HZ and Z- refer to. With two ionizable groups, there are really four species, $\ce{HZYH, HZY^-, ^-ZYH and ^-ZY-}$. If the pKa values are far apart, one of the mixed species would be very unlikely in aqueous solution (in your example, protonated carboxylate with deprotonated amino group). At pH 1.00, the amine would be 100% protonated, and the carboxylic acid would be mostly protonated (comparing its pKa of 2.33 to a pH of 1.00). At pH 11, the carboxylic acid would be 100% deprotonated, and the amine would be mostly deprotonated (comparing its pKa of 9.74 to a pH of 11.00).

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If you want to calculate with $\mathrm{pH} = 1$ that means the solution is predominantly acidic hence use $\mathrm{p}K_\mathrm{a} = 2.33,$ but if you want to calculate with $\mathrm{pH} = 11,$ the solution is predominantly basic so you solve with $\mathrm{p}K_\mathrm{a} = 9.74.$

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  • $\begingroup$ Unfortunately, this doesn't seem to add much to the already existing answer. Also, please note that the official language of Chemistry.SE is English. $\endgroup$ – andselisk Sep 15 at 10:05
  • $\begingroup$ @andselisk Sometimes, the perfect answer is not the one, where you cannot add anything, but the one where you cannot omit anything. Karsten's answer is very well elaborated, but less experienced chemist can get lost. Oo65's answer is rather laconic, and that is its added value. $\endgroup$ – Poutnik Sep 15 at 10:59
  • $\begingroup$ @Poutnik I agree, and that's why I didn't delete or converted this answer to a comment. The thing is, this would make much more sense if this were the first answer and Karsten's were the second. Currently it just looks redundant despite its simplicity. $\endgroup$ – andselisk Sep 15 at 11:10

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