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I'm trying to help a friend of mine with this chemistry question:

What is the pH of a $1.0M$ solution of glycine?

They're given that the pKa of COOH is $2.4$ and the pKa of the amine group is 9.6. The solution they were given says the pH is just $\frac{2.4+9.6}{2} = 6.1$, but I don't see any reason why this should be true especially since this answer initially appears independent of the given concentration of glycine.

Now if I try to solve this the long way: If I initially just have a solution of glycine, I believe it would function like a weak acid, and hence the equation is:

$$Ka = \frac{[H^+][A-]}{[HA]}$$

or

$$10^{-9.6} = \frac{x^2}{1-x} \approx x^2$$

Thus pH = $4.8$

At the very least, I found this document: https://www.ch.ntu.edu.tw/~ccchan/course/genchem2003/Notes/final_ans.pdf

Which showed that the pH of a $0.05 M$ solution of glycine has a pH of $5.54$, so with a much higher concentration of glycine functioning like a weak acid, my pH should be less than $5.54$ (and hence definitely less than the $6.1$ the professor put as his solution). Am I missing something?

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    $\begingroup$ The mean value of pKas is not pH of solution in general, but an isoelectric point. $\endgroup$ – andselisk Aug 28 '17 at 0:55
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    $\begingroup$ @andselisk Yea I'm familiar with that. That's why I didn't understand why the professor claimed that was the pH of a 1.0M solution $\endgroup$ – Brenton Aug 28 '17 at 0:58
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I think the professor is right. I would solve it like any other acid-base problem. Glycine (G, neutral as zwitterion) can either protonate to HG+; or deprototonate to G-. The mass balance and charge balance equations are:

C$_G$=1.0=[HG+]+[G]+[G-] (Mass balance for glycine)

[H+]+[HG+]=[OH-]+[G-] (Charge balance)

Given these two equations, and the three for $K_{a1}$, $K_{a2}$ and $K_{w}$, I could solve exactly the system of five equations with five unknowns. But that would be overkill, since we can do some approximations:

  • Since G is both a (not so) weak acid and a weak base, we can assume that it doesn't dissociate that much to HG+ or G-. So, the predominant form would be G, which (given the $pK_a$'s) is predominant in the 3.5 to 8.5 pH range (aprox, it depends on the errors that we tolerate).

  • In this pH range, and given the high glycine concentration and $pK_a$ values, [HG+]>>[H+] and [G-]>>[OH-]; so that, from the charge balance:

[HG+]=[G-]

This equality solves the problem since, combining the equations for $K_{a1}$, $K_{a2}$, we get:

$K_{a1}K_{a2}=\frac{[G-][H+]^2}{[HG+]}=[H+]^2$ so $[H+]=\sqrt{K_{a1}K_{a2}}$

$pH=\frac{1}{2}log(pK_{a1}+pK_{a2})=6.0$

You can check that at this pH, [G]=1,[HG+]=$10^{-3.2}$,[G-]=$10^{-3.6}$, so that the approximations [HG+]>>[H+] and [G-]>>[OH-] are valid.

(For more details see 6.7.5 at https://chem.libretexts.org/Textbook_Maps/Analytical_Chemistry_Textbook_Maps/Map%3A_Analytical_Chemistry_2.0_(Harvey)/06_Equilibrium_Chemistry/6.07%3A_Solving_Equilibrium_Problems)

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  • $\begingroup$ Maybe I missed it, but where in solving this problem did you use that the concentration is 1.0M? Because if that doesn't factor into solving this, then we're literally saying that the pH is 6.0 regardless of how much glycine is used which seems problematic. The other question I would have then is exactly what is wrong with the way I solved it which gave 4.8 $\endgroup$ – Brenton Aug 29 '17 at 5:47
  • $\begingroup$ @Brenton .The concentration is there in the mass balance equation. But it does not appear in the solution due to approximations I made. Check the more exact equation in the link provided, and if you calculate the pH with it, you'll see that the aaproximation (pH=6) is veeeery good. Anyway, the fact that the pH is independent of total concentration (as long as C does not get too small) is not problematic, but expected. It is quite common: pH of a buffer, pH of amphoteric form (which is this case),... $\endgroup$ – Zamu Aug 29 '17 at 9:44
  • $\begingroup$ @Brenton . About the way you solved it. You considered zwiterionic glycine (+)H3N-CH-COO(-) as a weak acid (from the ammonium group); but it is also a weak base (from the carboxylate). You assumed (correctly) that it is an stronger acid than base (that is the reason the pH is below 7). But you did not considered that -COO(-) is a stronger base than water; so as soon as H+ is liberated, (+)H3N-CH-COOH starts forming. You just cannot ignore the other pKa. The overall reaction would be: 2HG -> H2G(+) + G-, which has K=10^(-7.2). So, at equilibrium [H2G+]=[G-]=10^(-3.6), which means [H+]=10^(-6) $\endgroup$ – Zamu Aug 29 '17 at 10:00
  • $\begingroup$ I have another question. Why do we consider only the zwitterion $\ce{NH3+-CH-COO-}$ but not the neutral molecule $\ce{NH2-CH-COOH}$ so there would be altogether $4$ species? Is it that the zwitterion is much more abundant than the neutral molecule at any pH once amino acids dissolve in water? $\endgroup$ – Zhuoran He Jan 14 '18 at 15:46
  • $\begingroup$ @ZhuoranHe Both $\ce{NH2-CH-COO}$ and $\ce{NH3+-CH-COO-}$ are neutral, they are just different tautomeric isomers of the same species. From the point of view of acid-base equilibria, the zwitterion and non-zwitterion forms of neutral glycine are the same species, since both have lost a proton from the cationic $\ce{NH3+-CH-COOH}$. $\endgroup$ – Zamu Jan 14 '18 at 16:43

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