Determine the concentrations of the three ionic forms of glycine present if 1.0 mol of glycine is used to make 1.00 L of aqueous solution. $\mathrm{p}K_1 = 2.34, \mathrm{p}K_2 = 9.60$. Do you need to make any other assumptions to simplify the calculation?

Source: David Ball's Physical Chemistry.

I tried to solve this problem but I think it's quite flaw and naive. So I want to put it here if anybody have an improvement.

The question asked us to determine the concentrations of 3 ionic forms of glycine in aqueous solution. Let me call

  • $\ce{Gly}$ = $\ce{NH2-CH2-COO^-}$ at high alkalinity
  • $\ce{GlyH}$ = $\ce{NH3^+-CH2-COO^-}$ at neutral
  • $\ce{GlyH2}$ = $\ce{NH3^+-CH2-COOH}$ at high acidity

I can calculate the isoelectric point $\mathrm{pI} = (\mathrm{p}K_1 + \mathrm{p}K_2)/2 = 5.97$

If I assume "used to make $\pu{1.00L}$ of aqueous solution" means putting glycine into water, $T = \pu{25^\circ C}$, thus $\mathrm{pH}$ must be 7.00. However, the isoelectric point is 5.97. That means the solution cannot contain completely GlyH but also an $x$ amount of Gly.

Therefore, I use the Henderson-Hasselbalch equation to determine the amount of Gly and GlyH in the solution

$$ \begin{align} \mathrm{pH} &= \mathrm{p}K_2 + \log\frac{[\text{Gly}]}{[\text{GlyH}]}\\ 7 &= 9.6 + \log \frac{x}{1-x}\\ x &= 0.0025 M = [\text{Gly}] (\text{rounded})\\ [\ce{GlyH}] &= \pu{0.9975 M}\\ \text{Obviously } [\ce{GlyH2}] &= \mathrm{0 M} \end{align} $$

I think the wrong with my solution is to assume a fixed value of water pH. In common sense I think when glycine is put into water, it would drag the pH of the solution down a little bit, but not over pI, it means $\mathrm{p}K_1 < \mathrm{pI} < \mathrm{pH} < 7 < \mathrm{p}K_2$. But I really don't know how to find the value. Please guide me because I tried my best but really don't know how to solve this.

  • $\begingroup$ "[GlyH2] obviously is 0 M" - Why? $\endgroup$ Nov 21, 2015 at 12:07
  • $\begingroup$ Because pH is higher than the pI. I don't think GlyH2 would be presented above pI. $\endgroup$ Nov 21, 2015 at 14:17


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