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


> 
> 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.


