# Amphoteric amino acid titration

Glycine $$(\ce{NH2CH2COOH})$$ is an amino acid with two $$\mathrm{p}K_a$$ values $$(\mathrm{p}K_{a,1}=2.0, \ \ce{COOH};\ \mathrm{p}K_{a,2}=10.0, \ \ce{NH2})$$.

(a) If $$\pu{0.01 mol}$$ of this amino acid is used to prepare one $$\pu{1 L}$$ solution. Calculate the $$\mathrm{p}\ce{H}$$ and the $$\ce{NH3+CH2COOH}$$ concentration in the solution.

(b) If $$\pu{10 mL}$$ of the above glycine amino acid solution is mixed with $$\pu{90 mL}$$ of basic buffer system. If the concentration of the $$\ce{NH2CH2COO-}$$ was identified to be $$\pu{1.05 \cdot 10^{-4} M}$$ indicate the $$\mathrm{p}\ce{H}$$ of the glycine solution.

(c) If $$\pu{10 mL}$$ of strong base $$\ce{NaOH}$$ ($$\pu{0.01 M}$$) was added to solution in (a) what is the final $$\mathrm{p}\ce{H}$$.

For part (a):
I used $$[\ce{H+}]=\sqrt{10^{-10}\cdot0.01} =10^{-6}$$.
Then I used $$\mathrm{p}\ce{H} =14-\log(10^{-6})$$ to get a $$\mathrm{p}\ce{H}$$ of $$8$$.
While my $$\ce{NH3+CH2COOH}$$ is $$\pu{10^{-6} L}$$.

For part (b):

I used the Henderson-Hasselbalch equation for a basic buffer $$\mathrm{p}\ce{H} =\mathrm{p}K_a +\log\left(\frac{\ce{[NH2CH2COO- ]}}{\ce{[NH2CH2COOH]}}\right)$$ I got a $$\mathrm{p}\ce{H}$$ of $$10.97$$. But I'm unsure of my answer.

For part (c): I suspect that it is a basic buffer but I'm not sure which value to use. could anyone help me out with this.

• note that pH is written with the "p" in lowercase because it means "p function" that is $-logx$ – G M May 2 '14 at 9:14

As in aqueous solution: $$\ce{NH3+CH2COOH_{[b]}<=> NH3+CH2COO-_{[z]} + H+_{[x]}\\NH3+CH2COO-_{[z]} <=> NH2CH2COO-_{[a]} + H+_{[x]}}$$ So, $$10^{-2}=\frac{[z][x]}{[b]}\\10^{-10}=\frac{[a][x]}{[z]}\\\implies 10^{-12}=[x]^2$$ Notice that a=b as the solution is always neutral; also z is a neutral compund
As, $$[H^+]=10^{-6}\implies p^H=6$$ $$\implies[b]=100[z][x]=10^{2-2-6}=10^{-6}$$
You should have done $p^H=-log[H^+]$, your [b] is correct.
For part (b), yes, using Henderson-Hasslebalch equation: $$p^H=10+\log\left(\frac{1.05\times10^{-4}}{10^{-6}}\right)\approx 10.97$$
In basic solution, the second equilibrium shifts well(as a result the first one too) to the right.As NaOH is a strong acid, we can neglect any other bases.In presence of NaOH, $$\ce{NH3+CH2COOH_{[b]} + 2NaOH_{[k]}<=> 2NH2CH2COONa_{[a]} + 2H2O_{[y]}}$$ In millimoles: $$\begin{array}{c|c|c|c|c} &entity&b&k&a&y\\\hline &initial&10&.1&0&0\\\hline &final&9.95&0&.1&.1+110(original)\\\hline \end{array}$$ Now you can similiarly apply agin the Henderson-Hasslebalch equation: $$p_H=12+log(.1/9.95)\approx 10.01$$
Note that I am treating the compund as a monoprotic acid of $pK_a =10(-NH_3^+)+2(-COOH)=12$