homework - 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 $0.01~\mathrm{mol}$ of this amino acid is used to prepare one $1~\mathrm{L}$ solution. Calculate the $\mathrm{p}\ce{H}$ and the $\ce{NH3+CH2COOH}$ concentration in the solution.

  
  


• 
  

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

  
  


• 
  
  

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

  
  

---

For part (a)
I used $[H^{+}]=\sqrt{10^{-10}\cdot0.01} =10^{-6}$.
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 $10^{-6}~\mathrm{L}$.

---

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.