4.0ml di KOH 0.50M
+
150ml di glicina 0.03M

Totale :
154ml di KOH = 0.50M*4/154 = 0.013M
Glicina = 0.03M*150/154 = 0.029M

-OOC-CH(R)-NH3+ + KOH --> -OOC-CH(R)-NH2 + H2O + K+
_____0.029________0.013_________0__________/___/_____inizio
0.029-0.013_________0__________0.013________/___/_____fine

___0.016_______________________0.013

tampone -OOC-CH(R)-NH3+ / -OOC-CH(R)-NH2
con
[H+] = Ka2 * [-OOC-CH(R)-NH3+]/[-OOC-CH(R)-NH2 ]

1,67*10^-10 * 0.016 / 0.013 = 2.06*10^-10

pH = 9.69