Answer
A. Between pH 8.6 and pH 10.6
B. 1/5 deprotonated and 4/5 protonated
C. 10 mL
D. pH = pKa - 2
Work Step by Step
A. *A buffer functions best between one pH unit below and one pH
unit above its own pKa. The pKa of glycine is 9.6 given. Through ionization of its amino group, glycine acts as a good buffer between pH 8.6 and pH 10.6.
Answer: Between pH 8.6 and pH 10.6
B. *Use Henderson-Hasselbalch equation
pH = pKa + log ([A-]/[HA])
9.0 = 9.6 + log ([A-]/[HA])
-0.6 = log ([A-]/[HA])
([A-]/[HA]) = 10^0.6 = 0.25
*Result 0.25 corresponds to ratio of 1/4, indicating the amino group of glycine is about 1/5 deprotonated and 4/5 protonated at pH 9.0.
Answer: 1/5 deprotonated and 4/5 protonated
C. *Amino group is about 1/5, or 20%, deprotonated at pH 9.0 from Answer B.
*Moving from pH 9.0 to pH 9.6 (where amino group is 50% deprotonated), 30%, or 0.3, of glycine is titrated. We can now calculate percent protonation at pH 10.0 using Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
10.0 = 9.6 + log ([A-]/[HA])
0.4 = log ([A-]/[HA])
([A-]/[HA]) = 10^0.4 = 2.5 = 5/2
*This 5/2 ratio indicates that glycine is 5/7, or 71%, deprotonated at pH 10.0; this is an additional 21%, or 0.21, deprotonation above that found at the pKa (50%, or 0.5).
*Total fractional deprotonation moving from pH 9.0 to 10.0 is:
0.30 + 0.21 = 0.51
*This deprotonation corresponds to:
0.51 x 0.1 mol = 0.05 mol of KOH
*The volume of 5 M KOH solution required is:
(0.5 mol) / (5 mol/L) = 0.01 L, or 10 mL.
Answer: 10 mL
D. *UseHenderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
pH = pKa + log ([-NH2]/[-NH3+])
pH = pKa + log (0.01/0.99)
pH = pKa + (-2) = pKa - 2
*In general, any group with ionizable hydrogen is almost completely protonated at pH at least two pH units below its pKa value.
Answer: pH = pKa - 2
