The amino acid glycine is often used as the main ingredient of a buffer in biochemical experiments. the amino group of glycine, which has a pka of 9.6, can exist either in the protonated form (—nh1 3) or as the free base (—nh2), because of the reversible equilibrium ronh1 3 δ ronh2 1 h1 (a) in what ph range can glycine be used as an effective buffer due to its amino group? (b) in a 0.1 m solution of glycine at ph 9.0, what fraction of glycine has its amino group in the —nh1 3 form? (c) how much 5 m koh must be added to 1.0 l of 0.1 m glycine at ph 9.0 to bring its ph to exactly 10.0? (d) when 99% of the glycine is in its —nh1 3 form, what is the numerical relation between the ph of the solution and the pka of the amino group?

a. The effective range of glycine as buffer is 10,6 - 8,6

b. The fraction that have NH₃⁺ group is 0,8

c. You need to add 10,4 mL 5M KOH

d. he numerical relation between the pH of the solution and the pKa of the amino group is 1,26

Explanation:

The glycine buffer equilibrium is:

RONH₃⁺ ⇄ RONH2 + H⁺

a. The effective range in a buffer is pka ± 1

Thus, effective range of glycine as buffer is 10,6 - 8,6.

b. This question could be answered using Henderson-Hasselbalch equation:

pH = pka + log [A⁻]/[HA]

For 0,1M glycine at pH = 9,0

9,0 = 9,6 + log [A⁻]/[HA]

[HA] + [A⁻] = 0,1 M

0,08 M = [HA]; 0,02 M = [A⁻]. The fraction that have NH₃⁺ group is 0,08 M/0,1 M = 0,8

c. Again, using Henderson-Hasselbalch formula:

10,0 = 9,6 + log [A⁻]/[HA]

2,51 = [A⁻]/[HA]

As you have 0,1 mol of glycine:

[HA] = 0,028 moles

When pH was 9,0, you were 0,8 moles, thus, you need to spend 0,08 mol - 0,028 mol = 0,052 mol

0,052 moles that comes from 5M KOH:

RONH₃⁺ +KOH → RONH2 + H₂O

Thus, the volume of 5M KOH you need to add is:

0,052 moles* (1L/5mol) = 0,0104L ≡ 10,4 mL 5M KOH

d. The pH when 99% of glycine is in its NH₃⁺ form is:

pH = 9,6 + log [1%]/[99%]

pH = 7,6

The relation between pH and pka is: 9,6/7,6 = 1,26

I hope it helps!

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