Question

Amino acids are the building blocks of proteins, and can act as both a Brønsted acid and a Brønsted base through intramolecular proton transfer (see Chapter 16, pg. 709 of the textbook). The simplest amino acid known is glycine, NH2CH2CO2H (Ka = 4.5 ×10–3 and Kb = 6.0 ×10–5 ), and it can exist in three forms in equilibrium with one another:

H2N––CH2––COOH +H3N––CH2––COOH +H3N––CH2––COO– H2N––CH2––COO–

Glycine cation zwitterion anion

(a) Write the equilibria corresponding to Ka and Kb of glycine.

(b) Estimate the value of the equilibrium constant, K, for the intramolecular proton transfer to form the zwitterion in glycine: H2N––CH2––COOH (aq) K ? +H3N––CH2––COO– (aq)

(c) If the pH of an aqueous solution containing glycine is 7.2, in what form is glycine most abundant?

Answer 1

(a) Write the equilibria corresponding to Ka and Kb of glycine.

Ka = [H2N––CH2––COO–][H+]/[ H2N––CH2––COOH]

Kb= [+H3N––CH2––COOH][OH-]/[ H2N––CH2––COOH]

(b) Estimate the value of the equilibrium constant, K, for the intramolecular proton transfer to form the zwitterion in glycine: H2N––CH2––COOH (aq) K ? +H3N––CH2––COO– (aq)

H2N––CH2––COOH (aq) ----à H3N––CH2––COO– (aq)

K = [H3N––CH2––COO–]/ [H2N––CH2––COOH]

K = Ka*Kb/Kw=4.5 ×10–3* 6.0 ×10–5/1*10^-14= 2.7*10^7

The value of K for the zwitterion in glycine is very large measn it will exist predominantly as Zwitterions.

(c) If the pH of an aqueous solution containing glycine is 7.2, in what form is glycine most abundant?

When pH>pKa then it will deprotonate means H3N––CH2––COO– predominant, while When pH<pKa then the H+ present on the amino acid thus +H3N––CH2––COOH present.

Ka = 4.5 ×10–3

pKa = -log Ka

pKa = 2.35

If the pH of an aqueous solution containing glycine is 7.2 means pH>pKa then it will deprotonate means H3N––CH2––COO– predominant.