Amino acids are the building blocks of proteins, and can act as both a Brønsted

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. 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:

Glycine - H2N––CH2––COOH

cation - +H3N––CH2––COOH

zwitterion - +H3N––CH2––COO–

anion - H2N––CH2––COO–

(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?

Explanation / Answer

a) H2N-CH2-COOH (aq) <----> H2N-CH2COO-(aq) + H+ (aq) is equilibrium corresponding Ka

   H2N-CH2-COOH(aq) + H2O (l) <----> +H3N-CH2-COOH (aq) + OH- (aq) is equilibrium correspodning Kb

b) K = Ka x Kb       since we need both COOH losing H+ and NH2 gaining H+

K = (4.5 x 10^-3) x ( 6x10^-5) = 2.7 x 10^-7

c)   at isolelctric point i.e pI = (pka + pka ofNH3+) /2   we have positive charge of NH3+ = negative charge of COO-

pka = -log Ka = -log ( 0.0045) = 2.347 , pkb = -log ( 6x10^-5) = 4.22 , pka of Nh3+ = 14-4.22 = 9.78

isolelectric point = ( 2.347+9.78) /2 = 6.06

hence at pH = 7.2 which is higher than isolectric point we have more of COO- than COOH and more of Nh2 than of Nh3+

hence anionic form i.e H2N-CH2-COO- is major form

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