Question

Given that pK1 = 3.0 and pK2 = 8.0 for an amino acid X. Total concentration of amino acid X in solution is 10 mM.

Calculate the concentrations of protonated form (P), dipolar form (Z) and de-protonated form (D) of amino acid X at pH 2, and 10.

Answer 1

Given that the pki of the amino acidis 3 and ply is 8. The General structure of an amino acid I Night - & - Coon R So, depending on the pk valores the amino acid will be ionized ALPH2, the structure will be a tots 14 Nugt- & -cooh (Protonated) or P And at pt 10, structure will be NH- -coo (de -protonated or And at the isoelectric pH, that means. (pk it Pk 2/2 - 3+8/2) the structure will be 1 2 .5.5 Nut - G-coo (Dipolar) or Z. So now the concentration of the protonated form, dipalaw for and deproto nated form at pH=2 using the Henderson-Hassel batch equation, pH pkapt log [?]/[p] -> 2 = 3+ log [2] / [e] >> -12 lug [z]/[] => [2] / [+] = 0.1 H

w is lomu => [2] 0.1 [] Since the total concentration of amino acid'y' is that means & [2] + [p] = 10 min => 0 []+ [1] -10 >> P - 1 => [P]- 9.09 9.4 mm [2]- 0.9 mm - And [o] will be zero. AtpH 10 pH = playt log [D]/[2] 27 10 = 8 + log [b]/[2] o) 22 -[]/[z] => 100 = [0] /[2] => [p] - 100 [2] since [D]+ [2] = 10 mm > 100 [2] + [2] = 10 => [2] = 10/101 [2] = 0.0g9mm 10-0.ogg a 9.90 mm