by pH = 1 (pK, + pK2). The isoionic pH of a polyprotic species is the pH that wo

Question

by pH = 1 (pK, + pK2). The isoionic pH of a polyprotic species is the pH that would exist in a solution containing only the ions derived from the neutral polyprotic species and from H,O. For a of diprotic amino acid whose amphiprotic form is neutral, the isoionic ate pH is found from [H V(K,K,F KK)K, F), where dic is the formal concentration of the amino acid. d 10-D. Calculate the pH of a 0.010 M solution of each amino acid in the form drawn here. 3, H,N NH2 NH2 CH2 CH2 CH2 H2NCHCOS NH2 CH2 0 CH2 CH2 (a)H3NCHCO, (b) H,NCHC0% (c) Cysteine Arginine Glutamine 195

Explanation / Answer

10-D
a) Use Henderson's Hasselbalch equations,
PH = Pka + log10 [A-] / [HA] ;
  Pka = 9.1 ; Ka = 7.9432*10-10
HA ---------> H+ + A-   
At equilibrium concentrations will be,
[HA] = 0.010-X ; H+ = X = A-
7.9432*10-10 = X2 / [0.010-X]
7.9432*10-12-7.9432*10-10X = X2
X2 + 7.9432*10-10X - 7.9432*10-12 = 0
X = 2.8179*10-6 M = [H+] = [A-]
[HA] = 9.9971*10-3
PH = 9.1 + log10 [2.8179*10-6 / 9.9971*10-3] = 5.55

(b) Pka = 8.18 ; Ka = 10-8.18 = 6.6069*10-9
6.6069*10-9 = X2 / [0.010-X]
6.6069*10-11 - 6.6069*10-9X= X2
X2 + 6.6069*10-9X - 6.6069*10-11 = 0
X = 8.1249*10-6 M = [A-]
[HA] = 0.010-(8.1249*10-6) = 9.9918*10-3 M
PH = 8.18 + log10 (8.1249*10-6 / 9.9918*10-3) = 5.09

c) Pka = 12.48 ; Ka = 10-12.48 = 3.3113*10-13
3.3113*10-13= X2 / [0.010-X]
X2 + 3.3113*10-13X -3.3113*10-15 = 0
X = 5.7543*10-8 M = [A-]
[HA] = 0.010-(5.7543*10-8) = 9.9999*10-3 M
PH = 12.48 - log (5.7543*10-8 / 9.9999*10-3) = 7.24

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