Question 2.
a. (i) Pur(aq) + H2O(l) <----> HPur+(q) + OH-(aq); pKb1 = 14 - pKa1 = 14 - 2.54 = 11.46
(ii) HPur+ + H2O(l) <----> H2Pur+(q) + OH-(aq); pKb2 = 14 - pKa2 = 14 - 8.93 = 5.07
b. Kb1 for purine = 10-pKb1 = 10-11.46 = 3.47*10-12
Kb2 for purine = 10-pKb2 = 10-5.07 = 8.51*10-6
c. Initial pH = 14 - {1/2 (pKb1 - Log[purine])} = 14 - {1/2 (11.46 - Log0.145)} = 7.85
no net ionic is given 7. Purine, a weak dibasic molecule, is present in some foods...
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