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HOBr HOCN HCN HSO4 HCO3 O 10.4 O 3.60 O 0.40 K = 3.5 10 O -3.60 K, -2.5 10.9 K, 3.5 2 10-4 K=
HOBr HOCN HCN HSO4 HCO3 O 10.4 O 3.60 O 0.40 K = 3.5 10 O -3.60 K, -2.5 10.9 K, 3.5 2 10-4 K= 4.0 10-10 Ka1 = very large K2 1.2 10- Ka1 = 4.2 ? 10-7 Ka2 = 4.8 10-11 [Fe(OH)61 [Fe(OH)61+ [Be(OH)4]2+ [Cu(OH)4]2+ HBO2 (COOH) 2 CH3NH2 Kg = 3.0 x 10 K = 4.0 x 10-3 K = 1.0 x 10-5 K = 1.0 x 10-8 K = 6.0 x 10-10 Refer to Ch. 18 Values. Calculate the pH of a solution in which [OH-] = 2.50 x 10-4 M. Ka1 = 5.9 x 10- K2= 6.4 x 10-5 Kb = 5.0 x 10-4
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