Question

0.1mol/L浓度的氨水溶液的PH值为12,求该溶液的氢离子浓度?

Answer 1

· NH3H2O = NH4+ + OH-

平衡时 0.1-x x x

所以x^2 / (0.1-x) = Kb = 1.8×10^-5

[OH-] = x = 1.34×10^-3 mol/L

[H+] = Kw / [OH-] = 7.45×10^-12 mol/L