محلول مائي لملح خلات البوتاسيوم تركيزه (0.2mol/L)، فإذا علمت أن له (PH=9) عند درجة الحرارة (25C)

CH3COOK → CH3COO- + K+     (معادلة الإماهة)

CH3COO- + H2O ↔ CH3COOH + OH-     (معادلة الحلمهة)

[H3O+] = 10^-PH = 10^-9 mol/L

CH3COO- + H2O ↔ CH3COOH + OH-

0.2                                     0               0

0.2-x                                  x               x

Kh = [CH3COOH].[OH-] / [CH3COO-]

Kh = x2/(0.2-x)

X = [OH-] = 10^-14/10^-9 = 10^-5 mol/L

Kh = x2/(0.2-x)

Kh = (10^-5)2/(0.2-x)

Kh = (10^-5)2/0.2 = 5*10^-10

Kh.Ka = 10^-14

Ka = 10^-14/Kh = 10^-14/(5*10^-10) = 2*10^-5

% = 10^-5/0.2*100 = 5*10^-3%