محلول مائي لمحل كلوريد الأمونيوم تركيزه (0.2mol/L) وقيمة (PH=5) له، والمطلوب:

NH4NO3 → NH4+ + Cl-

NH4+ + H2O ↔ NH3 + H3O+

NH4+ + H2O ↔ NH3 + H3O+

0.2                         0        0

0.2-x                       x        x

Kh = [NH3]3[H3O+]/[NH4+]

Kh = x2/(0.2-x)

PH = H

[H3O+] = 10-5 mol/L = x

Kh = x2/0.2 = 10-10/0.2 = 5*10-10

Kh.Kb = 10-14

Kb = 10-14/Kh = 10-14/(5*10-10) = 2*10-5

[H3O+] = Ca = 0.01 mol/L

NH4+ + H2O ↔ NH3 + H3O+

0.2                         0      0.01

0.2-x                       x    0.01+x

Kh = x(0.01+x)/(0.2-x)

5*10-10 = x(0.01+x)/(0.2-x)

5*10-10 = x(0.01)/0.2

x = 10-8

y = 10-8/0.2*100 = 5*10-6 %