NH3 is a weak base (Kb = 1.8 × 10-5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.018 M in NH4Cl at 25 C?
The dissociation of ammonium ion in water is as
NH4+ + H2O --> NH3 + H3O+
The dissociation constant, Ka can be calculated as
Ka*Kb = Kw
Ka = Kw/Kb
= (1.0 x 10-14)/(1.8 x 10-5)
Ka = 5.5 x 10-10
The Ka for the above reaction can be calculated as
Ka = ([NH3][H3O+])/[NH4+]
(5.5 x 10-10) = [(x)(x)]/(0.018-x)
x2 + (5.5 x 10-10)x - (9.9 x 10-12) = 0
solving this equation, x = 3.15 x 10-6 M
[H3O+] = x = 3.15 x 10-6 M
Calculate the pH of the solution as follows
pH = -log[H3O+] = -log(3.15 x 10-6)
pH = 5.50 (Ans)
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