Question

Calculate the [OH^-] pH and percent ionization for a 0.2 M aqueous solution of NH₃. K_b = 1.8 x10^-5

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Answer 1

NH₃ dissociate as given below

NH₃ + H₂O NH₄⁺ + OH^-

K_b  = [NH₄⁺] [OH^-] / [NH₃]

but [NH₄⁺] = [OH^-] = x then

K_b  = x² / [NH₃]

x² = K_b [NH₃]

substtute value

x² = 1.8 10^-5 0.2 = 3.6 10^-6

x = 1.897 10^-3

but [NH₄⁺] = [OH^-] = x = 1.897 10^-3M

[OH^-] = 1.897 10^-3 M

pOH = -log[OH^-] = -log(1.897 10^-3) = 2.72

pH = 14 - pOH = 14 - 2.72 = 11.28

pH = 11.28

percent ionization = [OH^-] 100 / [NH₃]

= (1.897 10^-3) 100 / 0.2

= 0.9485 %

percent ionization = 0.9485 %