Consider a 0.78 M solution of NH3 (Kb = 1.8×10-5). calculate the pH

Answer 1

Look at the hydrolysis of NH3

NH₃ + H₂O ==> NH₄⁺ + OH^-

Kb = 1.8x10^-5 = [NH₄⁺][OH^-] / [NH3]

1.8x10^-5 = (x)(x) / 0.78 - x (assume x is small relative to 0.78 M and ignore it in denominator)

1.8x10^-5 = x² / 0.78

x² = 1.40x10^-5

x = 3.75x10^-3 M = [OH-] (note: above assumption was valid)

pOH = -log 3.75x10^-3

pOH = 2.43

pH = 14 - pOH

pH = 11.6