Question

Good day,
Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5
I calculated The pH of a 0.425 M ammonium solution is 11.44
What is the percent ionization of ammonia at this concentration?
Express your answer with the appropriate units.
Thanks

Answer 1

The percent ionization of the ammonium ion is calculated by estimating the fraction of ammonia dissociated from the initial concentration. The percentage ionization has been calculated as follows:

Percent ionization = ([NH₃]_dissociated / [NH₃]_initial)×100

[NH₃]_initial = 0.425 mol/L

[NH₃]_dissociated = [NH₄⁺] = [OH^-]

pH+pOH=14

pH=11.44

11.44 + pOH = 14

pOH= 2.56

pOH=−log[OH⁻]

2.56 = −log[OH⁻]

10^−(2.56) = [OH⁻]

[OH⁻] = 10^−(2.56) = 0.00275 mol/L

Percent ionization = (0.00275 / 0.425)×100 = 0.64 %

Answer: 0.64 %