How the calculation of the [OH-], pH and % ionization for 0.619 M ammonia (NH3) NH3 + H2O (liq) rightwards harpoon over leftwards harpoon NH4+ + OH- Kb = 1.8 x 10-5

[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52

Explanation:

Kb of the reaction:

NH3 + H2O(l) ⇄ NH4+ + OH-

Is:

Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]

As all NH₄⁺ and OH⁻ comes from the same source we can write:

[NH₄⁺] = [OH⁻] = X

And as [NH₃] = 0.619M

1.8x10⁻⁵ = [X] [X] / [0.619M]

1.11x10⁻⁵ = X²

3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]

% ionization:

[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%

pH:

As pOH = -log [OH-]

pOH = 2.48

pH = 14 - pOH

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the answer is:  

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