Calculate the molar concentration of OH ions in a 0.066 M solution of ethylamine (C2H5NH2: Kb= 6.4 x 10-) Express your answer to two significant figures.

6.5x10⁻³M = [OH⁻]

Explanation:

The Kb of a Weak base as ethylamine is expressed as follows:

Kb = [OH⁻] [C₂H₅NH₃⁺] / [C₂H₅NH₂]

As the equilibrium of ethylenamine is:

C₂H₅NH₂(aq) + H₂O(l) ⇄ C₂H₅NH₃⁺(aq) + OH(aq)

The concentration of C₂H₅NH₃⁺(aq) + OH(aq) is the same because both ions comes from the same equilibrium. Thus, we can write:

Kb = [OH⁻] [C₂H₅NH₃⁺] / [C₂H₅NH₂]

6.4x10⁻⁴ = [X] [X] / [C₂H₅NH₂]

Also, we can assume the concentration of ethylamine doesn't decrease. Replacing:

6.4x10⁻⁴ = [X] [X] / [0.066M]

4.224x10⁻⁵ = X²

6.5x10⁻³M = X

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