If kb for nx3 is 2.5×10−6, what is the poh of a 0.175 m aqueous solution of nx3? express your answer numerically

pOH = 3.18

Explanation:

The equilibrium of a weak base as NX3 is:

NX3(aq) + H2O(l) ⇄ HNX3⁺(aq) + OH-(aq)

Where the equilibrium constant, Kb, is:

Kb = 2.5x10⁻⁶ = [HNX3⁺] [OH-] / [NX3]

As both HNX3⁺ and OH- ions comes from the same equilibrium, their concentrations are the same, that is:

[HNX3⁺] = X

[OH-] = X

And: [NX3] = 0.175M:

2.5x10⁻⁶ = [X] [X] / [0.175M]

4.375x10⁻⁷ = X²

X = 6.61x10⁻⁴M = [OH-]

As pOH = -log [OH-]

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