If kb for nx3 is 1.5*10-6, what is the poh of a 0.175 m aqueous solution of nx3?

Question

Izayah Olsen

Chemistry

7 November, 23:03

If kb for nx3 is 1.5*10-6, what is the poh of a 0.175 m aqueous solution of nx3?

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Koen Odonnell · Answer 1

Answer is: pOH = 3,29.

Kb (NH₃) = 1,5·10⁻⁶.

c (NH₃) = 0,175M.

pOH = ?

Chemical reaction: NH₃ + H₂O ⇄ NH₄⁺ + OH⁻.

Kb = c (NH₄⁺) · c (OH⁻) : c (NH₃).

c (NH₄⁺) = c (OH⁻) = x.

x² = Kb · c (NH₃)

x² = 1,5·10⁻⁶ · 0,175 = 2,625 ·10⁻⁷.

x = c (OH⁻) = √2,625 ·10⁻⁷ = 5,12 · 10⁻⁴.

pOH = - log (c (OH⁻)) = 3,29.