If kb for nx3 is 1.5*10-6, what is the percent ionization of a 0.325 m aqueous solution of nx3?

Question

Chemistry

Julio Kent

5 October, 18:13

If kb for nx3 is 1.5*10-6, what is the percent ionization of a 0.325 m aqueous solution of nx3?

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Amelia Vaughan · Answer 1

Chemical reaction: NX₃ + H₂O ⇄ NX₃H⁺ + OH⁻.

c (NX₃) = 0,325 M.

Kb = 1,5·10⁻⁶.

[NX₃H⁺] = [OH⁻] = x.

[NX₃] = 0,325 M - x.

Kb = [NX₃H⁺] · [OH⁻] / [NX₃].

1,5·10⁻⁶ = x² / (0,325 M - x).

x = 0,0007 M.

percent of ionization:

α = 0,0007 M : 0,325 M · 100% = 0,215%.