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1 July, 20:38

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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  1. 2 July, 00:11
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    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%.
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