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Today, 01:16

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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  1. Today, 05:13
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    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.
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