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Help Plz! Calculate pH of a 0.5 M NH4+ (aq) solution. (Kb (NH3) = 1.8E-5).?

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Calculate pH of a 0.5 M NH4+ (aq) solution. (Kb (NH3) = 1.8E-5).

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  1. NH4+ --> NH3 & H+

    0.5   --->  x   &    x

    Ka = [NH3] [H+] / [NH4+]

    Ka = Kw / Kb = 1e-14 / 1.8e-5 = 5.56e-10

    5.56e-10 = [x] [x] / [0.5]

    x = [H+] = 1.67e-5

    your answer :  pH = 4.78

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