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Calculate the pH of 0.48M KNO2?and the concentration HNO2?

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Calculate the pH of 0.48M KNO2?and the concentration HNO2?

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  1. pKa of HNO2 = 3.35, so Ka = 4.47x10^-4

    Kb = Kw / Ka, so Kb = 2.24x10^-11

    The reaction is:

    NO2- (aq) + H2O ↔ HNO2 (aq) + OH- (aq)

    [OH-] ≈ √[Kb*c] = √[(2.24x10^-11) (0.48) = 3.28x10^-6 M.

    By the reaction equation, [HNO2] = [OH-] = 3.28x10^-6 M.

    pOH = -log(3.28x10^-6) = 5.48, so pH = 8.52.

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