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The pH of a 0.002 50 mol/L solution of benzoic acid is 3.65. Calculate the Ka for benzoic acid.

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  1. pH = 3.65 = -log(H^+).
    (H^+) = 2.24E-4

    Call benzoic acid (C6H5COOH) HBz.
    ...........HBz ==> H^+ + Bz^-
    initial..2.24E-4...0.....0
    change.....-x.......x.....x
    equil....2.24E-4-x...x....x

    Ka = (H^+)(Bz^-)/(HBz)
    Substitute from the ICE chart above into the Ka expression and solve for Ka.

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