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7 March, 18:00

Determine the ph of a 0.461 m c6h5co2h m solution if the ka of c6h5co2h is 6.5 x 10-5.

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  1. 7 March, 21:02
    0
    By using ICE table:

    C6H5CO2H ↔ C6H5CO2 - + H +

    intial 0.461 0 0

    change - X + X + X

    Equ (0.461 - X) X X

    when Ka = [H+] [C6H5CO2-] / [C6H5CO2H]

    6.5 x 10^-5 = X * X / (0.461-X) by solving for X

    ∴ X = 0.0054

    ∴[H+] = 0.0054

    ∴ PH = - ㏒[H+]

    = - ㏒0.0054

    = 2.27
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