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22 December, 20:33

The Ka for benzoic acid is 6.5 * 10-5. Calculate the pH of a 0.12 M benzoic acid solution.

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  1. 23 December, 00:08
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    pH=4.05

    Explanation:

    C7H6O2 - > C7H6O - + H+

    ka = [C7H6O-] [H+]/[C7H6O2]

    During equilibrium

    [C7H6O-] = [H+] = x^2

    [C7H6O2]=0.12-x

    Replace

    ka = x^2/0.12-x

    6.5 x10^-5 = x^2/0.12-x

    7.8x10^-6 - 6.5 x10^-5x=x^2

    x^2 + 6.5 x10^-5x - 7.8x10^-6

    Solution of quadratic equation

    x=8.8 x10^-5

    pH = - log [H+] = - log 8.8 x10^-5=4.05
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