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Calculate the [ H + ] and pH of a 0.0040 M butanoic acid solution. The K a of butanoic acid is 1.52 * 10 - 5. Use the method of successive approximations in your calculations.

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  1. Today, 02:26
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    Ka = [H+][But-]/H[Hbut]

    Let k moles of H + be formed

    1.52*10^-5=k^2/0.004-k

    K^2=0.004*1.52*10^-5-1.52*10^-5k

    K^2 - 1.52*10^-5k-6.08*10^-8=0

    Solving this quadratic equation.

    K=7.6*10^-6

    PH=-log (7.6*10^-6) = 5.2
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