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Help with pH ..... ?

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the pH of a solution that contains 0.818 M acetic acid (Ka=1.76 x 10-5) and 0.172 M sodium acetate is..........

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  1. pKa = 4.75

    We use Handerson - Hasselbalch equation

    pH = 4.75 + log 0.172 / 0.818 =  4.07


  2. pH = pKa + log ([acetate]/[acid acetic]) =  pKa+log 0.172/0.818

    pKa = log 1/Ka= log 1/1.76*10^-5=4.755

    pH=4.755+log0.21=4.755-0.677=4.078
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