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    Please help with the working for this:

    The acid dissociation constant, ka, for propanoic acid has the value 1.35 x10-5 mol DM-3 at 25 degrees c.

    Ka = (h+)(ch3ch2coo-)/(ch3ch2cooh)

    Calculate ph of a 0.169 mol DM-3 solution of propanoic acid.
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    ph = 2.47
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    Thanks but can you show me the working please.


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    Ka= 1.35 x 10^-5
    CHCH2COOH= 0.169

    Therefore :

    1.35 x 10^-5 = [H+][CH3CH2COO-]/0.169

    You can assume that the concentrations of the ions are the same, so 1.35 x 10^-5 = [H+]2/0.169

    Rearrange it to get 1.35 x 10^-5 x 0.169 = [H+]2

    Square root the answer to get [H+]

    ph = -log [H+]

 
 
 
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