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    A buffer solution has a pH of 4.69 and contains 0.15 mol of propanoic acid and 0.10 mol of sodium propanoate. Use these data to calculate a value of Ka for propanoic acid.

    [H+] = pH^4.69 = 2.04x10^-5
    I get that.

    I know you have to use Ka= [H+][CH3CH2COO-] / [CH3CH2COOH]

    But how the heck do i know the conc. of [CH3CH2COO-] and [CH3CH2COOH], they have only given the number of moles

    http://store.aqa.org.uk/qual/gceasa/...W-QP-JUN09.PDF
    http://store.aqa.org.uk/qual/gceasa/...W-MS-JUN09.PDF
    3e

    They do Ka =(2.04 × 10)(0.10) / (0.15)
    but 0.10 and 0.15 are the moles and conc at all?
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    It's because the volumes cancel in the equation, so only the moles are needed
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    It will have probally meant Mol/dm^-3
 
 
 
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