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Buffer question Watch

1. 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?
2. It's because the volumes cancel in the equation, so only the moles are needed
3. It will have probally meant Mol/dm^-3

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Updated: December 6, 2010
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