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# Ka calculation Watch

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.
2. ph = 2.47
3. Thanks but can you show me the working please.

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4. 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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Updated: April 9, 2013
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