Chem 2

A 0.15 M solution of butanoic acid, C3H7COOH, has a pH of 2.8210. Find the Ka of butanoic acid.

Call buanoic acid HB.
Then the ionization is
HB ==> H^+ + B^-
To start (HB) = 0.15
(H^+) = 0 and (B^-) = 0
After ionization, (H^+) = x but that is found by converting pH to (H^+)
(B^-) = x = same as (H^+)
(HB) = 0.15-x = 0.15 - (H^+) from above.
Set up Ka expression, plug in x and 0.15-x in the appropriate places and solve for Ka. Post your work if you get stuck.

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asked by Jess

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