# Chem 2

posted by
**Jess**
.

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.