Propanoic acid has a Ka of 1.3 x 10^-5. What is the % ionization in a 3.0M solution?

Answer 1

Let's call propanoic acid HPr. Then
...........HPr ==> H^+ + Pr^-
Initial....3.0......0.....0
Change......-x......x......x
Equil.....3.0-x.....x.......x

Ka = (H^+)(Pr^-)/(HPr)
Substitute from the ICE chart into the Ka expression and solve for x = (H^+).
Then %ionization = [(H^+)/3.0)]*100 = ?

Answer 2

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