Question

Need help with AP chemistry, specifically Acids and Bases

The acid ionization constant, Ka, for propanoic acid, C2H5COOH, is 1.3  10
-5
.
(a) Write the expression for the acid-dissociation constant, Ka for propanoic acid.

(b) Calculate the hydrogen ion concentration, [H
+
], in a 0.300 M propanoic acid.

(c) Calculate the percentage of propanoic acid molecules that are ionized in 0.300 M propanoic
acid.

Answer 1

Let's call propanoic acid HP.
.........HP ==> H^+ + P^-
initial.0.3M.....0.....0
change...-x.......x.....x
equil....0.3-x....x......x

Ka = (H^+)(P^-)/(HP)
Substitute from the ICE chart above and solve for x = (H^+).

%ion = [(H^+)/(HP)]*100 = [(?H^+)/0.3]*100

Post your work if you get stuck.