Asked by Sam
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.
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.
Answers
Answered by
DrBob222
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.
.........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.
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