Propionic acid = HPr
..........HPr ==> H^+ + Pr^-
initial.0.150......0.....0
change....-x.........x.....x
equil....0.150-x
Ka = (H^+)(Pr^-)/(HPr)
Substitute from the ICE chart above into Ka expression and solve for x = (H^+).
Then % ion = 100*(H^+)/0.15
acetic acid = HAc
..........HAc==> H^+ + Ac^-
You want to know (HAc) = x when
(H^+) = (%ion/100)*x and
(Ac^-) = (%ion/100)*x.
Substitute those into Ka for HAc and solve for x = (HAc).
What must be the molarity of an acetic acid solution if it has the same percent ionization as 0.150M HC3H5O2 propionic acid,Ka=1.3*10^-5 )?
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