Calculate the percent ionization of a 0.15 formic acid solution in a solution containing 0.11 potassium formate.

Question

I have the formic acid Ka=1.8 *10^-4
When I solve for "x" I'm getting 1.1.44*10^-2 When I plug that into equation (x/0.15-x)*100, I'm getting 10.6% which is incorrect. I getting lost in the math, plz help. Thanks

Answer 1

I don't believe it's the math. Let's simplify the formic acid and call it HF (I know that isn't right but it saves typing).
HF ==> H^+ + F^- and KF ==>K^+ + F^-
HF = 0.15M at the start. F^- from KF = 0.11
(H^+)(F^-)/(HF) = Ka
(H^+) = x
(F^-) = x + 0.11 but since x is small we can make this equal to 0.11.
(HF) = 0.15-x but since x is small we can make this equal 0.15.
Solve for x = (H^+). I get an answer of 2.45E-4 = (H^+). By the way, I solved the quadratic we get if we keep x+0.11 and 0.15-x and I get 2.445E-4 so there is almost no error in making the assumptions.
Then % ion = (ions/0.15)*100 = (2.45E-4/0.15)*100 =??
Note that I've carried too many significant figures, you should correct that, especially if you are keying the answer into a database.

Answer 2

Thank you so much! I see what I was doing wrong now. I need more practice with these type problems.