chemistry
posted by Saunders on .
For a solution that is 0.280M HC3H5O2 (propionic acid, Ka= 1.3*10^5) and 0.0894M HI , calculate the following:
[H3O+]
[OH]
[C3H5O2]
[I]

You should write out the equations to understand what is going on. Let's call propionic acid HP.
.............HP ==> H+ + P
initial.....0.280M...0....0
change......x.......x.....x
equil.....0.280x....x......x
...........HI ==> H+ + I
initial..0.0894....0....0
equil.......0..0.0894..0.0894
Ka = (H+)(P)/(HP)
Substitute from the ICE charts and solve for (P). (H+) = 0.0894+x and (HP) = 0.280x