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.....0.280-x....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) = 0.280-x
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-]
1 answer