Question

Q: The hydroxide ion concentration in a .157M solution of NaC3H5O2 is found to be 1.1x10^-5M Calculate the base ionization constant for propanoate ion.

my attempt:
[OH] = 1.1x10^-5M
[NaC3H5O2] = 0.157M

NaC3H5O2 --> Na = C3H5O2
1:1:1
]C3H5O2] = [NaC3H5O2]

pOH = -log[OH]
pOH= 4.96
pOH + pH = 14
pH = 9.04
pH=-log[H3O]
[H3O] = 10^-9.04
[H3O]= 9.12x10^-10M

ka = [H3O]/[C3H5O2]

I don't know what I'm doing...

How do I approach this question?