...A 0.10 molar solution of sodium benzoate has a pH of 8.60 at room temperature.

A. Calculate the [OH-] in the sodium benzoate solution described above.
B. Calculate the value for the equilibrium constant for the reaction: C6H5COO- + H2O <--> C6H5COOH + OH-
C. Calculate the value of Ka, the avid dissociation constant for benzoic acid.

I'm not really sure what to do..
So, for A, is the equation x^(2) over 0.10 - x?
For B, I think the equation is:
[C6H5COOH] [OH] over [C6H5COO-]
But I don't know how to solve for its value.

Yeah, I don't know what I'm doing, can you help me?
I asked this question earlier, but I didn't get any answers.