Question

Acid & Base Equilibrium Question
Q: Benzoic acid in a .100mol/L solution is found to be 2.5% ionized. How do I calculate the Ka for benzoic acid?


My attempt:
C6H5COOH <---> C6H5COO + H

2.5% =[C6H5COO]/[C6H5COOH] x 100%

2.5%/100% = [C6H5COO]/[.100mol/L<--I don't understand this]

.025(.100mol/L) = [C6H5COO]

.0025mol/L = [C6H5COO]

I don't know what I'm doing