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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
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