Benzoic Acid (HC6H5OO, 122.0 g/mol) has a Ka = 6.3 x 10-5. What is the pH of a solution that has 3.050 g of benzoic acid in enough water to make a 0.250 L solution?

Answer 1

Let's call benzoic acid HB.
mols HB = grams/molar mass
M HB = mols/L. I calculate approximately 0.1M
............HB ==> H^+ + B^-
initial....0.1.....0.....0
change.......x....x.......x
equil.....0.1-x....x.......x

Ka = (H^+)(B^-)/(HB)
Subsitute into the Ka expression and solve for H^+, then convert to pH.