posted by Betsy .
How many grams of benzoic acid, a monoprotic acid with a Ka=6.4x10 to the negative 5, must be dissolved in 500.0 mL of water to produce a solution with a pH=2.50?
HBz ==> H^+ + Bz^-
Ka = [(H^+)(Bz^-)]/(HBz)
Use pH = 2.50 = - log(H^+) to determine (H^+). Set that in the Ka expression. (Bz^-) is the same as (H^+).
(HBz) = unknown. Solve for that to determine molarity. Change to mols/500 mL, then to grams. Post your work if you get stuck.
Ok, so I'm at 1.56x10^-1 M, would I divide that by .5 L right away?
0.156 is what I have also. No, don't divide by 0.5.
M = mols/L. If you want 0.156 molar in 0.5 L, then you want 1/2 that amount in 500 mL or 0.5 x 0.156 = 0.078 mols in 500 mL.
mols = g/molar mass
0.078 mols = grams/molar mass.
You know molar mass, calculate grams.