Asked by Maria

The pH of a 0.002 50 mol/L solution of benzoic acid is 3.65. Calculate the Ka for benzoic acid.

Answers

Answered by DrBob222
pH = 3.65 = -log(H^+).
(H^+) = 2.24E-4

Call benzoic acid (C6H5COOH) HBz.
...........HBz ==> H^+ + Bz^-
initial..2.24E-4...0.....0
change.....-x.......x.....x
equil....2.24E-4-x...x....x

Ka = (H^+)(Bz^-)/(HBz)
Substitute from the ICE chart above into the Ka expression and solve for Ka.
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions