A salicyclic acid (MW= 138.123 g/mol) solution has a pH of 2.43. When 100 ml of this solution was evaporated 0.22g of dry salicylic acid was recovered. What is the Ka of salicylic acid?

Answer 1

mols SA = grams/molar mass = about 0.0015 but that's approximate.
M = mols/L = 0.0015/0.1 = about 0.015 approximaately.
pH = 2.43, pH = -log(H^-) so (H^+) = about 0.004 and that's approximate.

..........SA ==> H^+ + A^-
I........0.015...0.....0
C.........-x.....x......x
E......0.015-x....x.....x

Ka = (H^+)(A^-)/(HA)
Substitute from the E line (you know x from the above information) and solve for Ka.

Answer 2

what is x?

Answer 3

I solved the pH part before starting work on the second part. It's approximately 0.004M as I posted.