Question

In a .25 M solution, a weak acid is 3.0% dissociated.

A. calculate the pH of the solution
B. calculate the Ka of the acid

Please explain steps.

Answer 1

..............HA ==>H^+ + A^-
initial....0.25M....0......0
change.......-x......x......x
equil......0.25-x....x.....x
If it is 3% dissociated, that means that x = 0.25*0.03 = ?
Substitute into the Ka expression and solve for Ka.
pH = -log(H^+)

Answer 2

Please help me solved this problem