Question
A 0.039 M solution of a weak acid (HA) has a pH of 4.28. What is the Ka of the acid?
Answers
DrBob222
pH = -log (H^+)
(H^+) = ?. I estimate 5E-5 but you need a better answer than that.
...........HA ==> H^+ + A^-
I........0.039....0......0
C.........-x......x......x
E......0.039-x....x......x
The pH tells you the value of x; i.e., approx 5E-5 so you can evaluate x and 0.039-x.
Plug those values into the Ka expression and evaluate Ka.
(H^+) = ?. I estimate 5E-5 but you need a better answer than that.
...........HA ==> H^+ + A^-
I........0.039....0......0
C.........-x......x......x
E......0.039-x....x......x
The pH tells you the value of x; i.e., approx 5E-5 so you can evaluate x and 0.039-x.
Plug those values into the Ka expression and evaluate Ka.