pH = 3.45
3.45 = -log(H^+)
(H^+) = approx 3.55E-4.
...........HA ==> H^+ + A^-
I.........0.1.....0.....0
C..........-x.....x.....x
E.........0.1-x...x.....x
where x = 3.55E-4
Substitute that into Ka expression and solve for Ka.
What is the Ka of a weak acid (HA) that produces a pH of 3.45 with 0.1M?
2.54E-3
3.45E-3
1.27E-6
3.56E-3
5.48E-4
1 answer