The PH of a 1.3M HF solution is 3.18. What % of the acid ionized?

Answer 1

pH = -log(H^+)
3.18 = -log(H^+)
(H^+) = 6.6E-4 but you should confirm that.

...........HF ==> H^+ + F^-
...........1.3..6.6E-4..6.6E-4
%ionization = (6.6E-4/1.3)*100 = ?

Answer 2

thank you!!