Posted by Stanley3000 on Tuesday, February 14, 2012 at 5:17pm.
pH = 3.42.
3.42 = -log(H^+)
(H^+) = 3.80
E-4
...........HF ==> H^+ + F^-
initial....x.......0.....0
change.....................
equil...x-3.8E-4..3.8E-4..3.8E-4
Ka = (H^+)(F^-)/(HF)
Substitute from the ICE chart above and solve for x.
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