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.
If a solution of HF (Ka = 6.8 10-4) has a pH of 3.42, calculate the total concentration of hydrofluoric acid.
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no
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