Ka for a 0.100 M hydrofluoric acid solution where whose pH is 2.23 is
a. 3.5x10-4
b. 3.5x10-2
c. 8.0x10-1
d. 8.0x10-2
^I'm not sure how to solve.
Just follow the usual ice diagram and it works out fine.
....................HF --> H^+ + F^-
I...................0.1M.....0.......0
C...................-x.........x........x
E................0.1-x........x........x
Note that (H^+) = x
Ka = (H^+)(F^-)/(HF) which is
Ka = (x)(x)/(0.1-x)
The problem tells you the pH is 2.23 and you know
pH = - log (H^+) or
-2.23 = log (H^+) = x
Solve this for (H^+) = x and substitute into the Ka expression and solve for Ka. Post your work if you get stuck.