Posted by **Zac** on Sunday, April 29, 2012 at 11:38pm.

0.200 moles of NaF are added to 1L 2.0 molar HF. The Ka = 7.2*10^-4. What is the pH?

a) 2.14

b) 6.70

c) 10.80

d) 3.00

e) 7.00

I got 4.14. pH = -log(Ka)-log([NaF]/[HF]) ==> pH = -log(7.2*10^-4)-log(.2/2) ==> pH = 4.14

Or did I do something wrong?

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