# Chemistry

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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?

• Chemistry -

No.
Its pKa = -log(Ka) = -log 7.2E-4 = 3.14
pH = 3.14 + log(0.2/2) = 3.14 - 1.00 = 2.14

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