Question

What would the pH of 0.56M HF be? How would I go about solving this

Answer 1

is there any other information given..?
pka value?

Answer 2

HF <=>H+ + F-

initial no of mols
HF = 0.56
H+ = 0
F- = 0

at equilibrium
HF = 0.56 - x
H+ = x
F- = x

ka = [H+] * [F-] / [HF]

pka = -log ka

from that you can find ka

and
ka = x * x / (0.56 - x)

from this x can be calculated.
x is the H+ mols and since we got to the 1000 cm3 (we simply got the concentration as no of mols)
x = [H+]
then ph can be calculated.
---------------------------------------...

ka = 10^ -3.14
ka = 7.244 * 10 ^-4
7.244 * 10 ^-4 = x^2/ [0.56-x]
x is very small so 0.56 - x can be take as 0.56

7.244 * 10 ^-4 = x^2/ 0.56

x^2 = 0.0004032
x = 0.0200798 = [H+]
pH = -log [H+]
pH = 1.697

*assumed the pka value 3.14