Group Work
HF H3O+ F-
Initial 0.900 0 0.600
Change -x +x +x
Equil. 0.900 - x x 0.600 + x
x(0.600 + x)/(0.900 - x) = 7.0 x 10- 4
Assume x << 0.600 and 0.900:
x(0.600)/0.900 = 7.0 x 10- 4
x = 1.05 x 10-3, so the assumption was okay
pH = -log(1.05 x 10-3) = 2.979 (started at 2.854)
In water, pH would change from 7 to 13.