Group Work
What is the pH of 1.00 M HF solution to which is added 0.500 M NaF? Ka = 7.0 x 10-4
HF H3O+ F-
Initial 1.00 0 0.500
Change -x +x +x
Equil. 1.00 - x x 0.500 + x
x(0.500 + x)/(1.00 - x) = 7.0 x 10- 4
Assume x << 0.500 and 1.00:
x(0.500)/1.00 = 7.0 x 10- 4
x = 1.40 x 10-3, so the assumption was okay
pH = -log(1.40 x 10-3) = 2.854