Chemistry
posted by Chris .
consider 1.0 L of a solution which is 0.55 M HF and 0.2 M NaF (Ka for HF = 7.2 x 104). Calculate the pH after 0.10 mol of HCl has been added to the original solution. Assume no volume change on addition of HCl.
Goes along with the last one i posted, i know i have to realize that the acid will go to completion and subtract it from the base but im a little lost on how to go on

0.55M HF
0.2M F^
millimols HF = 0.55 x 1000 mL = 550 mmoles
mmols NaF = 0.2M x 1000 mL = 200 mmols.
0.1 mol HCl/1000 = 0.1M = 100 mmols in 1000 mL.

..........F^ + H^+ ==> HF
I........200....0.......550
add...........100.............
C......100...100.......+100
E........100....0........650
pH = pKa + log (base)/(acid)
Solve for pH. I obtained approximately 2.3 but check that carefully.