Chemistry
posted by Sandy on .
A buffer solution is prepared by dissolving 0.400 mol of CH3COOH and 0.200 mol of CH3COONa in 1.00 L of water. 1.00 mL of 10.0 M HCl is added to a 100 mL portion of this solution. What is the final pH of the resulting solution?
Ka of CH3COOH = 1.8 x 10¯5

You want to make an ICE chart and use the HendersonHasselbalch equation.
Let's simply our typing by calling CH3COOH just HAc and CH3COONa will be NaAc.
mmoles HAc = 0.400 M x 100 mL = 40.0
mmoles Ac^ = 0.200M x 100 mL = 20.0
mmoles HCl added = 1.00 mL x 10.0 M = 10.0
...............Ac^ + HCl ==> HAc + Cl^
initial.......20.0.....0......40.0
added................10.0..........
change.......10.0...10.0.....+10.0
equil.........10.0....0.........+50.0
Now plug all of that into the HH equation and solve for pH. Post your work if you get stuck.