What is the pH of a buffer solution that is prepared by treating 600.0 mL of 1.5 M sodium acetate with 240.0 mL of 0.25 M HCl?
I tried everything, I have no clue how to start nor what to do; very lost
Sodium acetate = NaAc
Acetic acid = HAc
millimols NaAc = mL x M = 600 x 1.5 = 900
millimols HCl = mL x M = 240 x 0.25 = 60
..............Ac^- + H^+ ==> HAc
I...........900.......0..............0
add.................60................
C.........-60.....-60.............+60
E..........840.....0................60
Ac^- is the base; HAc is the acid.
Purists will tell you that in the HH equation the numbers that go in are (base) and (acid) and not millimols. You can take care of that problem easily by (base) = (Ac^-) = millimols Ac^-/total mL and (acid) = (HAc) = millimols HAc/total mL. You get the same answer either way.
Plug the E line into the Henderson-Hasselbalch eqution and calculate pH.