Asked by Sandy
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
Ka of CH3COOH = 1.8 x 10¯5
Answers
Answered by
DrBob222
You want to make an ICE chart and use the Henderson-Hasselbalch 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.
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.
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