mL of 0.0100 M NaOH are added to 25.0 mL of 0.0100 M acetic acid. What is the pH of the resulting solution?
3 answers
You didn't copy all of the problem.
sorry it's:
5 mL of 0.0100 M NaOH are added to 25.0 mL of 0.0100 M acetic acid. What is the pH of the resulting solution?
5 mL of 0.0100 M NaOH are added to 25.0 mL of 0.0100 M acetic acid. What is the pH of the resulting solution?
5 mL x 0.0100M = 0.0500 millimoles NaOH.
25.0 mL x 0.01M HAc = 0.250 mmoles HAc.
.............NaOH + HAc ==> NaAc + H2O
initial...0.0500..0.25.....0......0
change...-0.0500..-0.0500.0.0500.0.0500
equil......0.......0.20....0.05..0.05
Substitute the equilibrium conditions into the Henderson-Hasselbalch equation and solve for pH.
25.0 mL x 0.01M HAc = 0.250 mmoles HAc.
.............NaOH + HAc ==> NaAc + H2O
initial...0.0500..0.25.....0......0
change...-0.0500..-0.0500.0.0500.0.0500
equil......0.......0.20....0.05..0.05
Substitute the equilibrium conditions into the Henderson-Hasselbalch equation and solve for pH.