Okay, so here's the question:
A 60.0 mL sample of 0.100M CH3COOH is titrated with 0.200M? What is the pH at each of the following points...
After 30.0 mL of NaOH:
This part...I'm getting confused on. I keep getting a number close to 13 and my friend is insisting that the pH is 8.79? HELP?
millimoles CH3COOH = 60.0 x 0.1M = 6,00 mmoles.
30 mL x 0.200 M NaOH = 6.00 mmols NaOH.
So at 30.00 mL you are at the equivalence point and the pH is determine by the hydrolysis of the salt, in this case, CH3COONa. The concn of the salt is 6.00 mmoles/(60+30)mL = 0.06667M. I'll call acetate Ac^-
..........Ac^- HOH ==> HAc + OH^-
initial...0.0667.......0......0
change.....-x...........x......x
equil..0.0667-x.........x......x
Kb = (Kw/Ka) = (HAc)(OH^-)/(Ac^-)
Substitute into Kb expression, solve for x, convert to H^+, then to pH.
Your friend is right. I obtained 8.79 for pH.