In a titration of 50 mL 1 Molar HC2H3O2 with 1 Molar NaOH, a student used bromcresol green as an indicator (Ka = 1 X 10^-5). About how many mL of NaOH would it take to reach the end point with this indicator? What would be a better indicator for this titration; why?

Answer 1

I would use the Henderson-Hasselbalch equation here. Have you had that in your course?
pH = pKa + log[(base)/(acid)]
Substitute pH = 5 (the middle point of where the indicator turns color) and calculate (base)/(acid). pKa for acetic acid (which I will call HAc) is 4.74.
I found about 2 but that's approximate and you need to do the calculation and be more exact about it. Let's just assume 2 is the number for the moment.
B/A = 2 and that means
(base) = 2*(acid).
We know that we have 50 x 1M or 50 mmoles HAc to neutralize and as we add NaOH, we will take away from th HAc and add to acetate (Ac^-); therefore,
HAc + Ac^- = 50 mmoles or
A for acid + B for base = 50
Then we can substitute the ratio from above.
A + 2*A = 50
3A = 50 and A = about 17 and B = about 33. Since the solutions are 1 M and since the volume for the acid and the base are the same (because they are in the same solution), these numbers are millimoles BUT they are the same as mL. You know that the equivalence point is at 50 mL of the NaOH; therefore, something in the neighborhood of 33 mL or so for NaOH doesn't even come close to matching the end point with the equivalence point. The end point comes out far short of the equivalence point. Phenolphthalein indicator (which turns between 8 and 10--midpoint about 9) would be a much better indicator. You can calculate the pH at the equivalence point from the hydrolysis of the salt.
Ac^- + HOH ==> HAc + OH^-
Set up an ICE chart and substitute into the Kb expression for the hydrolysis.
Kb = Kw/Ka and the pH should be about 9 for the equivalence point.