Question

The pH of 0.02M solution of a weak acid was measured at 7.6. What is the [OH-] in this
solution?
B) The Ka for benzoic acid is 6.4 x 10-5M. 150 ml of 0.1 M NaOH is added to 200 ml of 0.1 M
benzoic acid, and water is added to give a final volume of 1 L. What is the pH of the final solution?

Answer 1

A). pH = -log(H^+)
7.6 = -log(H^+). Solve for (H^+). You should get approximately 2.5E-8.

B)
150 mL x 0.1M NaOH = 15 millimiles.
200 mL x 0.1M benzoic acid = 20 mmols.

Let's call benzoic acid HB, then
............HB + NaOH ==> NaB + H2O
initial....20......0........0.....0
add................15..............
change.....-15...-15........15...15
equil.......5......0.........15
pH = pKa + log (base)/(acid)
pH = 4.19 + log (15/5); solve for pH.

Note and this is important:
I used millimoles (the equivalent of mol) INSTEAD of concentration. The HH equation calls for concentration. I used millimoles as a shortcut. The answer will be the same; however, some profs will count off if you use mols (or millimoles) instead of concentration. So you need to do one of three things if your prof is picky (as I am).
(base) = mmols/mL = 15/1000 = ?
(acid) = mmols/mL = 5/1000 = ?
#1.
pH = 4.19 + log(0.015/0.005) = ?
#2.
pH = 4.19 + log(15/1000/5/1000) =
= cancel the denominator since both are the same (by the way they will ALWAYS be the same) so
= 4.19 + log(15/5) = ?
or third,
pH = 4.19 + log(15/volume/(5/volume) =
cancel v and
pH = 4.19 + log (15/5) = ?