'A solution is prepared by titrating a 100.0 mL sample of 0.10 M HF (Ka = 7.2 x 10^-4) with 0.1 M NaOH. What is the pH after 20.0 mL of the 0.10 M NaOH is added?'

Question

I'm a little bit confused but here's my ICE chart (in moles):

HF________OH-_____F-
1.0_______0.002___0
-0.002____-0.002__+0.002
0.998______0______0.002

So if the total volume is now 120 mL (=0.12 L), that means the concentration of HF is about 8.32 M and the concentration of F- is about 0.017. I found pKa using -log(Ka) and then I plugged everything into the pH = pkA + log [A-]/[HA] equation, but that gave me 0.424, and I know that's not correct. Where did I go wrong?

Thanks!

Answer 1

You're on the right track but I don't like your ICE chart. That 1.0 for HF isn't right. Where did you go wrong. Two places. I think the HF is (0.1*0.1/0.12) = 0.0833. I agree with the 0.017(I used 0.01667) for F^-. The second error is that you didn't subtract the added NaOH from the HF beginning; i.e., 0.0833 - 0.0167 = ?. Then plug those new numbers into the Henderson-Hasselbalch equation. I get something like 2.57. Hope this helps. Here is the way I do it and I think it makes it much simpler; however, some profs frown on it. To be honest about it, I always took points off when my students did it but the answer comes out the same. Work in millimols.
mmols HF = 100 x 0.1 = 10
mmols OH added = 20 x 0.1 = 2

........HF + OH^- ==> F^- + H2O
I.......10....0.......0......0
add...........2...............
C......-2....-2.......+2
E.......8.....0........2

pH = pKa + log (b)/(a) = 3.17 + log 1/4 = about 2.57.

Here is the technical problem. The HH equations says that (base) and (acid) are in M = mols/L and my shortcut I use is in mols and not mol/L (actually mmols and not mmols/mL). So why does it come out the same? Because the volume in a titration is ALWAYS the same for base and acid; i.e., if we write it right we have pH = 3.17 + log (2/0.12)/(8/0.12) and the 0.12 cancels (every time) so I just remember this.
pH = pKa + log (mmols base/mL)/(mmols acid/mL) = ? The mL cancels every time and mmols stay and the answer always works. It also saves the time of dividing that 2/0.12 and 8/0.12.