[Jiro]

suppose you have 100mL of a solution that is 0.1M in HF and 0.10M in NaF, to which your lab instructor now adds 10mL of 0.10M NaOH.

what is final pH?
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I think first you go like this:
    HF  ->  H^+ + F^-
I   .1        0        0
C  .1-x      +x      +x
E  .1-x      x        x

with HF's ka=7.1E-4 i got [H+]=.008M

    NaOH  -> Na^+ + OH^-
I   .1            0        0
C   0          +.1      +.1
E   0            .1        .1

so therefore [OH-]=.1M find [H+] here using Kw i got 1E-13

Now tricky part... adding the two H+ concentrations... remember you can't add M units together you can only add mol units. remember you only compare mols.

change (.008mol / L) * .1L = 8E-4mol
change (1E-13mol/L) * .01 = 1E-15mol

8E-4 mol + 1E-15mol = 8.0E-4mol

now divide by the whole volume...

8.0E-4 mol / 0.110 L

= .00727M = [H+]

find pH=-log[H+]
=2.14

Correct me if im wrong, cause it looks fishy... haha thanks...

[Borek]

That's a buffer question, you should use Henderson-Hasselbalch equation. Assume that NaOH reacted quantitatively with HF.