[Nikolai]

I was wondering of some of you guys could confirm my answer to a titration curve problem.

20.00mL of 0.100M HF is titrated with 0.0856M NaOH. Calculate the pH at the end point.


First, I solved for V₂ at equilibrium.
V₂= (20.00mL)(0.100M) = 23.36 mL of NaOH at equivalence point.
                             (0.0856M)
V_total= 20.00mL + 23.36mL = 40.36mL or 0.04036L at equivalence point.

Second, [HA] = [A^-] at equivalence points too, so
0.0200 L X .100 M = 0.002 moles of A^-

Third, solved for molarity of [A^-] using the moles found at the second step and volume found at the first step
M= 0.002 moles =0.0496 M
           0.04036 L

I set up the normal ice table

A^-           + H₂0                                HA + OH^-
0.0496                                          0       0
-x                                              +x     +x
0.0496 - x                                     x       x

Since I was solving for Kb I converted from Kw/6.8 x 10^-4 = 1.5 x 10^-11

Ignoring the x in the A- the answer came out to x = 8.63 x 10^-7
pH= 14 + log (8.63 x 10^-7) = 7.94

Thanks for taking your time, guys.

[Borek]

20.00mL + 23.36mL = 40.36mL

Close.

Final answer is close too.

[Nikolai]

I just noticed that! How embarrassing  ! Thanks alot!