I tried this question twice got different answers (Chemistry question)!

Question

A 25.0-mL sample of 1.00 M HF is titrated with 0.50 M NaOH. Calculate the pH of the solution after 25.0 mL of NaOH is added. (Ka for HF is 7.2x10^-4)

First time i got 0.60 (knew it was wrong)

Then i tried a different way and got 2.84. Which is wrong too. Not sure what im doing wrong.

Answer 1

The reaction taking place is...

HF + NaOH ==> NaF + H2O and since HF is a weak acid, the addition of NaOH produces F^- which is the conjugate base. This makes a buffer of HF + F^-

To find the pH, you can use the Henderson Hasselbalch equation: pH = pKa + log [F^-]/[HF]

We can find the pKa by taking -log Ka = -log 7.2x10-4 = 3.14

Now, we find the [F^-] and [HF]

HF + OH- ===> F- + H2O

0.025...0.0125.........0.........0.......Initial

-0.0125..-0.0125...+0.0125.........Change

0.0125......0............0.0125.........Equilibrium

pH = 3.14 + log (0.0125/0.0125)

pH = 3.14 +0

pH = 3.14