A buffer containing 0.2832 M of acid, HA, and 0.1504 M of its conjugate base, A−, has a pH of 2.36. What is the pH after 0.0015 mol NaOH is added to 0.5000 L of this solution?

Answer 1

mols HA = 0.2832M x 0.5L = about 0.142
moles A^- = 0.1504 x 0.5L = 0.0752

.............HA + OH^- ==> A^- + H2O
initial...0.142....0....0.0752
...add..........0.0015................
change..-0.0015.-0.0015...+0.0015
equil.....0.1405...0.......0.0015
Now substitute into the HH equation and solve for pH.

Answer 2

I forgot to note that you need to confirm all of the numbers as I just wrote the first or second significant figures.