need help with a pH question

Question

In a beaker, there is 40 mL of 0.1 M weak acid, HA (Ka=2.5 x 10^-8). Calculate the pH of this acid then calculate the pH after 25 mL of 0.1 M NaOH has been added.

Answer 1

Initial moles HA = 40 ml x 1 L / 1000 ml x 0.1 mol/L = 0.004 mols

Moles NaOH added = 25 ml x 1 L / 1000 ml x 0.1 mol / L = 0.0025 mols

Final volume = 40 ml + 25 ml = 65 ml = 0.065 L

HA + OH^- ==> A^- + H2O

0.004.....0.0025.....0...................Initial

-0.0025..-0.0025...+0.0025.......Change

0.0015........0.........0.0025.........Equilibrium

So at equilibrium we have 0.0015 mols HA and 0.0025 mols A^- and this produces a BUFFER.

Final concentrations are as follows:

[A^-] = 0.0025 mol / 0.065 L = 0.038 M

[HA] = 0.0015 mol / 0.065 L = 0.023 M

To find the pH of an acidic buffer solution, we can use the Henderson Hasselbalch equation:'

pH = pKa + log [A^-] / [HA]

pH = ?

pKa = -log Ka = -log 2.5x10^-8 = 7.60

log log [A^-] / [HA] = log (0.038 / 0.023) = 0.22

pH = 7.60 + 0.22

pH = 7.82

(be sure to check all of the math)