Question

A 250mL buffer solution is 0.250M in acetic acid and 0.250M in sodium acetate.

A) What is the pH after the addition of 0.005 mol of HCl?

B) What is the pH after the addition of 0.005 mol of NaOH?

Answer 1

When the ratio of salt/acid is 1, then the pH = pKa. Since both salt and acid are present at 0.25 M, the pH = pKa = 4.74.
b). HCl reacts with the sodium acetate or more specifically with the acetate anion...
CH3COO^- + H+ ---> CH3COOH
0.005 mol HCl will react with 0.005 mol CH3COO^-. Since the [CH3COO^-] is 0.25 M, that is 0.25 mol/L
or 0.0625 moles in 250 ml. When you react this with 0.005 mol H^+, you are left with 0.0625 - 0.005 =
0.0575 mol/0.25 L (assuming no change in vol upon addition of HCl) = 0.23 M. So concentration of CH3COO^- went from 0.25 M to 0. 23 M and of course the [CH3COOH] increased by the same amount.
pH = pKa + log [salt]/[acid] = 4.74 + log [0.23]/[0.27]
pH = 4.74 + (-0.069)
pH = 4.67
NOTE: upon addition of acid, the pH changed from 4.74 to 4.67. Not much, but more acidic as expected.
c). Do it the same way, but as follows: NaOH reacts with the CH3COOH species, i.e. the acid species to make the salt, CH3COO^- + Na+