A volume of 500.0 mL of 0.130 M NaOH is added to 605 mL of 0.200 M weak acid (𝐾a=3.87×10−5). What is the pH of the resulting buffer?

Question

HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)

pH=?

Answer 1

HA = a weak acid

NaOH = a strong base

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

Initial moles HA present = 605 ml x 1 L / 1000 ml x 0.200 mol / L = 0.121 mols HA

Initial moles OH^- added = 500 ml x 1 L / 1000 ml x 0.130 mol / L = 0.065 mols OH^-

Set up an ICE (BCA) table:

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

0.121...0.065.........0......0.........Initial

-0.065..-0.065...........+0.065....Change

0.056......o...................0.065....Equilibrium

So, at the end of the reaction, we have 0.056 mols HA and 0.065 mols A^-. Since this represents a buffer (a weak acid + the conjugate base), we can use the Henderson Hasselbalch equation:

pH = pKa + log [conj.base] /[acid]

pH = ?

pKa = -log Ka = -log 3.87x10^-5 = 4.41

log [conj,base]/[acid] = log (0.065 / 0.056) = 0.065

Solving for pH, we have...

pH = 4.41 + 0.065

pH = 4.47