A volume of 500.0 mL of 0.110 M NaOH is added to 545 mL of 0.200 M weak acid (𝐾a=8.45×10−5). What is the pH of the resulting buffer? HA(aq)+OH−(aq)⟶H2O(l)+A−(aq) pH=

Answer 1

moles OH- = 500.0 ml x 1 L/1000 ml x 0.110 mol/L = 0.055 mols OH-

moles HA = 545 ml x 1 L/1000 ml x 0.200 mol/L = 0.109 mols HA

Let the weak acid = HA

HA + OH- ==> A- + H2O

0.109....0.055.......0...........Initial

-0.055...-0.055...+0.055...Change

0.054.......0.........0.055....Equilibirium

Henderson Hasselbalch equation

pH = pKa + log (A-/HA)

pKa = -log Ka = -log 8.45x10-5

pKa = 4.07

pH = 4.07 + log (0.055/0.054) = 4.07 + 0.008

pH = 4.08