A volume of 500.0 mL of 0.130 M NaOH is added to 585 mL of 0.200 M weak acid (𝐾a=2.92×10−5). What is the pH of the resulting buffer? HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)

Answer 1

Final volume = 500.0 ml + 585 ml = 1085 ml = 1.085 L

Initial moles HA = 585 ml x 1 L / 1000 ml x 0.200 mol/L = 0.117 moles

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

HA + OH- ==> H2O + A-

0.117...0.065..................0..............Initial

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

0.053.......0...................0.065.......Equilibrium

Final [A-] = 0.065 mol/1.085 L = 0.0599 M

Final [HA] = 0.053 mol / 1.085 L = 0.0488 M

Henderson Hasselbalch equation:

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

pKa = -log Ka = -log 2.92x10^-5

pKa = 4.53

pH = 4.53 + log (0.0599/0.0488) = 4.53 + 0.089

pH = 4.62