A volume of 500.0 mL of 0.160 M NaOH is added to 625 mL of 0.250 M weak acid (𝐾a=1.95×10−5). What is the pH
of the resulting buffer?
HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)
First you must determine what you have in the final solution. It could be predominately excess strong base, or excess weak acid, or a buffer solution. Initially you have this.
millimoles NaOH = mL x M = 500 mL x 0.160 M = 80
millimoles HA = 625 mL x 0.25 M = 156.25
............HA(aq) + OH−(aq) ⟶ H2O(l) + A−(aq)
Initial....156.25.......0...................0...........0..........................
add........................80...............................................
change...-80........-80..................+80......+80
E...........76.25........0....................+80......+80
After reaction you have a buffer in a final solution of 625 mL + 500 mL = 1125 mL.
Final concentrations:
(HA) = 76.25 mmoles/1125 mL = ?
(A^-) = 80 mmoles/1125 mL = ?
Plug the final values for HA and A^- into the Henderson-Hasselbalch equation and solve for pH.
The HH equation is pH = pKa + log (A^-)/(HA)
For the pKa value remember pKa = -log Ka = -log 1.95E-5
Post your work if you get stuck.