Question

500.0 mL of 0.140 M NaOH is added to 625 mL of 0.200 M weak acid (Ka = 1.37 × 10-5). What is the pH of the resulting buffer?

Answer 1

500 mL x 0.140 = 70 millimols NaOH
625 mL x 0.200 HA = 125 mmols HA.

.......NaOH + HA ==> NaA + H2O
I......70.....125.....0.....0
C.....-70.....-70.....+70...+70
E.......0......55......70....70

Plug E line values in to HH equation and solve for pH.
pH = pKa + log(base)/(acid)