5000 mL of 0130 M NaOH is added to 595 mL of 0250 M weak aci

500.0 mL of 0.130 M NaOH is added to 595 mL of 0.250 M weak acid (Ka = 1.14 × 10-5). What is the pH of the resulting buffer?

Solution

Final concentration of NaOH = 0.13*500/(500+595) = 0.0594 M ( = concentration base)
Final concentration of acid = 0.25*595/(500+595) - 0.0594= 0.1359 M - 0.0594 M = 0.0764

pH = -log(1.14*10^-5) + log(0.0594/0.0764)
= 4.83 Answer