Question

500.0 mL of 0.170 M NaOH is added to 575 mL of 0.250 M weak acid (Ka = 1.50 × 10-5). What is the pH of the resulting buffer?

Answer 1

Calculate moles of NaOH was used = 500 x 0.17 /1000 = 0.085 Moles

Calculate moles of weak acid was used = 575 x 0.25 /1000 = 0.14375 Moles

Once these solution were mixed

Moles of sodium salt formed = 0.085 Moles

unreacted acid = 0.14375 - 0.085 = 0.05875 Moles

Total volume = 500 + 575 = 1075 ml

Moles of salt = 0.085 /1.075 = 0.079 Moles

Moles of acid = 0.05875 /1.075 = 0.05465 Moles

pKa of acid = -log Ka = -log (1.5 x10^-5) = 4.82

pH = pKa + log ([salt]/[acid])

pH = 4.82 + log 0.079 / 0.05465

pH = 4.98