Question

500.0 mL of 0.120 M NaOH is added to 605 mL of 0.250 M weak acid (Ka = 4.44 × 10-5). What is the pH of the resulting buffer?

Answer 1

Mol NaOH in 500mL of 0.120M solution = 500/1000*0.120 = 0.06 mol NaOH
Mol Weak acid in 605mL of 0.250M solution = 605/1000*0.250 = 0.151mol

When these are mixed they react to form 0.06 mol of the Na salt and 0.151 - 0.06= 0.09125 mol of unreacted acid remain . The total volume = 1105mL = 1.105L
Molarity of salt = 0.06/1.085 = 0.055M
Molarity of acid = 0.09125/1.105 = 0.0825M
pKa of acid = -log Ka = -log (4.44*10^-5) = 4.35

Use Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.35 + log (0.055/0.0825)
pH = 4.35 + log 0.6029
pH = 4.35 +(-0.176)
pH = 4.17