QuestionIonic equilibrium / buffer
A solution is obtained by mixing 10 ml of 0.2 N KOH solution with 30 ml of 0.1 N CH₃COOH. The pH of the solution would be: (Given: pKa of CH₃COOH = 4.75; log 2 = 0.30)
- (a) 5.301
- (b) 5.05
- (c) 4.699
- (d) 4.398
Answer: (b) 5.05
KOH = 10 × 0.2 = 2 meq; CH₃COOH = 30 × 0.1 = 3 meq. After reaction: salt 2, acid 1 → buffer. pH = pKa + log(salt/acid) = 4.75 + log 2 = 4.75 + 0.30 = 5.05 → (b).