Question

=Assume you have prepared 100.0 mL of a buffer solution using 0.400 mol of acetic acid (pKa = 4.74) and 0.400 mol of sodium acetate. The pH of this buffer solution is initially 4.74. After preparing this buffer solution, you added 55.0 mL of a 1.10 M NaOH solution to your buffer to see what would happen. What will the pH of this new solution be?

Answer 1

pH of a buffer

pH = pKa + log(A-/HA)

before addition of base

mol of acid = 0.4

mol of conjguate = 0.4

then after addition of base

mmol of base added = M*V = 55*1.1 = 60.5 mmol = 0.0605 mol of base

mmol of conjugate = 0.4+0.0605 = 0.4605

mmol of acid = 0.4-0.0605 = 0.3395

substitue in pH of buffer

pH = pKa + log(A-/HA)

pH = 4.75 + log(0.4605/0.3395) = 4.882