Question

Suppose you have 1.00 l of an aqueous buffer containing 60.0 mmol acetic acid (pka = 4.76) and 40.0 mmol acetate. what volume of 3.50 m naoh would be required to increase the ph to 4.93?

Answer 1

Required pH = 4.93
- OH⁻ from NaOH reacts with CH₃COOH giving CH₃COO⁻ and H₂O 
- Let the volume of 3.5 M NaOH be x ml
Moles of NaOH = Moles of OH⁻ = Molarity * x ml = 3.5x mmol
- The reaction table for moles is as follows:
                   CH₃COOH + OH⁻ →  CH₃COO⁻ + H₂O
Initial               60             3.5x          40
Change         -3.5x         -3.5x         +3.5x
Final            (60-3.5x)       0           (40+3.5x)
- Substitute in Henderson equation and solve for x:
pH = pKa + log $\frac{[CH_{3}COO^-]}{[CH_3COOH]}$
4.93 = 4.76 + log $\frac{(40+3.5x)}{(60-3.5x)}$
0.17 = log $\frac{(40+3.5x)}{(60-3.5x)}$
$\frac{(40+3.5x)}{(60-3.5x)} = 1.479$
x = 5.62 ml NaOH required