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]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BCH_%7B3%7DCOO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D%20)
4.93 = 4.76 + log
0.17 = log
x = 5.62 ml NaOH required
- 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
4.93 = 4.76 + log
0.17 = log
x = 5.62 ml NaOH required