20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X)? (pKa value of acetic acid is 4.75).
- 7.0
- 4.75
- 3.5
- 4.82
Answer
(2) 4.75
Solution
Molarity of acetic acid: M1V1 = M2V2
⇒ M1 × 20 = 0.1 × 28.4 ⇒ M1 = 0.142 M
In solution (X):
mmol of CH3COOH = 0.142 × 20 = 2.84
mmol of NaOH = 0.1 × 14.2 = 1.42
CH3COOH + NaOH CH3COONa + H2O
⇒ CH3COOH left = 2.84 − 1.42 = 1.42 mmol; CH3COONa formed = 1.42 mmol (acidic buffer, half-neutralised)
pH = pKa + [CH3COO-][CH3COOH]
⇒ pH = 4.75 + log 1.421.42 = 4.75 + 0 = 4.75
Answer: (2)