Question

What is the difference in pH for 1/3 and 2/3 stages of neutralisation of 0.1 M CH3COOH with 0.1 M NaOH.

• A. 2 log 3
• B. 2 log (1/4)
• C. 2 log (2/3)
• D. 2 log 2

Answer 1

Answer: (D)

2 log 2

Answer 2

CH3COOH + OH– $\longrightarrow$ CH3COO– + H2O

Difference in pH between $\frac{1}{3}$ &$\frac{2}{3}$ stages of neutralisation = $\left\lbrack pK_{a} + \log\left( \frac{2/3}{1/3} \right) \right\rbrack$–$\left\lbrack pK_{a} + \log\left( \frac{1/3}{2/3} \right) \right\rbrack$= 2 log 2.