Question: 72: The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is [Given \mathrm{pK}_{\mathrm{a}} of \mathrm{CH}_{3}\mathrm{COOH}=4.57 ]
(1) 4.57
(2) 2.57
(3) 5.57
(4) 3.57
Answer: Option (3)
Explanation:
The given solution contains acetic acid and its conjugate base sodium acetate, so it is a buffer solution.
The pH of a buffer is given by the Henderson–Hasselbalch equation
\mathrm{pH}=\mathrm{p}K_{\mathrm{a}}+\log\frac{[\text{salt}]}{[\text{acid}]}.
Number of moles of sodium acetate is 0.10\times0.050=0.005 mol.
Number of moles of acetic acid is 0.01\times0.050=0.0005 mol.
The ratio of salt to acid is \frac{0.005}{0.0005}=10.
Substituting the values in the equation gives
\mathrm{pH}=4.57+\log 10.
Since \log 10=1,
The pH of the solution is 4.57+1=5.57.
Hence, the correct answer is Option (3).