First slide

Acidic buffer

Question

20 ml of 0.2 M NaOH is added to 50 ml, of 0.2 M ${CH}_{3} COOH$ to give 70 ml, of the solution. What is the pH of the solution ? The ionization constant of acetic acid is $2 \times 10^{- 5}$

Moderate

A

4.522
B

5.568
C

6.522
D

7.568

Solution

The addition of NaOH converts equivalent amount of acetic acid into sodium acetate. Hence,
Concentration of acetic acid after the addition of
$NaOH = \frac{30}{70} \times 0.2 M$
Concentration of ${CH}_{3} COONa$ after the addition of
$NaOH = \frac{20}{70} \times 0.2 M$
Hence, using the expression
$pH = {pK}_{a} + \log \frac{[Salt]}{[Acid]}$
$= - \log (2 \times 10^{- 5}) + \log (\frac{2}{3}) = 4.699 - 0.177 = 4.522$

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