The number of moles of CH 3 COONa added to 1 litre of 0.1 M CH 3 COOH solution to prepare a buffer solution of pH = 4 ( K a of CH 3 COOH = 2 × 1 0 − 5 ) :

Question

The number of moles of CH 3 ​ COONa added to 1 litre of 0.1 M CH 3 ​ COOH solution to prepare a buffer solution of pH = 4 ( K a ​ of CH 3 ​ COOH = 2 × 1 0 − 5 ) :

TestHub · Accepted Answer

Explanation

We need to determine the number of moles of sodium acetate $(CH_{3} COONa)$ required to prepare a buffer solution of $pH = 4$ using 1 liter of 0.1 M acetic acid $(CH_{3} COOH)$ .

Given Data

pH of buffer = 4

$K_{a}$ of $C H_{3} COO H = 2 \times 1 0^{- 5}$

Concentration of $C H_{3} COO H = 0.1 M$

Volume of solution = 1 L

Concept:

For a buffer solution, the Henderson-Hasselbalch equation is used:

$p H = p K_{a} + lo g (\frac{[ S a lt ]}{[ A c i d ]})$

Where:

$p K_{a} = - lo g (K_{a})$

$[S a lt]$ is the concentration of $C H_{3} COON a$

$[A c i d]$ is the concentration of $C H_{3} COO H$

Calculation:

1. Calculate $p K_{a}$ :

$p K_{a} = - lo g (2 \times 1 0^{- 5}) = 4.7$

2. Apply the Henderson-Hasselbalch equation:

$4 = 4.7 + lo g (\frac{[ S a lt ]}{0.1})$

3. Solve for salt concentration:

$lo g (\frac{[ S a lt ]}{0.1}) = 4 - 4.7 = - 0.7$

$\frac{[ S a lt ]}{0.1} = 1 0^{- 0.7} \approx 0.2$

$[S a lt] = 0.1 \times 0.2 = 0.02 M$

4. Since the volume of the solution is 1 L, the number of moles of $C H_{3} COON a$ required is:

$M o l es = M o l a r i t y \times V o l u m e = 0.02 \times 1 = 0.02 m o l es$

Answer option B, (0.02 moles).

Chemistry - Ionic Equilibrium Question with Solution | TestHub

Options:

Doubts & Discussion