Question

To prepare 1.0 L of a buffer that is 0.10 M CH3COOH and 0.10 M CH3COo, you can i. Add 0.10 mol of NaCH3COO (8.20 g NaCH3COO) to 1 L of 0.10 M CH3COOH ii. Add 0.10 mol of NaOH (4.00 g NaOH) to 1 Lof 0.20 M CH3COOH ili. Add 0.20 mol NaCH3COO (28.7 g NaCH3COO) to 1 L of 0.10 M HOI A. Only i B. Only ii C. Only ii D. i, ii, and ii

Answer 1

The correct answer is D. We can prepare buffer solution by i, ii and iii.

We just need the weak acid (present in unionized form) and its salt.

In the method i, We are directly taking the required amount of acid and the salt, this will make the buffer.

In method ii, NaOH is a strong base and hence it will completely dissociate in Na⁺ and OH^-. Therefore 0.1 mol of sodium acetate will be obtained and 0.1 mol of acetic acid will be left unconsumed.

$NaOH \; +\; CH_{3}COOH\rightarrow CH_{3}COONa\; + \; H_{2}O$

Similarly, in method iii, HCl is a strong acid and 0.1 M HCl will give 0.1 mol H⁺ and 0.1 mol OH^-. Hence, when 0.2 mol CH₃COONa is added, 0.1 mol of CH₃COOH is formed and 0.1 mol CH₃COONa is left unconsumed.

$HCl \; +\; CH_{3}COONa\rightarrow CH_{3}COOH\; + \; NaCl$

NaCl is formed in this reaction which will not affect the buffer solution.