Question

Which of the following solutions, when mixed, will not form a buffer solution?

• A. 100 mL 0.1 M NaOH + 50 mL 0.1 M CH3COOH
• B. 50 mL 0.1 M NaOH + 100 mL 0.1 M CH3COOH
• C. 50 mL 0.1 M NH4OH + 50 mL 0.1 M CH3COOH
• D. 50 mL 0.1 M HCl + 100 mL 0.1 M CH3COONa

Hint:

For buffer solutions, the amount of acid and conjugate base should be in similar concentrations. A strong acid or base will disrupt the buffering capacity.

Answer 1

The Correct Option is A

To form a buffer solution, we need: 

• A weak acid and its conjugate base
• or a weak base and its conjugate acid

Now let's analyze each option: (a) 100 mL 0.1 M NaOH + 50 mL 0.1 M CH₃COOH 
- CH₃COOH is a weak acid, NaOH is a strong base.
- Moles of CH₃COOH = $0.1 \times 0.05 = 0.005$ mol
- Moles of NaOH = $0.1 \times 0.1 = 0.01$ mol
- NaOH is in excess, completely neutralizes acid ⇒ no buffer 

(b) 50 mL 0.1 M NaOH + 100 mL 0.1 M CH₃COOH 
- CH₃COOH (weak acid), NaOH (strong base)
- Moles of CH₃COOH = $0.1 \times 0.1 = 0.01$ mol
- Moles of NaOH = $0.1 \times 0.05 = 0.005$ mol
- Partial neutralization ⇒ forms CH₃COONa ⇒ buffer 

(c) 50 mL 0.1 M NH₄OH + 50 mL 0.1 M CH₃COOH 
- Weak base (NH₄OH) and weak acid (CH₃COOH)
- Can act as a buffer system depending on strength balance ⇒ buffer 

(d) 50 mL 0.1 M HCl + 100 mL 0.1 M CH₃COONa 
- HCl is a strong acid, CH₃COONa is a conjugate base source
- Partial neutralization ⇒ forms CH₃COOH and CH₃COONa ⇒ buffer 

Answer: (a) 100 mL 0.1 M NaOH + 50 mL 0.1 M CH₃COOH (no buffer formed)