Test: Buffer Solutions (27 July) - JEE MCQ

If the addition of a strong acid or base changes the pH of a buffer by one unit, the buffer solution is assumed to be destroyed, meaning the new pH = pKa ± 1; that means [salt]/[acid] or [acid]/[salt] = 10 or 1/10.

We have the Henderson-Hasselbalch equation as pH = pKa + log[salt]/[acid]. So by substituting the concentrations of salt and acid along with the acid dissociation constant, we get pH = -log[1.8 × 10^-5] + log [0.50 mol/L]/[0.20 mol/L] = 5.14.

Concept:

Basic buffer solution:

• As the name suggests, a basic buffer maintains the basic pH in the solution.
• It is also called an alkaline buffer solution.
• It is prepared by mixing a weak base with a salt of strong acid.
• These buffers have a pH > 7.
• For example - a buffer of NH₄OH and NH₄Cl.

Explanation:

The basic buffer solution is made by mixing of a weak base with strong acid in such a way to maintain the pH of the solution basic or >7.

⇒ So, a basic buffer has a weak base and a salt of strong acid in the solution. 

⇒In 1st solution we have - 50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH₃COOH

total moles of NaOH initially 50 × 0.1 = 5 mmol
total moles of CH3COOH initially 25 × 0.1 = 2.5 mmol
2.5 mmole of NaOH neutralize the 2.5 mmol of CH₃COOH and formed 2.5 mmole of sodium salt of acetic acid.
2.5 mmol of NaOH remains in the solution.
So, the solution has 2.5 mmol NaOH + 2.5 mmol CH₃COONa, which is only a basic solution as NaOH is a strong base, not a weak base.
 ⇒ In 2nd solution we have - 100 mL of 0.1 M CH₃COOH + 100 mL of 0.1 M NaOH

10 mmol of CH₃COOH and 10 mmol of NaOH neutralize each other.
Same amount of acid, base, and salt will not form a basic buffer.
⇒ In the 3rd solution we have - 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH₄OH

10 mmol of HCl reacts with only 10 mmol of NH4OH and forms 10 mmol of NH₄Cl.
10 mmol of NH₄OH remains in the solution.
Hence, the solution will have 10 mmol of NH₄OH + 10 mmol of NH₄Cl. 
It makes a basic buffer solution, as the solution has a weak base with a salt of strong acid.
⇒ 4th solution - 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH

It is a neutral solution as the same amount of acid and base neutralize each other.

Hence only, option c (100 mL of 0.1 M HCl + 200 mL of 0.1 M NH₄OH) makes a basic buffer solution.

∴ the correct answer is option C.