Question

Which of the following mixture gives a buffer solution with \( \text{pH} = 9.25 \)?
Given: \( \mathrm{p}K_b(\mathrm{NH_4OH}) = 4.75 \)

• A. \(0.2\,\text{M } \mathrm{NH_4OH}\,(0.5\,\text{L}) + 0.1\,\text{M } \mathrm{HCl}\,(0.5\,\text{L})\)
• B. \(0.4\,\text{M } \mathrm{NH_4OH}\,(1\,\text{L}) + 0.1\,\text{M } \mathrm{HCl}\,(1\,\text{L})\)
• C. \(0.2\,\text{M } \mathrm{NH_4OH}\,(0.4\,\text{L}) + 0.1\,\text{M } \mathrm{HCl}\,(1\,\text{L})\)
• D. \(0.5\,\text{M } \mathrm{NH_4OH}\,(0.2\,\text{L}) + 0.2\,\text{M } \mathrm{HCl}\,(0.5\,\text{L})\)

Hint:

For a basic buffer, when \([\text{salt}] = [\text{base}]\), the pH equals \(14 - \text{p}K_b\).

Answer 1

The Correct Option is A

Since the buffer consists of a weak base and its conjugate acid, we use the Henderson equation for basic buffer: \[ \text{pOH} = \text{p}K_b + \log\left(\frac{[\text{salt}]}{[\text{base}]}\right). \]

Step 1: Calculate pOH.
Given: \[ \text{pH} = 9.25 \Rightarrow \text{pOH} = 14 - 9.25 = 4.75. \]

Step 2: Apply Henderson equation.
\[ 4.75 = 4.75 + \log\left(\frac{[\text{salt}]}{[\text{base}]}\right). \] This implies: \[ \log\left(\frac{[\text{salt}]}{[\text{base}]}\right) = 0 \Rightarrow \frac{[\text{salt}]}{[\text{base}]} = 1. \]
Thus, moles of \( \mathrm{NH_4^+} \) formed must equal the remaining moles of \( \mathrm{NH_4OH} \).

Step 3: Check option (A).
Moles of \( \mathrm{NH_4OH} \): \[ 0.2 \times 0.5 = 0.1 \text{ mol}. \] Moles of \( \mathrm{HCl} \): \[ 0.1 \times 0.5 = 0.05 \text{ mol}. \] After neutralization: \[ \text{Remaining } \mathrm{NH_4OH} = 0.05 \text{ mol}, \quad \mathrm{NH_4^+} = 0.05 \text{ mol}. \] Thus, \[ \frac{[\text{salt}]}{[\text{base}]} = 1, \] which gives the required pH.

Step 4: Conclusion.
Option (A) forms a buffer of pH \(9.25\).

Final Answer: \[ \boxed{\text{Option (A)}} \]