Question:medium

Solution of 0.1 N \(NH_4OH\) and 0.1 N \(NH_4Cl\) has pH 9.25. Then p\(K_b\) of \(NH_4OH\) is :

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For equal buffer components, pH = pKa or pOH = pKb.
Updated On: Apr 18, 2026
  • 9.25
  • 4.75
  • 3.75
  • 8.25
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The given solution is a basic buffer because it contains a weak base (\( \text{NH}_{4}\text{OH} \)) and its salt with a strong acid (\( \text{NH}_{4}\text{Cl} \)).
: Key Formula or Approach:
For a basic buffer, the Henderson-Hasselbalch equation is:
\[ \text{pOH} = \text{pK}_{\text{b}} + \log \left( \frac{[\text{salt}]}{[\text{base}]} \right) \]
Also, \( \text{pH} + \text{pOH} = 14 \).
Step 2: Detailed Explanation:
1. Calculate the pOH of the solution:
Given pH \( = 9.25 \).
\[ \text{pOH} = 14 - \text{pH} = 14 - 9.25 = 4.75 \]
2. Identify the concentrations:
\( [\text{salt}] = [\text{NH}_{4}\text{Cl}] = 0.1 \text{ N} \)
\( [\text{base}] = [\text{NH}_{4}\text{OH}] = 0.1 \text{ N} \)
3. Substitute the values into the buffer equation:
\[ 4.75 = \text{pK}_{\text{b}} + \log \left( \frac{0.1}{0.1} \right) \]
\[ 4.75 = \text{pK}_{\text{b}} + \log(1) \]
Since \( \log(1) = 0 \):
\[ \text{pK}_{\text{b}} = 4.75 \]
Step 3: Final Answer:
The \( \text{pK}_{\text{b}} \) of \( \text{NH}_{4}\text{OH} \) is \( 4.75 \).
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