Question

What is the pH of buffer solution prepared by mixing 0.2 M \(NH_4OH\) and 0.4 M \(NH_4Cl\) solution (\(pK_b\) of \(NH_4OH\) = 4.620)

• A. 4.025
• B. 7.035
• C. 9.079
• D. 8.450

Hint:

For a basic buffer, pOH = pKb + log([salt]/[base]).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
A mixture of weak base \(\text{NH}_4\text{OH}\) and its salt \(\text{NH}_4\text{Cl}\) is a basic buffer.
\(\text{pOH} = \text{p}K_b + \log\dfrac{[\text{salt}]}{[\text{base}]}\)

Step 2: Calculation
\(\text{pOH} = 4.620 + \log\dfrac{0.4}{0.2} = 4.620 + 0.301 = 4.921\)

Step 3: Result
\(\text{pH} = 14 - 4.921 = 9.079\)
This matches option (C).

Final Answer:
The pH of the buffer is 9.079. \[ \boxed{\text{(C) }9.079} \]