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MHT CET 202625 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
25 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

MHT CET 2026 · 25 April · Evening Shift · Chemistry · Ionic Equilibrium

The Basic buffer solution is made with 0.1\ M\ NH_4OH and 0.2\ M\...

The Basic buffer solution is made with 0.1 M NH4OH0.1\ \text{M}\ \text{NH}_4\text{OH} and 0.2 M NH4Cl0.2\ \text{M}\ \text{NH}_4\text{Cl}. If KbK_b for the base is 1.8×10−51.8 \times 10^{-5}, what is the pH of the buffer?

Options

  1. A4.744.74
  2. B9.269.26
  3. C8.968.96
  4. D5.055.05

Correct answer

C. 8.968.96

Step-by-step solution

Given:
[NH4OH]=0.1 M[\text{NH}_4\text{OH}] = 0.1\ \text{M}
[NH4Cl]=0.2 M[\text{NH}_4\text{Cl}] = 0.2\ \text{M}
Kb=1.8×10−5K_b = 1.8 \times 10^{-5}

First, calculate pKb\text{p}K_b:
pKb=−log⁡(1.8×10−5)=5−log⁡(1.8)=4.74\text{p}K_b = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 4.74

Using the Henderson-Hasselbalch equation for a basic buffer:
pOH=pKb+log⁡([Salt][Base])<br><br>Substitutingthegivenvalues:<br>pOH=4.74+log⁡(0.20.1)\text{pOH} = \text{p}K_b + \log \left( \dfrac{[\text{Salt}]}{[\text{Base}]} \right)<br><br>Substituting the given values:<br>\text{pOH} =4.74 + \log \left( \dfrac{0.2}{0.1} \right)

pOH=4.74+log⁡2\text{pOH} = 4.74 + \log 2

pOH=4.74+0.30=5.04\text{pOH} = 4.74 + 0.30 = 5.04

Now, calculate the pH:
pH=14−pOH\text{pH} = 14 - \text{pOH}

pH=14−5.04=8.96\text{pH} = 14 - 5.04 = 8.96

Answer: 8.968.96
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