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JEE Main 201910 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2019 · 10 April · Evening Shift · Chemistry · Ionic Equilibrium

The pH of a 0.02M N H _ 4 Cl solution will be [Given: K _ b (N H _...

The pHpH of a 0.02M0.02M NH4ClN{H}_{4}Cl solution will be [Given: Kb(NH4OH)=10−5{K}_{b}\left(N{H}_{4}OH\right) = {10}^{- 5} and log⁡2=0.301\log 2 = 0.301]

Options

  1. A4.654.65
  2. B2.652.65
  3. C4.354.35
  4. D5.355.35

Correct answer

D. 5.355.35

Step-by-step solution

NH4ClN{H}_{4}Cl is a salt of a strong acid and a weak base.

pH=12(pKw−pKb−logC)pH = \dfrac{1}{2}\left({pK}_{w} - {pK}_{b} - logC\right)

=12[14−5−log⁡2×10−2]= \dfrac{1}{2}\left[14 - 5 - \log 2 \times {10}^{- 2}\right]

=12[9−log⁡2−log10−2]= \dfrac{1}{2}\left[9 - \log 2 - log {10}^{- 2}\right]

=12[9−log⁡2+2]= \dfrac{1}{2}\left[9 - \log 2 + 2\right]

=10.72=5.35= \dfrac{10.7}{2} = 5.35
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