JEE Main · 2019 · Shift-IImediumIEQ-042

The pH of a 0.02 M NH4Cl solution will be: [Kb(NH4OH) = 10-5,\ 2 = 0.301]

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

The pH of a 0.02 M NH4Cl\mathrm{NH_4Cl} solution will be:

[Kb(NH4OH)=10−5, log⁡2=0.301][K_b(\mathrm{NH_4OH}) = 10^{-5},\ \log 2 = 0.301]

Options
  1. a

    4.65

  2. b

    2.65

  3. c

    4.35

  4. d

    5.35

    ✓
Correct Answerd

5.35

Detailed Solution

Step 1 — NH4Cl\mathrm{NH_4Cl} is a salt of weak base + strong acid →\rightarrow acidic solution:

NH4++H2O⇌NH4OH+H+Kh=KwKb=10−1410−5=10−9\mathrm{NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+} K_h = \frac{K_w}{K_b} = \frac{10^{-14}}{10^{-5}} = 10^{-9}

Step 2 — pH formula: pH=7−12(pKb+log⁡C)=7−12(5+log⁡0.02)=7−12(5+log⁡2×10−2)=7−12(5+0.301−2)=7−12(3.301)=7−1.65=5.35\mathrm{pH} = 7 - \frac{1}{2}(\mathrm{pK_b} + \log C) = 7 - \frac{1}{2}(5 + \log 0.02) = 7 - \frac{1}{2}(5 + \log 2 \times 10^{-2}) = 7 - \frac{1}{2}(5 + 0.301 - 2) = 7 - \frac{1}{2}(3.301) = 7 - 1.65 = 5.35

Answer: Option (4) — 5.35

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