Verified Solution • 2025–26 Session moderate

QJEEM19Ch7eb5: JEE Main 2019 Chemistry - Class 11 JEEM19Ch7eb5 Solution: The pH of a 0.02 M NH 4 Cl solution will be : [given K b (NH 4…

The pH of a 0.02 M NH4Cl solution will be :
[given Kb (NH4OH) = 10–5 and log 2 = 0.301]

Multiple Choice Options

  • A 2.56
  • B 5.35
  • C 4.35
  • D 4.65

Step-by-Step Verified Solution

NH4+ + H2O ⇋ NH4OH + H+

[H+] = c α\alpha

= ka(NH4+)×c\sqrt {{k_a}\left( {NH_4^ + } \right) \times c}

= kwkb(NH4OH)×c\sqrt {{{{k_w}} \over {{k_b}\left( {N{H_4}OH} \right)}} \times c}

= 10−1410−5×0.02\sqrt {{{{{10}^{ - 14}}} \over {{{10}^{ - 5}}}} \times 0.02}

= 20×10−6\sqrt {20} \times {10^{ - 6}}

∴\therefore pH = −log⁡(20×10−6)- \log \left( {\sqrt {20} \times {{10}^{ - 6}}} \right) = 5.35

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