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JEE Main201910 Apr 2019Evening ShiftChemistryIonic EquilibriumActual

The p H of a 0.02 M N H 4 C l solution will be [Given: K b N H 4 O H = 10 - 5 and log ⁡ 2 = 0.301 ]

Options

  1. A4.65
  2. B2.65
  3. C4.35
  4. D5.35

Correct answer

D. 5.35

Step-by-step solution

N H 4 C l is a salt of a strong acid and a weak base. pH = 1 2 pK w - pK b - logC = 1 2 14 - 5 - log ⁡ 2 × 10 - 2 = 1 2 9 - log ⁡ 2 - log ⁡ 10 - 2 = 1 2 9 - log ⁡ 2 + 2 = 10.7 2 = 5.35

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