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
- A4.65
- B2.65
- C4.35
- 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