JEE Main 2019 — Ionic Equilibrium Question with Solution
From: JEE Main 2019 (Online) 10th April Evening Slot
Question
The pH of a 0.02 M NH4Cl solution will be :
[given Kb (NH4OH) = 10–5 and log 2 = 0.301]
[given Kb (NH4OH) = 10–5 and log 2 = 0.301]
Choose an option
Show full solutionCorrect option: B
Correct answer
B5.35
Step-by-step explanation
NH4+ + H2O ⇋ NH4OH + H+
[H+] = c
=
=
=
=
pH = = 5.35
[H+] = c
=
=
=
=
pH = = 5.35
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