JEE Main 2017ChemistryIonic EquilibriumHardMCQ

JEE Main 2017 — Ionic Equilibrium Question with Solution

JEE Main 2017 (09 Apr Online)

Question

50 mL of 0.2M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia solution is 4.75, the pH of the mixture will be:

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Show full solutionCorrect option: B
Correct answer
B9.25

Step-by-step explanation

                                          NH3          +                      HCl                     ⇌      NH4Clmoles at ​​​​​​​ t=0         50×.2×10−3            25×.2×10−3           −moles at ​​​t=t         25×.2×10−3                       0                           25×.2×10−3

∴ salt & base present in soluton, so it is basic Buffer solution.

pOH=pkb  NH3 +log⁡saltbase=4.75∴ pOH=Pkb=4.75

pH=14-pOH=14-4.75=9.25.

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About this question

This is a previous-year question from JEE Main 2017, covering the Ionic Equilibrium chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.