JEE Main 2023 — Ionic Equilibrium Question with Solution
From: JEE Main 2023 (Online) 11th April Morning Shift
Question
of silver nitrate solution (1M) is added dropwise to of potassium iodide solution. The ion(s) present in very small quantity in the solution is/are :
Choose an option
Show full solutionCorrect option: C
Correct answer
C and both
Step-by-step explanation
The reaction between silver nitrate (AgNO3) and potassium iodide (KI) forms silver iodide (AgI), which is practically insoluble in water. The reaction is as follows :
AgNO3(aq) + KI(aq) → AgI(s) + KNO3(aq)
Although the reaction stoichiometry indicates that iodide ions (I-) and silver ions (Ag+) are produced, silver iodide (AgI) precipitates out of the solution due to its low solubility, and very little Ag+ and I- ions remain in the solution. Hence, their concentration in the solution will be negligible.
So, the correct answer should be :
Option C : Ag+ and I- both
AgNO3(aq) + KI(aq) → AgI(s) + KNO3(aq)
Although the reaction stoichiometry indicates that iodide ions (I-) and silver ions (Ag+) are produced, silver iodide (AgI) precipitates out of the solution due to its low solubility, and very little Ag+ and I- ions remain in the solution. Hence, their concentration in the solution will be negligible.
So, the correct answer should be :
Option C : Ag+ and I- both
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This is a previous-year question from JEE Main 2023, 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.