JEE MAIN - Chemistry (2023 - 11th April Morning Shift - No. 7)
$$25 \mathrm{~mL}$$ of silver nitrate solution (1M) is added dropwise to $$25 \mathrm{~mL}$$ of potassium iodide $$(1.05 \mathrm{M})$$ solution. The ion(s) present in very small quantity in the solution is/are :
$$\mathrm{I^-}$$ only
$$\mathrm{K^+}$$ only
$$\mathrm{Ag^+}$$ and $$\mathrm{I^-}$$ both
$$\mathrm{NO_3^-}$$ only
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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