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JEE Chemistry 2023 Question with Solution

25 mL25 \, \text{mL} of silver nitrate solution (1 M1 \, \text{M}) is added dropwise to 25 mL25 \, \text{mL} of potassium iodide (1.05 M1.05 \, \text{M}) solution. The ions(s) present in very small quantity in the solution is/are:

  • A

    NO3−\text{NO}_3^- only

  • B

    Ag+\text{Ag}^+ and I−\text{I}^- both

  • C

    Ag+\text{Ag}^+ only

  • D

    I−\text{I}^- only

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: 25 mL25 \, \text{mL} of silver nitrate solution of concentration 1 M1 \, \text{M} is mixed with 25 mL25 \, \text{mL} potassium iodide solution of concentration 1.05 M1.05 \, \text{M}.

Find: Which ion is present in very small quantity in the final solution.

On adding AgNO3\text{AgNO}_3 to KI\text{KI}, AgI\text{AgI} will form, and the solubility of AgI\text{AgI} is very low. As a result, only Ag+\text{Ag}^+ will remain in very small quantities in the solution. The majority of the I−\text{I}^- ions will precipitate out as AgI\text{AgI}, leaving very few ions in solution.

Therefore, the correct option is C, that is, Ag+\text{Ag}^+ only.

Common mistakes

  • Assuming both Ag+\text{Ag}^+ and I−\text{I}^- remain equally small in solution is incorrect because the solution states that AgI\text{AgI} is highly insoluble and identifies only Ag+\text{Ag}^+ as the ion left in very small quantity. Use the precipitation of AgI\text{AgI} to decide the answer.

  • Choosing NO3−\text{NO}_3^- is incorrect because nitrate ions are spectator ions here and do not form an insoluble salt with the given ions. Focus on the insoluble product AgI\text{AgI} rather than ions that remain freely soluble.

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