$25 \mathrm{~mL}$ 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 :

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 : <br/><br/>AgNO₃(aq) + KI(aq) → AgI(s) + KNO₃(aq) <br/><br/>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. <br/><br/>So, the correct answer should be : <br/><br/>Option C : Ag+ and I- both