25 mL of silver nitrate solution ( 1 M ) is added dropwise to 25 mL of potassium iodide ( 1 . 05 M ) solution. The ion(s) present in very small quantity in the solution is/are

Step-by-step solution

To determine the ion(s) present in very small quantity in the solution, we need to consider the chemical reaction that occurs when silver nitrate reacts with potassium iodide. The reaction between AgNO 3 and KI is a double replacement reaction and can be represented as follows: AgNO 3   +   KI   →   AgI   +   KNO 3 Millimoles of AgNO 3   =   25 Millimoles of KI   =   25   ×   1 . 05 = 26 . 25 ∴ KI is in excess & AgI forms negatively charge