25mL of silver nitrate solution ( 1M ) is added dropwise to 25mL of...

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:

${AgNO}_{3} + KI \to AgI + {KNO}_{3}$

Millimoles of ${AgNO}_{3} = 25$

Millimoles of $KI = 25 \times 1 . 05 = 26 . 25$

$∴ KI$ is in excess & $AgI$ forms negatively charged colloid. (Some ${Ag}^{+}$remains in solution) lons ${Ag}^{+}$& ${F}^{-}$are therefore, present in very small quantity.