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 ${\mathrm{AgNO}}_3$ and $\mathrm{KI}$ is a double replacement reaction and can be represented as follows:

${\mathrm{AgNO}}_3 + \mathrm{KI} \to \mathrm{AgI} + {\mathrm{KNO}}_3$

Millimoles of ${\mathrm{AgNO}}_3 = 25$

Millimoles of $\mathrm{KI} = 25 \times 1.05=26.25$

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