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 AgNO3 and KI is a double replacement reaction and can be represented as follows:
AgNO3+KI→AgI+KNO3
Millimoles of AgNO3=25
Millimoles of KI=25×1.05=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.