Question

Which mixture of the solutions will lead to the formation of negatively charged colloidal $[AgI{]}^{-1}$ sol ?

• A. $50mL$ of $1M$ $AgN{O}_3$ + $50mL$ of $0.5M$ $KI$
• B. $50mL$ of $1M$ $AgN{O}_3$ + $50mL$ of $2M$ $KI$
• C. $50mL$ of $2M$ $AgN{O}_3$ + $50mL$ of $1.5M$ $KI$
• D. $50mL$ of $0.1M$ $AgN{O}_3$ + $50mL$ of $0.1M$ $KI$

Answer 1

The correct option is B $50mL$ of $1M$ $AgN{O}_3$ + $50mL$ of $2M$ $KI$

Generally, the charge present on the colloid is due to the adsorption of common ion from the dispersion medium. Since the millimoles of $KI$ formed is maximum in option (2) $(50\times 2=100)$, it acts as the solvent and the anion $I^{–}$ is adsorbed by the colloid $AgI$ formed.

$AgN{O}_3$ + $KI$ $⟶$ $AgI$ + $KN{O}_3$
excess negatively charged colloid

Note: In offical answer key given by NTA both (a) and (b) were the answers as option (a) in original question paper was
$50mL$ of $1M$ $AgN{O}_3$ + $50mL$ of $1.5M$ $KI$