Question
On adding $$AgN{O_3}$$ solution to $$KI$$ solution, a negatively charged colloidal sol will be formed in which of the following conditions?
A.
$$100\,mL\,\,{\text{of}}\,\,0.1\,M\,AgN{O_3}$$ $$ +\, 100\,mL\,\,{\text{of}}\,\,0.1\,M\,KI$$
B.
$$100\,mL\,\,{\text{of}}\,\,0.1\,M\,AgN{O_3}$$ $$ + \,50\,mL\,\,{\text{of}}\,\,0.2\,M\,KI$$
C.
$$100\,mL\,\,{\text{of}}\,\,0.2\,M\,AgN{O_3}$$ $$ + \,100\,mL\,\,{\text{of}}\,\,0.1\,M\,KI$$
D.
$$100\,mL\,\,{\text{of}}\,\,0.1\,M\,AgN{O_3}$$ $$ + \,100\,mL\,\,{\text{of}}\,\,0.15\,M\,KI$$
Answer :
$$100\,mL\,\,{\text{of}}\,\,0.1\,M\,AgN{O_3}$$ $$ + \,50\,mL\,\,{\text{of}}\,\,0.2\,M\,KI$$
Solution :
If colloidal sol of $$AgI$$ is prepared by adding $$KI$$ solution to $$AgN{O_3}$$ till $$KI$$ is in slight excess, iodide ion $$\left( {{I^ - }} \right)$$ will be adsorbed on the surface of $$AgI$$ thereby giving a negative charged sol.
$$AgI + \mathop {{I^ - }}\limits_{\left( {{\text{From}}\,\,KI} \right)} \to \mathop {AgI:{I^ - }}\limits_{{\text{Negative}}\,\,{\text{sol}}} $$