Solid `AgNO_3` is slowly added to a solution that is 0.001 M each in NaCl, NaBr and Nal.
(a) Calculate the concentration of `Ag^+` required to initiate the precipitation of each of AgCl, AgBr and Agl.
(b) Calculate the percentage of `I^-` precipitated before AgBr precipitates.
(c) Calculate the percentage of `I^-` precipitated before `CI^-` precipitates.

Explanation

(a) NaCl, NaBr and NaI are completely ionised,
`[Cl^(-)]=[Br^(-)]=[I^(-)] =0.001M`
`K_(sp) (AgI) =[Ag^(+)] [I^(-)]=1.5 xx 10^(-16)`
`[Ag^(+)]` to start preicpitation of AgI `=(1.5 xx 10^(-16))/(0.001) =1.5 xx 10^(-13)M`
Similarly, `[Ag^(+)]` to initiate precipitation of AgBr and AgCl may be calculated.
`[Ag^(+)]` to initiate precipitation of AgBr and AgCl may be calculated.
`[Ag^(+)]` to start precipitation of AgBr `=(K_(sp) (AgBr))/([Br^(-)])=(3.3 xx 10^(-13))/(0.001)`
`=3.3 xx 10^(-10)M`
`=3.3 xx 10^(-10)M`
And,
`[Ag^(+)]` to start precipitation of AgCl `=(K_(sp) (AgCl))/([Cl^(-)]) =(1.8 xx 10^(-10))/(0.001)`
`=1.8 xx 10^(-7)M`
(b) When AgBr just starts precipitating `[Ag^(+)]=3.3 xx 10^(-10)M`
At this state `[I^(-)]` may be calculated as
`K_(sp) (AgI) =[Ag^(+)] [I^(-)]`
or `[I^(-)]=(K_(sp) (AgI))/([Ag^(+)]) =(1.5 xx 10^(-16))/(3.3 xx 10^(-10))`
`=4.5 xx 10^(-7)M`
perentage of `I^-` not precipitated `=(4.5 xx 10^(-7))/(0.001) xx 100=0.045%`.
percentage of `I^-` precipitated before AgBr begins to precipitate
=99.955%
(c ) Similarly, `[I^(-)]` may be calculated when AgCl begins to precipitate Here `[Ag^(+)]=1.8xx 10^(-7)`
`K_(sp) (AgI)=[Ag^(+)] [I^(-)]`
`[I^(-)]= (K_(sp) (AgI))/([Ag^(+)])=(1.5 xx 10^(-16))/(1.8 xx 10^(-7))=8.3 xx 10^(-10)M`
perentage of `I^-` not precipitated `=(8.3 xx 10^(-10))/(0.001) xx 100=0.000083%`
perentage of `I^-` precipitated before AgCl begins to precipitate
=99.999917%