Question

Determine the number of mole of AgI which may be dissolved in `1.0` litre of `M CN^(-)` solution. `K_(SP)` for AgI and `K_(C)` for `Ag(CN)_(2)^(-)` are `1.2xx10^(-17)M^(2)` and `7.1xx10^(19)M^(-2)` respectively.

Answer 1

Given, `AgI_((s))hArrAg_((aq.))^(+)+I_((aq.))^(-)`
`K_(SP)=[Ag^(+)][I^(-)1.2xx10^(-17)` ….(1)
`Ag_((aq.))^(+)+2CN_((aq.)hArr[Ag(CN_(2)]_((aq.))^(-)`,
`K_(f)= ([Ag(CN)_(2)^(-)])/([Ag^(+)][CN^(-)]^(2))=7.1xx10^(19)` ….(2)
Let x mole of AgI be dissolved in `CN^(-)` solution.
`{:(,AgI_((s))+2CN^(-)hArr,[Ag(CN)_(2)^(-)]+,I^(-)),("Mole before reaction",1,0,0),("Mole after reaction",(1-2x),x,x):}`
By eqs. (1) and (2), `K_(eq)=K_(SP)xxK_(f)`
`=([Ag(CN)_(2)^(-)][I^(-)])/([CN^(-)]^(2))= 1.2xx10^(-17)xx7.1xx10^(19)`
`K_(eq)=8.52xx10^(2)`
`:. K_(eq)=8.52xx10^(2)=(x.x)/((1-2x^(2)))=(x^(2))/((1-2x)^(2))`
or `(x)/(1-2x)=29.2`
Thus, `x=29.2-58.4x`
or `x=0.49` mole