Determine the number of mole of AgI which may be dissolved in `1.0` litre of `1M 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.

Explanation

Given, `Agl(s) hArr Ag^(+) (aq) + I^(-) (aq)` , `K_(sp)=[Ag^+][I^-]=1.2xx10^(-17)`…(i)
`Ag^+(aq) + 2CN^(-)(aq) hArr [Ag(CH_2)]^(-)(aq) , K_f=([Ag(CH)_2^-])/([Ag^+][CN^-]^2)=7.1xx10^19`….(ii)
Let ‘x’ mole of Agl be dissolved in `CN^–` solution, then
Now `Agl(s) + 2CN^(-) hArr [Ag(CN)_2^(-)] + I^-`
`{:("Mole before reaction",1,0,0),("Mole at equilibrium" , (1-2x) , x, x) :}`
By equations (i) and (ii) . `K_(eq)=K_(sp)xxK_f`
`K_(eq)=([Ag(CN)_2^-][I^-])/([CN^-]^2)=1.2xx10^(-17)xx7.1xx10^19`
`K_(eq)=8.52xx10^2` ...(iii)
`therefore K_(eq)=8.52xx10^2=x^2/((1-2x)^2)` or `x/(1-2x)=29.2`
`therefore ` x = 0.49 mole.