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

Suppose number of moles of AgI which may be dissolved in 1.0 litre of 1.0 M `CN^(-)` solution = x. Then
`{:(,AgI (s) ,+ ,2CN^(-) (aq) ,hArr,[Ag(CN)_(2)]^(-),+,I^(-)),("Initial moles " ,x,,1,," "0,,0),("Moles after reaction",0,,1-2x,," "x,,x),(,,,,,,,),(,,,,,,,):}`
`K_(eq)=([Ag(CN)_(2)]^(-)[I^(-)])/([CN^(-1)]^(2))=(x^(2))/((1-2x)^(2))` ...(i)
Further, `AgI (s) hArr Ag^(+) (aq) + (Aq) + I^(-) (aq)`
`K_(sp) = [Ag^(+)][I^(-)] = 1.2xx10^(-17) ` (Given) ...(ii)
`Ag^(+) (aq) + 2 (CN^(-) (aq)) hArr [Ag(CN)_(2)]^(-)`
`K_(c)=([Ag(CN)_(2)]^(-))/([Ag^(+) ] [ CN^(-) ] ^(2) ) = 7.1xx10^(19) ` (Given)
From eqn. (i) , (ii) and (iii)
`K_(eq)=K_(sp)xxK_(c)=(1.2xx10^(-17))xx(7.1xx10^(19))= 8.52xx10^(2)`
`:. (x^(2))/((1-2x)^(2)) = 8. 52xx10^(2) or (x)/(1-2x) = 29.2 or x = 29.2- 58.4x or x = 0.49 ` mole