For the reaction
`Ag(CN)_(2)^(ɵ)hArr Ag^(o+)+2CN^(ɵ)`, the `K_(c )` at `25^(@)C` is `4 xx10^(-19)` Calculate `[Ag^(o+)]` in solution which was originally `0.1 M` in `KCN` and `0.03 M` in `AgNO_(3)`.

Explanation

`underset((0.1-0.06))underset(0.1)(2KCN)+underset(0)underset(0.03)(AgNO_(3))tounderset(0.03)underset(0)(Ag(CN)_(2)^(-))+underset(0.03)underset(0)(KNO_(3))+underset(0.03)underset(0)(K^(+))`
`:. [Ag(CN)_(2)^(-)]=0.03 M`
`{:(Now use,(Ag(CN)_(2)^(-)),hArr,(Ag^(+)),+,2CN^(-)),(,0.03,,0,,0.04),(,(0.03-a),,a,,0.04+2a~~0.04):}`
Since `K_(c)` is too small and dissociation of `Ag(CN)_(2)^(-)` is very less and thus,
`:. 0.04+2a~~0.04` and `0.03-a~~0.03`
`:. [Ag(CN)_(2)^(-)]=0.03`, `[Ag^(+)]=a`, `[CN^(-)]=0.04`
Now `K_(c)=([Ag^(+)][CN^(-)]^(2))/([Ag(CN)_(2)^(-)])=(axx(0.04)^(2))/(0.03)`
`:. a=7.5xx10^(-18)`