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

`0.1M KCN` and `0.03 M AgNo_(3)` are mixed.
Since `K_(eq)` of `Ag^(o+) +2CN^(Θ) hArr Ag(CN)_(2)^(Θ)` is very-very high `(K =(1)/(4)xx10^(-19))`, first assume that whole of `Ag^(o+)` is converted to `Ag(CN)_(2)^(Θ)`
`{:(Ag^(o+)+,2CN^(Θ)rarr,Ag(CN)._(2)^(Θ),,),(0.03M,0.1M,0,,),(0,0.1-2xx0.03M,0.03M(K_(eq)=4xx10^(-19)),,):}`
Now, `0.03Ag(CN)_(2)^(Θ)` dissociate as follows :
`{:(Ag(CN)_(2)^(Θ)hArr,Ag^(o+)+,2CN^(Θ),,),(0.03M,-,0.04M,,),(0.03-x,x,(0.04+2x),,):}`
`k_(c) = ([Ag^(o+)][CN^(Θ)]^(2))/([Ag(CN)_(2)^(Θ)]) = 4xx10^(-19)`
`rArr [Ag^(o+)] = (K_(c)[Ag(CN)_(2)(Θ)])/([CN^(Θ)]^(2)) = (4xx10^(-9) (0.03-x))/((0.04+2x)^(2))`
Solve the above equation by assuming 'x' to be very small
`rArr [Ag^(o+)] = (4xx10^(-19)(0.03))/((0.04)^(2)) = 7.5 xx 10^(-18)M`
[Verify the approximation yourself]