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)`.

Answer

A

Explanation

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