How much `AgBr` could dissolve in `1.0L` of `0.4M NH_(3)`? Assume that `Ag(NH_(3))_(2)^(o+)` is the only complex formed. Given: the dissociation constant for
`Ag(NH_(3))_(2)^(o+) hArr Ag^(o+) xx 2NH_(3)`,
`K_(d) = 6.0 xx 10^(-8)` and `K_(sp)(AgBr) =5.0 xx 10^(-13)`.

Options

A. `2.82xx10^(-3)M`
B. `2.22xx10^(-4)`
C. `2.22xx10^(-3)`
D. `2.82xx10^(-4)`

Answer

A

Explanation

`AgBr hArrAg_(("aq."))^(+)+Br_(("aq"))^(-)`
`K_(sp)=[Ag^(+)][Br^(-)]`
`Ag^(+) +2NH_(3) hArr[Ag(NH_(3))_(2)]^(+)`
`K_(f)=([Ag(NH_(3))_(2)]^(+))/([Ag^(+)][NH_(3)]^(2))=1xx10^(8)`
Let `s` be the solubility of `AgBr` in `NH_(3)` then `[Br^(-)]=[Ag^(+)]+[Ag(NH_(3))_(2)]^(+)=S`
Since, almost all `Ag^(+)` in solution state passes to `[Ag(NH_(3))_(2)]^(+)` as `K_(f)` is very high. Thus,
`[Ag^(+)ltltlt[Ag(NH_(3))_(2)]^(+)`
`:.[Br^(-)]=[Ag(NH_(3))_(2)]^(+)=S`
Now `K_(sp)=[Ag^(+)][Br^(-)]=([[Ag(NH_(3))_(2)]^(+)])/(K_(f)xx [NH_(3)]^(2))xx[Br^(-)]`
Since `[NH_(3)]=0.4M`
`5xx10^(-13)=(SxxS)/(1.0xx10^(8)xx(0.4)^(2))`
`S^(2)=8xx10^(-6)`
`S=2.82xx10^(-3)M`