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How much AgBr could dissolve in 1.0L of ...

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+) + 2NH_(3)`,
`K_(d) = 6.0 xx 10^(-8)` and `K_(sp)(AgBr) =5.0 xx 10^(-13)`.

A

`2.82xx10^(-3)M`

B

`2.22xx10^(-4)`

C

`2.22xx10^(-3)`

D

`2.82xx10^(-4)`

Text Solution

Verified by Experts

The correct Answer is:
A
`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`
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