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