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Determine the concentration of `NH_(3)` solution whose one litre can dissolve `0.10` mole AgCI. `K_(SP)` of AgCI and `K_(f)` of `Ag(NH_(3))_(2)^(+)` are `1.0xx10^(-10)M^(2)` and `1.6xx10^(7)M^(-2)` respectively. |
Answer» `AgCI+2NH_(3)hArr Ag(NH_(3))_(2)^(+)+CI^(-)` `AgCI_((s))hArrAg^(+)+CI^(-)` `K_(SP)=[Ag^(+)][CI^(-)]` ………..(1) Given `Ag^(+)+2NH_(3)hArr Ag(NH_(3))_(2)^(+)` `K_(f)=([Ag(NH_(3))_(2)^(+)])/([Ag^(+)][NH_(3)]^(2))` ………….(2) By eqs. (1) and (2) `K_(SP)xxK_(f)=([Ag(NH_(3))_(2)^(+)][CI^(-)])/([NH_(3)]^(2))` or `1xx10^(-10)xx1.6xx10^(7)=(axxa)/([NH_(3)]^(2))` Given solubility of `AgCI= 0.1M` `:. a=0.1M` for `Ag(NH_(3))_(2)^(+)` and `CI^(-)` `:. [NH_(3)]^(2)=(0.1xx0.1)/(1.6xx10^(-3))=6.25` `[NH_(3)]=2.5M` Also `0.2M NH_(3)` is needed to dissolve `0.1M Ag^(+)` ionsthus `[NH_(3)]=2.5+0.2=2.7M` |
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