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.

Explanation

`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`