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

`{: (AgC(s) hArr Ag^(+)+Cl^(-)" "K_(1)=K_(sp)),(Ag^(+)+2NH_(3)hArrAg^(+)(NH_(3))_(2)" " K_(2)-K_(f)),(bar(AgCl(s)+2NH_(3)hArrAg^(+)(NH_(3))_(2)+Cl^(-)" " K=K_(sp)xxK_(f))):}`
` :. K= ([Ag(NH_(3))_(2)^(+)][Cl^(-)])/([NH_(3)]^(2))` (Given solubility of AgCl = `0.10` )
` :. [Ag(NH_(3))_(2)^(+) ] = 0.10 M `
Also , `[Cl^(-)] = 0.1 `
` :. [NH_(3)]^(2) = 6.25 rArr [NH_(3)] = 2.5 M `
Thus , `(NH_(3))` at equilibrium = 2.5 M
Also `0.2 ` of `NH_(3)` must have been used to dissolve `0.1` M AgCl
` :. [NH_(3)]_("Total") = 2.5 + 0.2 = 2.7 `