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

`AgCl rarr Ag^(+) + Cl^(-)`
`:. 0.1 ` mole AgCl on dissolution will give 0.1 mole `Ag^(+) and 0.1` mole `Cl^(-) ` ions . `Ag^(+)` ions will combine with `NH_(3)` as follows :
`Ag^(+) + 2 NH_(3) hArr Ag(NH_(3))_(2)^(+)`
Hence, `[Ag^(+)] + [Ag(NH_(3))_(2)^(+)] = 0.1 ` mole ...(i)
Also for the above formation reaction, we are given
`K_(f) = 1.6xx10^(7) = ([Ag(NH_(3))_(2)^(+)])/([Ag^(+)][NH_(3)]^(2))` ...(ii)
Further, `[Ag^(+) ] [ Cl^(-)]= K_(sp) = 10^(-10) or [Ag^(+)](0.1) = 10^(-10) or [Ag^(+) ] = 10^(-9)`
Neglecting this conc. in eqn. (i) , we get
`[Ag(NH_(3))_(2)^(+)]=0.1`
Putting these values in eqn. (ii) , we get
`(0.1)/(10^(-9)xx[NH_(3)]^(2))=1.6xx10^(7)`
or `[NH_(3)]^(2) = (0.1)/(10^(-9)xx1.6xx10^(7))=(1)/(16xx10^(-2)) or [NH_(3)] = (1)/(4xx10^(-1))=2.5 M`
Also 0.2 M ` NH_(3)` is needed to dissolve 0.1 M `Ag^(+)` ions. Thus,
Total `[NH_(3)]=2.5+0.2 = 2.7 M`