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

The reaction is
`AgCl+2NH_3 hArr [Ag(NH_3)_2]^(+) + Cl^(-)`
`K_(eq)=([Ag(NH_3)_2]^+[Cl^-])/[NH_3]^2`
`K_(sp)` of (AgCl)=`[Ag^+][Cl^-]`
`K_f=([Ag(NH_3)_2]^+)/([Ag^+][NH_3]^2)`
`K_(eq)=K_(sp)xxK_f =1.0xx10^(-10)xx1.6xx10^7`
`=1.6xx10^(-3)`
When 0.1 mole of AgCl is dissolved in 1.0 L of solution, then at equilibrium,
`[Ag(NH_3)_2]^(+)` =0.1 M
`[Cl^-]` = 0.1 M
`K_(eq)=((0.1)xx(0.1))/[NH_3]^2`
`1.0xx10^(-3)=(1xx10^(-2))/[NH_3]^2`
`[NH_3]^2 =(1.0xx10^(-2))/(1.0xx10^(-3))=6.25 M^2`
`[NH_3]` =2.5 M
Thus, the equilibrium concentration of `NH_3`=2.5 M . However , 0.2 mol of `NH_3` are used in dissolving AgCl, so that the conc. of `NH_3` is 2.5+0.2 =2.7 M