Determine the concentration of NH_(3) solution whose one litre can dissolve 0.10 mole AgCl. K_(sp) of AgCl and K_(f)of Ag(NH_(3))_(2)^(+) are 1.0xx10^(-10)M^(2) and 1.6xx10^(7)M^(-2) respectively .

Determine the concentration of NH_(3) solution whose one litre can dis

1.	Determine the concentration of NH_(3) solution whose one litre can dissolve 0.10 mole AgCl. K_(sp) of AgCl and K_(f)of Ag(NH_(3))_(2)^(+) are 1.0xx10^(-10)M^(2) and 1.6xx10^(7)M^(-2) respectively .
Answer» Solution :`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`