Given: `Ag(NH_(3))_(2)^(+)hArrAg^(+)2NH_(3), K_(C)=6.2xx10^(-8)` and `K_(SP)` of `AgCI=1.8xx10^(-10)` at 298 K. Calculate the concentration of the complex in `1.0M` aqueous ammonia.

Explanation

For AgCl
`AgCl hArr Ag^(+)+Cl^-`
`K_(sp)=[Ag^+][Cl^-]`
It is given that
`[Ag(NH_3)_2]^(+) hArr Ag^(+) + 2NH_3`
`K_c=6.2xx10^(-8)`
`Ag^(+) + 2NH_3 hArr [Ag(NH_3)_2]^+`
`K_f=1/K_c=1/(6.2xx10^(-8))=1.61xx10^7`
Now `K_f=([Ag(NH_3)_2]^+)/([Ag^+][NH_3]^2)`
`[Ag^+]=([Ag(NH_3)_2]^+)/(K_f[NH_3]^2)`
Since the formation constant of the complex is very high , most of the `Ag^+` which dissolves must be converted into complex and each `Ag^+` dissolves also requires dissolution of `Cl^-` .
`therefore [Ag(NH_3)_2]^(+) =[Cl^-]` =C (assume)
Now `K_(sp)=[Ag^+][Cl^-]`
`=([Ag(NH_3)_2]^(+))/(K_f[NH_3]^2)xx[Cl^-]`
`=C/(K_f(1)^2)xxC`
`C^2=K_(sp).K_f =1.8xx10^(-10)xx1.61xx10^7`
`=2.898xx10^(-3)`
`C=5.38xx10^(-2)` M