The solubility product of AgCl at `25^@C" is "1 xx 10^(-10)`. A solution of `Ag^(+)` at a concentration of `4 xx 10^(-3)M` just fails to yeild a precipitate of AgCl with concentration of `1 xx 10^(-3)M` of `Cl^(-)` when the concentration of `NH_(3)` in the solution is `2 xx 10^(-2)M`. Calculate the equilibrium constant for
`[Ag(NH_(3))_(2)]^+ Leftrightarrow Ag^(+) +2NH_(3)`.

Explanation

`K_(sp)=[Ag^(+)] [Cl^(-)]`
`10^(-10)=[Ag^(+)] (10^(-3)), [Ag^(+)]=10^(-7)`
Now,
`4 xx 10^(-3)M" "10^(-7)M" "2 xx 10^(-2)M" Initially"`
`Ag(NH_(3))_(2)^(+) Leftrightarrow Ag^(+) +2NH_(3)`
`(4 xx 10^(-3)-x)M " "(x+10^(-7)M " "(2x+2 xx 10^(-2))M" At eqb."`
`=4 xx 10^(-3) " "=10^(-7)" "=2 xx 10^(-2)`
Thus, `K=([Ag^(+)] [NH_(3)]^(2))/([Ag(NH_(3))_(2)^(+)])=(10^(-7) xx (2 xx 10^(-2))^2)/(4 xx 10^(-3))=10^(-8)`.