If the solution `0.5 M NH_(3)` and stability constant for `A^(+)(NH_(3))_(2)` is `K_(stb)=([Ag(NH_(3))_(2)]^(+))/([Ag^(+)(aq)][NH_(3)]^(2))=6.4xx10^(7),`
then find the solubility of AgCl in the above solution `K_(sp)of AgCl=2xx10^(-10)`

Explanation

`K_(eq)=K_(sp)xxK_(stb)`
`implies2xx10^(-10)xx6.4xx10^(7)=([Ag(NH_(3))_(2)][Cl^(-)])/([NH_(3)]^(2))=(SxxS)/((0.5-2S)^(2))`
`implies(S)/((0.5-2S))=sqrt(2xx64xx10^(-4))`
`impliesS=0.046`