What is the molar solubility of AgCl(s) in 0.1 M `NH_(3)(aq)?K_(sp)(AgCl)=1.8xx10^(-10), K_(f)[Ag(NH_(3))_(2)]^(+)=1.6xx10^(7).`

Explanation

Let the solubility of `AgCl(s)` is S mol/litre
`AgCl(s)hArrunderset((S-X)M)(Ag^(+)(aq))+underset(SM)(Cl^(-))(aq)`
`{:(,Ag^(+)+,2NH_(3), hArr,[Ag(NH_(3))_(2)]^(+)),("Initial conc.",S,0.1,,0),("Conc. at equlibrium",(S-x)M,(0.1-2x)M,,xM):}`
`K_(sp)=[Ag^(+)][Cl^(-)]=S(S-x)`
`K_(f)=([Ag(NH_(3))_(2)]^(+))/([Ag^(+)][NH_(3)]^(2))=(x)/((S-x)(0.1-2x)^(2))`
We can assume here `0.1-2x~~0.1` because the solubility product of `K_(sp)` is very how so it is obvious that x will be less than 0.1
So, `0.1-2x~~0.1M`
`S(S-x)=1.8xx10^(-10)" "...(i) `
`(x)/((S-x)(0.01))=1.6xx10^(7)" "...(ii)`
`(x)/(S-x)=1.6xx10^(5)`
`1.6xx10^(5)S-(1.6xx10^(5))X=X`
`X=(1.6xx10^(5)S)/((1+1.6xx10^(5)))`
On putting the value of x in equation (i), we get,
`S[S-(1.6xx10^(5)S)/(1+1.6xx10^(5))]=1.8xx10^(-10)`
`S^(2)[(1+1.6xx10^(5)-1.6xx10^(5))/(1+1.6xx10^(5))]=1.8xx10^(-10)`
`S^(2)=1.8xx10^(-10)xx1.6xx10^(5)" "(1+1.6xx10^(5))~~1.6xx10^(5)`
`S=5.36xx10^(-3)M`
Solubility of `AgCl=5.36xx10^(-3)M`