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Given: Ag(NH(3))(2)^(+)hArrAg^(+)2NH(3),...

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.

Text Solution

Verified by Experts

`K_(c) = ([Ag^(+)][NH_(3)]^(2))/([Ag(NH_(3))_(2)^(+)])` ...(i)
`[NH_(3)]=1.0 M ` (as `NH_(3)` obtained from dissociation of the complex is negligible)
To know `[Ag^(+)]`, we have
`K_(sp) = [Ag^(+)][Cl^(-)]`
But `[Cl^(-)]= "Total" [Ag^(+)]` in the free and combined state `= [Ag^(+)]+[Ag(NH_(3))_(2)^(+)]`
`:. (K_(sp))/([Ag^(+)])=[Ag^(+)]+[Ag(NH_(3))_(2)^(+)] or [Ag(NH_(3))_(2)^(+)]=(K_(sp))/([Ag^(+)])-[Ag^(+)] ...(ii)`
Substituting the values in eqn. (i), we get
`6.2xx10^(-8)=([Ag^(+)][1.0]^(2))/((1.8xx10^(-10))/([Ag^(+)])-[Ag^(+)])`
Takiing `[Ag^(+)]=C "mol" L^(-1)`, we have
`6.2xx10^(-8)=(C^(2))/(1.8xx10^(-10)-C^(2))~=(C^(2))/(1.8xx10^(-10))`
or `C^(2)=11.16xx10^(-18)` ltbr. or `C=3.34 xx 10^(-9) M, i.e., [Ag^(+)]=3.3410^(-9)M`
Substituting this value in eqn. (ii), we get
`[Ag(NH_(3))_(2)^(+)]=(1.8xx10^(-10))/(3.34xx10^(-9))-3.34xx10^(-9)~=(1.8xx10^(-10))/(3.34xx10^(-9))=0.054 M`
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