1.

The solubility of AgSCN in 0.002 M NH_(3) is (K_(sp) for AgSCN=1.0xx10^(-12),K_(d) for Af(NH_(3))_(2)^(+)=1.0xx10^(-8))

Answer»

`3xx10^(-5)M`
`4XX10^(-4)M`
`4xx10^(-5)M`
`2xx10^(-5)M`

Solution :`AG(NH_(3))_(2)^(+)hArrAg^(+)+2NH_(3)`
`K_(d)=1.0xx10^(-8)=([Ag^(+)][NH_(3)]^(2))/([Ag(NH_(3))_(2)^(+)])`
If S mol `L^(-1)` = solubility of AgSCN, `[Ag(NH_(3))^(2)]^(+)`
`=S,[SCN^(-)]=S`
Hence, `([Ag^(+)][0.002]^(2))/(S)=1.0xx10^(-8)`
`implies[Ag^(+)]=(1.0xx10^(-8)S)/(4xx10^(-6))=2.5xx10^(-13)S`
Also, `K_(sp)[AgSCN]=1.0xx10^(-12)=[Ag^(+)][SCN^(-)]`
`=2.5xx10^(-3)SxxSimpliesS=2.0xx10^(-5)molL^(-1)`


Discussion

No Comment Found

Related InterviewSolutions