InterviewSolution
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InterviewSolution
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The solubility proudct of AgCl at `25^(@)C` is `1xx10^(-10)` A solution of `Ag^(+)` at a concentration `4xx10^(-3)` M just fails to yield a prenciitate of AgCl with concentration of `1xx10^(-3) M Cl^(-)` when the concentration of `NH_(3)` in the solution is `2xx10^(-2)M.` Calculate the equlibrium constant for `[Ag(NH_(3))_(2))hArrAg^(+)+2NH_(3)` |
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Answer» `AgCl(s)hArrAg^(+)(aq)+Cl^(-)(aq)` `K_(sp)=[Ag^(+)][Cl^(-)]` `10^(-10)=[Ag^(+)][10^(-3)]` `[Ag^(+)]=10^(-7)M` Since the initial concentration of `Ag^(+)is 4xx10^(-3)M,` therefore, ionic product of AgCl will be `[Ag^(+)][Cl^(-)]=4xx10^(-3)xx10^(-3)=4xx10^(-6)` which is greater than `K_(sp)` Thus precipitation will take place but in the question at the concentration of `Ag^(+)` ion `4xx10^(-3)` M there is no precipitation, it means the concentration of `Ag^(+)` will be equal to `10^(-7)` M or less than this. But the precipitation just fails, therefore, we assume that the equlibrium concentration of `Ag^(+)is 10^(-7)` M. `Ag^(+)+2NH_(3)hArr[Ag(NH_(3))_(2)]^(+)` `{:("Initial conc.",4xx10^(-3)M,CM,0),("At equlibrium",(4xx10^(-3)-x),(c-2x)M,xM):}` Since at equlibrium `[Ag^(+)]=10^(-7)M` and conc. of `NH_(3)=2xx10^(-2)M` `4xx10^(-3)-x~~10^(-7)" "(C-2x)=2xx10^(-2)` `thereforex~~4xx10^(-3)` `K_(f)([Ag(NH_(3))_(2)]^(+))/([Ag^(+)][NH_(3)]^(2))=(4xx10^(-3))/((10^(-7))(2xx10^(-2)))=10^(5)` `therefore` Therefore equlibrium constant for the reaction `[Ag(NH_(3))_(2)]^(+)hArrAg^(+)+2NH_(3)K_(f)` `K_(f)=(1)/(K_(f))=(1)/(10^(8))=10^(-8)` |
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