InterviewSolution
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InterviewSolution
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What is `[Cd^(2+)]` in `1.0L` of solution prepared by dissolving `0.001mol Cd(NO_(3))_(2)` and `1.5 mmol. NH_(3) ? K_(d)` for the dissociation of `Cd(NH_(3))_(4)^(2+)` into `Cd^(2+)` and `4NH_(3)` is `1.8 xx 10^(-7)`. Neglect the amount of `Cd` in complexes containing fewer than `4` ammonia molecules. |
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Answer» `{:(,Cd^(2+)+,4NH_(3)hArr,Cd(NH_(3))_(4)^(2+),),("Initial",rArr0.001,1.5,0,),("Used up",rArr0.001-x,4(0.001-x),0.001-x,),("Eq.",rArr~~x,~~1.5,~~0.001,):}` `Cd(NH_(3))_(4)^(2+) hArr Cd^(2+) + 4NH_(3)`. `K_(d) = ([Cd^(2+)][NH_(3)]^(4))/([Cd(NH_(3))_(4)^(2+)])` `1.8 xx 10^(-7) = ((x)(1.5)^(4))/(0.001)` `:.x = 3.6 xx 10^(-16) = [Cd^(2+)]` |
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