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Calculate the [Fe^(2+)] in a solution co...

Calculate the `[Fe^(2+)]` in a solution containing `0.2M [Fe(CN)_(6)]^(4-)` and `0.1 M CN^(Theta). K_(f)Fe (CN)_(6)^(4-) = 1 xx 10^(24)`.

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The correct Answer is:
A, B
`Fe^(2+) + 6CN^(Theta) hArr Fe(CN)_(6)^(4-)`
`K_(f) = ([Fe(CN)_(6)^(4-)])/([Fe^(2+)][CN^(Theta)]^(6))`
`10^(24) = (0.2)/((x)(0.1)^(6)) , x = 2xx 10^(-19)M`.
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