Calculate the `[Fe^(2)]` in a solution prepared by mixting `75.0mL` of `0.03M FeSO_(4)` with `125.0mL` of `0.2 "M KCN" K_(f) Fe (CN)_(6)^(4-) = 1 xx 10^(24)`.

Answer

A, C

Explanation

`{:(Fe^(2+)+6CN^(Theta)hArr Fe(CN)_(6)^(4-)),("Initial"[{:(75xx0.03,125xx0.2),(=2.25 mmol,=25.0 mmol),(=(2.25 mmol)/((125+75) mL),),(=0.01125 M,):}):}`
Since `K_(f)` value is very large, `Fe^(2+)` ion is essentially all in the form of its complex ion.
`:. [Fe(CN)_(6)^(4-)] = 0.01125M`.
Initial mmol of `CN^(Theta) = 25.0 m mol`.
Complexed mmol of `CN^(Theta) = 6 xx 2.25 = 13.5 mmol`
`[CN^(Theta)]` left in solution `= ((25-13.5))/((125 +75))`
` = (11.5mmol)/(200mL) = 0.0575 M`
`:. K_(f) = ([Fe(CN)_(6)^(4-)])/([Fe^(2+)][CN^(Theta)]^(6))`
`10^(24) = ((0.01125))/(x(0.0575)^(6)) :. [Fe^(2+)] = x = 3 xx 10^(-19)M`