Question

What is the concentration of ferrous ion?

A solution is prepared by mixing 75 ml of 0.030 M `FeSO_4` with 125 ml of 0.20 M KCN?The stability constant for the complex`[Fe(CN)_6]^-4` is`1* 10^24`.[Given:`(0.0575)^6`=`3.62*10^-8`].

Answer 1

I got `["Fe"^(2+)] = 3.2 xx 10^(-19) "M"`.

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Well, the reaction is:

`"FeSO"_4(aq) + 6"KCN"(aq) -> "K"_4["Fe"("CN")_6] (aq) + "K"_2"SO"_4(aq)`

For simplicity, we include only the complexation participants:

`"Fe"^(2+)(aq) + 6"CN"^(-)(aq) -> ["Fe"("CN")_6]^(4-)(aq)`

The concentration of iron at the start of the reaction is

`"0.030 M" xx "75 mL"/"75 + 125 mL" = "0.0113 M"`,

due to dilution by mixing.

The concentration of cyanide at the start of the reaction is

`"0.20 M" xx "125 mL"/"75 + 125 mL" = "0.125 M"`,

again due to dilution by mixing.

Therefore, we construct the ICE table using molar concentrations:

`"Fe"^(2+)(aq) " "+" " 6"CN"^(-)(aq) -> ["Fe"("CN")_6]^(4-)(aq)`

`"I"" "0.0113" "" "" "" "0.125" "" "" "" "" "0`
`"C"" "-x" "" "" "" "" "-6x" "" "" "" "+x`
`"E"" "0.0113-x" "" "0.125-6x" "" "" "x`

The equilibrium formation of the complex is then given by:

`K_f = 1 xx 10^24 = ([["Fe"("CN")_6]^(4-)])/(["Fe"^(2+)]["CN"^(-)]^6)`

`= x/((0.0113 - x)(0.125 - 6x)^6)`

Since `K_f` is so large, we approximate that `x ~~ 0.0113`, and notice that iron is the limiting reactant. For the purposes of maximizing accuracy, consider the following solving method.

`ln K_f = ln(10^24) = 24ln10`

`= ln[x/((0.0113 - x)(0.125 - 6x)^6)]`

`= lnx - ln(0.0113 - x) - 6ln(0.125 - 6x)`

Now, we can plug in `x = 0.0113` EXCEPT for where we would get `ln -> -oo`. This gives us:

`24ln10 = ln(0.0113) - ln(0.0113 - x) - 6ln(0.125 - 6(0.0113))`

Evaluate the right-hand side to get:

`24ln10 = -4.4830 - ln(0.0113 - x) - 6ln(0.0572)`

If you notice, this would be where that hint came from. Without the `ln`, we would have gotten that `(0.0575)^6 = 3.62 xx 10^(-8)`, since `lna^c = clna`. We proceed to get:

`24ln 10 = -4.4830 - ln(0.0113 - x) + 17.1672`

`= 12.6842 - ln(0.0113 - x)`

Solving for `ln(0.0113 - x)`, we obtain:

`ln(0.0113 - x) = 12.6842 - 24ln10`

`= -42.5778`

As a result,

`color(blue)(["Fe"^(2+)]) = 0.0113 - x`

`= e^(-42.5778)`

`= color(blue)ul(3.2 xx 10^(-19) "M")` to two significant figures.

And just to verify that this is correct... `x = 0.0113 - e^(-42.5778)`, and so we do indeed get `K_f` back:

`K_f = (0.0113 - e^(-42.5778))/((0.0113 - 0.0113 + e^(-42.5778))(0.125 - 6(0.0113 - e^(-42.5778)))^6)`

`~~ (0.0113)/(e^(-42.5778)(0.125 - 6(0.0113))^6)`

`= ul(1.0_(0001349) xx 10^24) = 1 xx 10^24` `color(blue)(sqrt"")`

where the subscripts indicate digits past the last significant figure.