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# ].
A solution is prepared by mixing 75 ml of 0.030 M
1 Answer
I got
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
#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
#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
#24ln 10 = -4.4830 - ln(0.0113 - x) + 17.1672#
#= 12.6842 - ln(0.0113 - x)#
Solving for
#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...
#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.