# For complex formation of iron(III) thiocyanate at a certain ionic strength, #K_f = 1099#. If its initial concentration is #"4.0 M"#, determine the equilibrium concentration of all species in solution for the dissociation of iron(III) thiocyanate in water?

##### 1 Answer

You should get:

#["Fe"^(3+)] = ["SCN"^(-)] = "0.060 M"# #["FeSCN"^(2+)] = "3.94 M"#

Unlike many equilibrium reactions, this one is actually the backwards reaction. Many transition metal complexation reactions are favorable forwards, towards forming the complex.

Some formation constants are listed here, involving reactions with multiple ligands in a row.

This is *one step* in one of those kinds of reactions, but since it's the backwards reaction, the complex formation constant (for this first step) is usually reported, which is therefore

You can treat this just like any other **equilibrium problem**. At its core, it's very similar to acid-base equilibrium, solubility equilibrium, and other kinds of equilibrium you've seen.

Recall that the definition of

#aA(aq) + bB(aq) rightleftharpoons cC(s) + dD(aq) + eE(l)#

is:

#K = ([D]^d)/([A]^a[B]^b)#

Similarly, we just have a *complex dissociation equilibrium* reaction:

#"FeSCN"^(2+)(aq) rightleftharpoons "Fe"^(3+)(aq) + "SCN"^(-)(aq)#

whose equilibrium **complex dissociation constant** is:

#K_D = (["Fe"^(3+)]["SCN"^(-)])/(["FeSCN"^(2+)])#

You can **set up an ICE table** just as for other equilibria.

#"FeSCN"^(2+)(aq) rightleftharpoons "Fe"^(3+)(aq) + "SCN"^(-)(aq)#

#"I"" ""4.0 M"" "" "" "" "" ""0 M"" "" "" ""0 M"#

#"C"" "-x" M"" "" "" "+x " M"" "" "+x " M"#

#"E"" "(4.0 - x) "M"" "" "x " M"" "" "" "x " M"#

So the equilibrium constant can be utilized:

#K_D = 9.1xx10^(-4) = (x^2)/(4.0 - x)#

If

The full quadratic would have been:

#x^2 + 9.1xx10^(-4)x - 4.0(9.1xx10^(-4)) = 0#

When you plug

#x = (-b pm sqrt(b^2 - 4ac))/(2a)# ,

you should get that a *physically reasonable value* for

However,

The small

#9.1 xx 10^(-4) ~~ x^2/4.0#

#=> x = sqrt(4.0 cdot 9.1 xx 10^(-4)) = "0.0603 M"#

Either way, the **equilibrium concentrations of each ion** in the solution are:

#color(blue)(["Fe"^(3+)] = ["SCN"^(-)]) = color(blue)("0.060 M")# #color(blue)(["FeSCN"^(2+)]) = 4.00 - 0.060 = color(blue)("3.94 M")#