# 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?

Jan 22, 2017

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 ${K}_{f} = \frac{1}{K} _ D \approx 1099$.

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 $K$ for the reaction

$a A \left(a q\right) + b B \left(a q\right) r i g h t \le f t h a r p \infty n s c C \left(s\right) + \mathrm{dD} \left(a q\right) + e E \left(l\right)$

is:

$K = \frac{{\left[D\right]}^{d}}{{\left[A\right]}^{a} {\left[B\right]}^{b}}$

Similarly, we just have a complex dissociation equilibrium reaction:

${\text{FeSCN"^(2+)(aq) rightleftharpoons "Fe"^(3+)(aq) + "SCN}}^{-} \left(a q\right)$

whose equilibrium complex dissociation constant is:

${K}_{D} = \left(\left[{\text{Fe"^(3+)]["SCN"^(-)])/(["FeSCN}}^{2 +}\right]\right)$

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

${\text{FeSCN"^(2+)(aq) rightleftharpoons "Fe"^(3+)(aq) + "SCN}}^{-} \left(a q\right)$

$\text{I"" ""4.0 M"" "" "" "" "" ""0 M"" "" "" ""0 M}$
$\text{C"" "-x" M"" "" "" "+x " M"" "" "+x " M}$
$\text{E"" "(4.0 - x) "M"" "" "x " M"" "" "" "x " M}$

So the equilibrium constant can be utilized:

${K}_{D} = 9.1 \times {10}^{- 4} = \frac{{x}^{2}}{4.0 - x}$

If ${K}_{D} < {10}^{- 5}$, then a reasonable approximation could be made that $x$ is small, but it isn't smaller than ${10}^{- 5}$.

The full quadratic would have been:

${x}^{2} + 9.1 \times {10}^{- 4} x - 4.0 \left(9.1 \times {10}^{- 4}\right) = 0$

When you plug $a = 1$, $b = {K}_{D}$, and $c = - \left[{\text{FeSCN}}^{2 +}\right] {K}_{D}$ into the quadratic formula

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$,

you should get that a physically reasonable value for $x$ is $\text{0.0599 M}$.

However, $\text{4.0 M}$ is rather big of a concentration, so the approximation is fine to make since percent dissociation decreases as concentration increases.

The small $x$ approximation would instead be:

$9.1 \times {10}^{- 4} \approx {x}^{2} / 4.0$

$\implies x = \sqrt{4.0 \cdot 9.1 \times {10}^{- 4}} = \text{0.0603 M}$

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")