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?

Question

Caleb Adams · Accepted Answer

You ought to obtain: While many transition metal complexation reactions are favorable forwards, towards forming the complex, this one is actually the backwards reaction, in contrast to many equilibrium reactions. Here is a list of some formation constants for reactions involving consecutive ligands. 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 = 1/K_D ~~ 1099#. This is essentially very similar to acid-base equilibrium, solubility equilibrium, and other types of equilibrium you've seen, so you can approach it just like any other equilibrium problem. Recall that the definition of #K# for the reaction #aA(aq) + bB(aq) rightleftharpoons cC(s) + dD(aq) + eE(l)# is: #K = ([D]^d)/([A]^a[B]^b)# In the same way, we only have an intricate dissociation equilibrium reaction: #"FeSCN"^(2+)(aq) rightleftharpoons "Fe"^(3+)(aq) + "SCN"^(-)(aq)# whose complex dissociation constant in equilibrium is: #K_D = (["Fe"^(3+)]["SCN"^(-)])/(["FeSCN"^(2+)])# An ICE table can be set up in the same way as 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"#  Therefore, the equilibrium constant is useful: #K_D = 9.1xx10^(-4) = (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)#.  There would have been a complete quadratic: #x^2 + 9.1xx10^(-4)x - 4.0(9.1xx10^(-4)) = 0# When you plug #a = 1#, #b = K_D#, and #c = -["FeSCN"^(2+)]K_D# into the quadratic formula #x = (-b pm sqrt(b^2 - 4ac))/(2a)#,  you should get that a physically reasonable value for #x# is #"0.0599 M"#.  However, #"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 xx 10^(-4) ~~ x^2/4.0# #=> x = sqrt(4.0 cdot 9.1 xx 10^(-4)) = "0.0603 M"# In any case, each ion's equilibrium concentrations in the solution are as follows:

Axel Alvarado · Answer

undefined $$ Fe^{3+} + 3SCN^- \rightleftharpoons Fe(SCN)_3 $$

Given:
Initial concentration of Fe(SCN)₃ = 4.0 M
K_f = 1099

Let x be the concentration of Fe(SCN)₃ dissociated.

At equilibrium:
$$ [Fe^{3+}] = 4.0 - x $$
$$ [SCN^-] = 4.0 - 3x $$
$$ [Fe(SCN)_3] = x $$

$$ K_f = \frac{[Fe(SCN)_3]}{[Fe^{3+}][SCN^-]^3} $$
$$ 1099 = \frac{x}{(4.0 - x)(4.0 - 3x)^3} $$

Solving the equation yields:
$$ x = 3.34 \times 10^{-4} \, M $$

At equilibrium:
$$ [Fe^{3+}] = 4.0 - 3.34 \times 10^{-4} \, M $$
$$ [SCN^-] = 4.0 - 3 \times 3.34 \times 10^{-4} \, M $$
$$ [Fe(SCN)_3] = 3.34 \times 10^{-4} \, M $$

Avery Adams · Answer

undefined To determine the equilibrium concentrations of all species in solution for the dissociation of iron(III) thiocyanate ($ \text{Fe(SCN)}^{2+} $) in water, we can use the equilibrium constant expression and solve the equilibrium concentrations.

The dissociation reaction can be represented as:

$$ \text{Fe(SCN)}^{2+} (\text{aq}) \rightleftharpoons \text{Fe}^{3+} (\text{aq}) + \text{SCN}^- (\text{aq}) $$

Given that the equilibrium constant ($ K_f $) for the complex formation is $ 1099 $, the equilibrium constant expression is:

$$ K_f = \frac{[\text{Fe}^{3+}][\text{SCN}^-]}{[\text{Fe(SCN)}^{2+}]} $$

Let's denote the equilibrium concentrations of $ \text{Fe(SCN)}^{2+} $, $ \text{Fe}^{3+} $, and $ \text{SCN}^- $ as $ x $, $ x $, and $ x $ respectively.

Substitute these into the equilibrium constant expression:

$$ 1099 = \frac{x \cdot x}{4.0 - x} $$

Now, solve for $ x $, which represents the equilibrium concentration of $ \text{Fe(SCN)}^{2+} $.

$$ x^2 = 1099 \cdot (4.0 - x) $$
$$ x^2 = 4396 - 1099x $$
$$ x^2 + 1099x - 4396 = 0 $$

Using the quadratic formula, solve for $ x $:

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

Where $ a = 1 $, $ b = 1099 $, and $ c = -4396 $.

$$ x = \frac{-1099 \pm \sqrt{(1099)^2 - 4(1)(-4396)}}{2(1)} $$

$$ x = \frac{-1099 \pm \sqrt{1210801 + 17584}}{2} $$

$$ x = \frac{-1099 \pm \sqrt{1228385}}{2} $$

$$ x \approx \frac{-1099 \pm 1108.92}{2} $$

$$ x \approx -554.46 \quad \text{or} \quad x \approx 553.46 $$

Since $ x $ represents a concentration, it cannot be negative. So, $ x \approx 553.46 \, \text{M} $.

Therefore, the equilibrium concentrations are approximately:

$$ [\text{Fe(SCN)}^{2+}] \approx 553.46 \, \text{M} $$
$$ [\text{Fe}^{3+}] \approx 553.46 \, \text{M} $$
$$ [\text{SCN}^-] \approx 553.46 \, \text{M} $$

Please note that the concentrations are rounded to two decimal places.