What happens to the equilibrium constant in a complexation reaction?

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What happens to the equilibrium constant in a complexation reaction?

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Answer 1 · 2016-02-19T08:54:48

A complexation reaction has a unique equilibrium constant called a formation constant, $K_{f}$ $K_f$ .

Generally, complexation reactions are favorable, and as such, $K_{f, overall}$ $K_(f,"overall")$ is usually large.

COMPLEXATION REACTIONS MAY BE MULTI-STEP

Take the following reaction for example:

${[Cu {(H_{2} O)}_{4}]}_{4}^{2 +} (a q) + 4 {NH}_{3} (a q) ⇌ {[Cu {({NH}_{3})}_{4}]}_{4}^{2 +} (a q) + 4 H_{2} O (l)$ $color(green)(["Cu"("H"_2"O")_4]^(2+)(aq) + 4"NH"_3(aq) rightleftharpoons ["Cu"("NH"_3)_4]^(2+)(aq) + 4"H"_2"O"(l))$

This, although it seemingly could occur in one step, actually, it happens in multiple steps, and there is one formation constant for the displacement of each aqua ligand on the tetraaquacopper(II) complex. This is possible because ${NH}_{3}$ $"NH"_3$ is a stronger-field ligand than $H_{2} O$ $"H"_2"O"$ (see the spectrochemical series).

(On the other hand, when a free ion transition metal complexes with a ligand, like ${Ag}^{+}$ $"Ag"^(+)$ with $2 {NH}_{3}$ $2"NH"_3$ , it tends to be one step.)

THERE EXIST FORMATION CONSTANTS AS EQUILIBRIUM CONSTANTS FOR EACH INDIVIDUAL STEP

We would separate the steps to incorporate the ammine ligands by displacing one aqua ligand at a time.

${[Cu {(H_{2} O)}_{4}]}_{4}^{2 +} (a q) + {NH}_{3} (a q) ⇌ {[Cu {(H_{2} O)}_{3} {NH}_{3}]}_{3}^{2 +} (a q) + H_{2} O (l)$ $["Cu"("H"_2"O")_4]^(2+)(aq) + "NH"_3(aq) rightleftharpoons ["Cu"("H"_2"O")_3"NH"_3]^(2+)(aq) + "H"_2"O"(l)$

$K_{f 1} = 1.9 \times 10^{4} = \frac{[{[Cu {(H_{2} O)}_{3} {NH}_{3}]}_{3}^{2 +}]}{[{[Cu {(H_{2} O)}_{4}]}_{4}^{2 +}] [{NH}_{3}]}$ $K_(f1) = 1.9xx10^4 = \frac([["Cu"("H"_2"O")_3"NH"_3]^(2+)])([["Cu"("H"_2"O")_4]^(2+)]["NH"_3])$

Next step, displace another:

${[Cu {(H_{2} O)}_{3} {NH}_{3}]}_{3}^{2 +} (a q) + {NH}_{3} (a q) ⇌ {[Cu {(H_{2} O)}_{2} {({NH}_{3})}_{2}]}_{2}^{2 +} (a q) + H_{2} O (l)$ $["Cu"("H"_2"O")_3"NH"_3]^(2+)(aq) + "NH"_3(aq) rightleftharpoons ["Cu"("H"_2"O")_2("NH"_3)_2]^(2+)(aq) + "H"_2"O"(l)$

$K_{f 2} = 3.9 \times 10^{3} = \frac{[{[Cu {(H_{2} O)}_{2} {({NH}_{3})}_{2}]}_{2}^{2 +}]}{[{[Cu {(H_{2} O)}_{3} {NH}_{3}]}_{3}^{2 +}] [{NH}_{3}]}$ $K_(f2) = 3.9xx10^3 = \frac([["Cu"("H"_2"O")_2("NH"_3)_2]^(2+)])([["Cu"("H"_2"O")_3"NH"_3]^(2+)]["NH"_3])$

It can be hard to keep track of these $K_{f}$ $K_f$ 's and all the parentheses, so let's keep going for practice. Next step, displace another:

${[Cu {(H_{2} O)}_{2} {({NH}_{3})}_{2}]}_{2}^{2 +} (a q) + {NH}_{3} (a q) ⇌ {[Cu (H_{2} O) {({NH}_{3})}_{3}]}_{3}^{2 +} (a q) + H_{2} O (l)$ $["Cu"("H"_2"O")_2("NH"_3)_2]^(2+)(aq) + "NH"_3(aq) rightleftharpoons ["Cu"("H"_2"O")("NH"_3)_3]^(2+)(aq) + "H"_2"O"(l)$

$K_{f 3} = 1.0 \times 10^{3} = \frac{[{[Cu (H_{2} O) {({NH}_{3})}_{3}]}_{3}^{2 +}]}{[{[Cu {(H_{2} O)}_{2} {({NH}_{3})}_{2}]}_{2}^{2 +}] [{NH}_{3}]}$ $K_(f3) = 1.0xx10^3 = \frac([["Cu"("H"_2"O")("NH"_3)_3]^(2+)])([["Cu"("H"_2"O")_2("NH"_3)_2]^(2+)]["NH"_3])$

And finally, the fourth step!

${[Cu (H_{2} O) {({NH}_{3})}_{3}]}_{3}^{2 +} (a q) + {NH}_{3} (a q) ⇌ {[Cu {({NH}_{3})}_{4}]}_{4}^{2 +} (a q) + H_{2} O (l)$ $["Cu"("H"_2"O")("NH"_3)_3]^(2+)(aq) + "NH"_3(aq) rightleftharpoons ["Cu"("NH"_3)_4]^(2+)(aq) + "H"_2"O"(l)$

$K_{f 4} = 1.5 \times 10^{2} = \frac{[{[Cu {({NH}_{3})}_{4}]}_{4}^{2 +}]}{[{[Cu (H_{2} O) {({NH}_{3})}_{3}]}_{3}^{2 +}] [{NH}_{3}]}$ $K_(f4) = 1.5xx10^2 = \frac([["Cu"("NH"_3)_4]^(2+)])([["Cu"("H"_2"O")("NH"_3)_3]^(2+)]["NH"_3])$

PATTERN IN COMPLEX FORMATION FAVORABILITY

Notice how $K_{f 4} < K_{f 3} < K_{f 2} < K_{f 1}$ $K_(f4) < K_(f3) < K_(f2) < K_(f1)$ . As each ammine ligand displaces one aqua ligand, there is one less aqua ligand on the complex; that is obvious.

What is less obvious is that with less aqua ligands after each step, it tends to be entropically more difficult to displace each consecutive aqua ligand, provided no other variables change (like the complex's structure).

Usually, barring exceptions, it is easier to find a ligand to displace when there are a lot present, and more and more difficult when there are less and less target ligands to find and displace.

THE OVERALL FORMATION CONSTANT

Now, is there one formation constant that describes the whole four-step process? Yes!

The overall formation constant can be acquired from multiplying those of each individual, consecutive step.

You can convince yourself that multiplying the four equilibrium constants cancels out all the intermediates and racks up exponents for ${NH}_{3}$ $"NH"_3$ to give:

$K_{f, overall} = β_{4} = \frac{[{[Cu {({NH}_{3})}_{4}]}_{4}^{2 +}]}{[{[Cu {(H_{2} O)}_{4}]}_{4}^{2 +}] {[{NH}_{3}]}_{3}^{4}}$ $color(blue)(K_(f,"overall") = beta_4 = \frac([["Cu"("NH"_3)_4]^(2+)])([["Cu"("H"_2"O")_4]^(2+)]["NH"_3]^4))$

where $β_{n}$ $beta_n$ is a convention to indicate an $n$ $n$ -step formation constant up to and including the $n$ $n$ th step for $n \geq 2$ $n >= 2$ .

$β_{n} = K_{f 1} K_{f 2} \cdot \cdot \cdot K_{f n}$ $beta_n = K_(f1)K_(f2)cdotcdotcdotK_(fn)$

Therefore, we have:

$β_{4} = K_{f 1} K_{f 2} K_{f 3} K_{f 4}$ $color(blue)(beta_4 = K_(f1)K_(f2)K_(f3)K_(f4))$

$= (1.9 \times 10^{4}) (3.9 \times 10^{3}) (1.0 \times 10^{3}) (1.5 \times 10^{2}) = 1.1 \times 10^{13}$ $= (1.9xx10^4)(3.9xx10^3)(1.0xx10^3)(1.5xx10^2) = color(blue)(1.1xx10^13)$

From here you should know how to determine... $β_{2}$ $beta_2$ , for instance. All you would do then is $K_{f 1} \times K_{f 2}$ $K_(f1)xxK_(f2)$ to model the formation through to and including the second intermediate (going through two reaction steps).

Make sure to keep track of your parentheses and the number of ligands you are writing in your formation constants!

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What happens to the equilibrium constant in a complexation reaction?

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