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Write a series of equations to show the stepwise displacement of \(\mathrm{H}_{2} \mathrm{O}\) ligands in \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) by ethylenediamine, for which \(\log K_{1}=4.34, \log K_{2}=3.31\), and \(\log K_{3}=2.05 .\) What is the overall formation constant, \(\beta_{3}=K_{f},\) for \(\left[\mathrm{Fe}(\mathrm{en})_{3}\right]^{3+} ?\)

Short Answer

Expert verified
The overall formation constant, \(\beta_{3}=K_{f}\), for \(\left[\mathrm{Fe}(\mathrm{en})_{3}\right]^{3+}\) is \(10^{9.7}\).

Step by step solution

01

Write the stepwise displacement equations

The first thing is to write the stepwise displacement equations. This can be achieved by replacing two water molecules with an ethylenediamine molecule at each step: \n\(1) \: [Fe(H_2O)_6]^{3+} + en \rightarrow [Fe(H_2O)_4(en)]^{3+} + 2H_2O,\ \log K_{1}=4.34\n 2) \: [Fe(H_2O)_4(en)]^{3+} + en \rightarrow [Fe(H_2O)_2(en)_2]^{3+} + 2H_2O,\ \log K_{2}=3.31\n 3) \: [Fe(H_2O)_2(en)_2]^{3+} + en \rightarrow [Fe(en)_3]^{3+} + 2H_2O,\ \log K_{3}=2.05
02

Calculate the overall formation constant

To find the overall formation constant \(\beta_{3}\), multiply the individual equilibria constants together. This is possible because \(\log K_{f} = \log K_{1} + \log K_{2} + \log K_{3}\). This gives \(\log K_{f} = 4.34 + 3.31 + 2.05 = 9.7\). To obtain \(K_f\), we convert from logarithmic form to exponential form which gives \(K_{f}=10^{9.7}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stepwise Displacement Equations
In the realm of coordination chemistry, the intriguing dance of ligand exchange is captured by stepwise displacement equations. These equations illustrate how a complex ion like \(\left[\mathrm{Fe}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{3+}\) reorganizes its surroundings by exchanging water molecules (\(\mathrm{H}_2\mathrm{O}\)) for other ligands—in this case, ethylenediamine (en). Imagine this exchange as a choreographed sequence, with each step replacing two water molecules with one en molecule.

To communicate this chemistry in a relatable manner, think of a 'friendship circle,' where the iron ion is the most popular individual in the social group, surrounded by water friends. Gradually, it becomes acquainted with ethylenediamine, a 'new friend,' ending up with a completely transformed inner circle represented by \(\left[\mathrm{Fe}(\mathrm{en})_3\right]^{3+}\). This sequential social interaction can be mirrored in equations that show a one-on-one exchange, as it naturally occurs until the initial set of water friends are fully replaced.
Formation Constant
Every friendship has a strength, a measure of how likely it is to form and endure. In the context of complex ions, this is quantified by the formation constant (\(K_f\) or \(\beta\)), an equilibrium constant that reflects the stability of the complex. Higher values signify a stronger 'bond' between the central ion and its ligands.

Understanding the formation constant is akin to assessing a relationship's resilience: a high \(\log K_f\) means a tight-knit group that's reluctant to part ways. The overall formation constant, \(\beta_3\), for the entire sequence of ligand substitutions is determined by multiplying the equilibrium constants of each individual 'friendship' or stepwise reaction. The logarithmic nature of this relationship allows us to sum these values, resulting in the cohesive strength of the new social chemistry within the complex ion.
Ligand Substitution
Ligand substitution is the process of one ligand replacing another in the sphere of a metal ion. In a real-world scenario, it's as if the metal ion is updating its preferences for different 'accessories' or ligands. The choice of whom to 'wear' isn't random; it's influenced by various factors, such as size, charge, and electronic properties of the prospective ligand.

In our textbook case, ethylenediamine is like a new, stylish accessory that iron finds more appealing compared to its old, plain water molecules. The affinity of the central ion towards various 'styles' (ligands) is a fundamental aspect of chemistry's fashion, dictated by principles that ensure the resulting compound is stable and favorable.
Chemical Equilibrium
Equilibrium lies at the very heart of numerous chemical processes, posing as a delicate balance where the speed of the forward reaction equals the speed of the reverse one. In the realm of complex ions, it's the point where the formation of the complex ion equals its breakdown into the original components.

Picture this as a popularity contest where the central metal ion, like the iron ion in our case study, maintains a constant group of ligand friends. At equilibrium, the rate at which iron makes new 'friendships' with ethylenediamine is perfectly matched by the rate at which these friendships 'fall apart,' resulting in a dynamic, yet stable, social circle. Each social gathering—or complex ion—has an equilibrium constant that signifies just how comfortable or preferred that particular group is.

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Most popular questions from this chapter

Write equations to represent the following observations. (a) A quantity of \(\mathrm{CuCl}_{2}(\mathrm{s})\) is dissolved in concentrated HCl(aq) and produces a yellow solution. The solution is diluted to twice its volume with water and assumes a green color. On dilution to ten times its original volume, the solution becomes pale blue. (b) When chromium metal is dissolved in \(\mathrm{HCl}(\mathrm{aq}), \mathrm{a}\) blue solution is produced that quickly turns green. Later the green solution becomes blue-green and then violet.

If the ion \(\mathrm{Co}^{2+}\) is linked with strong-field ligands to produce an octahedral complex, the complex has one unpaired electron. If \(\mathrm{Co}^{2+}\) is linked with weak-field ligands, the complex has three unpaired electrons. How do you account for this difference?

Without performing detailed calculations, show why you would expect the concentrations of the various ammine-aqua complex ions to be negligible compared with that of \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) in a solution having a total \(\mathrm{Cu}(\mathrm{II})\) concentration of \(0.10 \mathrm{M}\) and a total concentration of \(\mathrm{NH}_{3}\) of \(1.0 \mathrm{M}\). Under what conditions would the concentrations of these ammine-aqua complex ions (such as \(\left.\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{NH}_{3}\right]^{2+}\right)\) become more significant relative to the concentration of \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} ?\) Explain.

Supply acceptable names for the following. (a) \(\left[\mathrm{Co}(\mathrm{OH})\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NH}_{3}\right)\right]^{2+}\) (b) \(\left[\mathrm{Co}(\mathrm{ONO})_{3}\left(\mathrm{NH}_{3}\right)_{3}\right]\) (c) \(\left[\operatorname{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]\left[\mathrm{PtCl}_{6}\right]\) (d) \(\left[\mathrm{Fe}(\mathrm{ox})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]^{-}\) (e) \(\mathrm{Ag}_{2}\left[\mathrm{HgI}_{4}\right]\)

Draw plausible structures corresponding to each of the following names. (a) pentamminesulfatochromium(III) ion (b) trioxalatocobaltate(III) ion (c) triamminedichloronitrito-O-cobalt(III)

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