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Chapter 16: Problem 20

The step wise formation constants for a complex ion usually have values much greater than 1. What is the significance of this?

Short Answer

Expert verified
The significance of having step-wise formation constants (\(K_{f}\)) much greater than 1 for complex ions is that it results in the formation of stable complexes with strong bonds, which are energetically favorable. It implies a high affinity of the central metal ion to bind with ligands, strong complex formation, and the generation of stable intermediate species during the step-wise formation process. Hence, larger formation constants lead to more stable complex ions and favor their formation.

Step by step solution

01

Understanding step-wise formation constants

Step-wise formation constants (\(K_{f}\)) are equilibrium constants for each step in the process of forming a complex ion. A complex ion generally forms by successive addition of ligands to a central metal ion. The magnitude of the step-wise formation constant can determine the stability of the intermediate species formed in each step.
02

Formation process and large step-wise formation constants

Complex ions with large step-wise formation constants show a strong tendency to form from their constituent ions. The larger the value of the formation constant, the more stable the complex ion. In other words, the equilibrium for each step lies far to the right, favoring the formation of the complex ion.
03

Significance of large step-wise formation constants

The significance of step-wise formation constants being much greater than 1 lies in the stability of the formed complex ions. Large step-wise formation constants are indicative of: 1. High affinity of the central metal ion to bind with ligands: This implies that the central metal ion can attract ligands and form strong bonds with them. 2. Strong complex formation: The energy required to break the bonds in the complex ion becomes larger when the formation constants are greater than 1, making the complex ion more stable. 3. Formation of stable intermediate species: During the step-wise formation process, stable intermediate species are formed, which promotes the continuation of the complex formation process. In conclusion, the significance of step-wise formation constants being much greater than 1 for complex ions is that it results in the formation of stable complexes with strong bonds, which are energetically favorable.

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Most popular questions from this chapter

Devise as many ways as you can to experimentally determine the \(K_{\mathrm{sp}}\) value of a solid. Explain why each of these would work.

Calculate the concentration of \(\mathrm{Pb}^{2+}\) in each of the following. a. a saturated solution of $\mathrm{Pb}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=1.2 \times 10^{-15}$ b. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13.00\) c. Ethylenediaminetetraacetate \(\left(\mathrm{EDTA}^{4-}\right)\) is used as a complexing agent in chemical analysis and has the following structure: Solutions of \(\mathrm{ED} \mathrm{TA}^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of \(\mathrm{EDTA}^{4-}\) with \(\mathrm{Pb}^{2+} \mathrm{is}\) $$\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) \quad K=1.1 \times 10^{18}$$ Consider a solution with 0.010 mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 \(\mathrm{L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing 0.050 \(\mathrm{M}\) Na_thion. Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?

If \(10.0 \mathrm{mL}\) of $2.0 \times 10^{-3} M \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\( is added to 10.0 \)\mathrm{mL}$ of a \(\mathrm{pH}=10.0 \mathrm{NaOH}\) solution, will a precipitate form?

Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the following solids a. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) b. \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) c. \(\mathrm{BaF}_{2}\)

Describe how you could separate the ions in each of the following groups by selective precipitation. a. \(\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cu}^{2+}\) b. \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\) c. \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\)
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