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What is a formation constant and what is an instability constant?

Short Answer

Expert verified
A formation constant (\(K_f\)) is an equilibrium constant for the formation of a complex ion from aqueous ions. An instability constant (\(K_d\)) is the inverse of the formation constant, representing the tendency of a complex ion to dissociate back into its constituents.

Step by step solution

01

Definition of Formation Constant

The formation constant, often represented as \(K_f\), is an equilibrium constant for the formation of a complex ion from the aqueous ions that compose it. In a generalized chemical reaction, a metal ion (M) binds with ligands (L) to form a complex ion (ML), the equilibrium can be represented as \(M^{n+} + xL^{m-} \rightleftharpoons ML_x^{(nx-mx)+}\). The formation constant is defined by the equation \(K_f = \frac{[ML_x^{(nx-mx)+}]}{[M^{n+}]^1[L^{m-}]^x}\), where the square brackets denote the molar concentrations of the species.
02

Definition of Instability Constant

The instability constant, also known as the dissociation constant and represented as \(K_{d}\), is the inverse of the formation constant. It is an equilibrium constant for the dissociation of a complex ion into its aqueous ions. For the above reaction, the instability constant would be given by \(K_{d} = \frac{1}{K_f} = \frac{[M^{n+}]^1[L^{m-}]^x}{[ML_x^{(nx-mx)+}]}\). It quantifies the tendency of the complex ion to dissociate back into its constituent ions.

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Key Concepts

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

Equilibrium Constant
Understanding the equilibrium constant is vital to grasping many chemical reactions, especially those that are reversible. At equilibrium, the rate of the forward reaction – the formation of products from reactants – equals the rate of the reverse reaction, where products revert to reactants. The equilibrium constant (\(K_{eq}\) ) is a quantitative measure of this balance.

For a generalized reaction, where reactants A and B form products C and D, the equilibrium constant is expressed as \(K_{eq} = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}\) where the letters in brackets represent the concentrations of the substances and the letters as exponents are their stoichiometric coefficients.

In the context of complex ion formation, the equilibrium constant is known as the formation constant (\(K_f\)). It signifies the stability of the complex ion; the higher the value of \(K_f\), the more stable the complex ion. This stability is key in various applications like metal extraction and drug delivery.
Complex Ion
A complex ion is an entity formed by the coordinate bonding between a central metal ion and surrounding molecules or ions called ligands. These ligands have lone pairs of electrons available and can donate them to the metal ion, forming dative covalent bonds.

The metal and ligands create a formation that is electrically charged. For example, when copper(II) ions react with four ammonia molecules, the complex ion [Cu(NH3)4]2+ is formed. The properties of complex ions, such as color, magnetic behavior, and solubility, are vastly different from those of the individual components.

Factors Influencing Complex Ion Stability

  • The charge on the metal ion: Higher charges often lead to stronger attractions with ligands.
  • The size of the metal ion: Smaller ions tend to form stronger bonds due to closer interactions with ligands.
  • The nature of the ligands: Certain ligands are known to form stronger bonds due to their electronic arrangements.
  • The number of ligands attached: A larger number of ligands can stabilize the metal ion through the 'chelate effect.'
Complex ions are not only interesting in inorganic chemistry but also play crucial roles in biological systems, such as in oxygen transport by hemoglobin.
Instability Constant
In contrast to the formation constant, the instability constant (\(K_{d}\)) offers insight into the likelihood of a complex ion to fall apart into its constituents – the metal ion and the ligands. It’s an essential concept when considering the reversibility of complex ion formation.

For the equilibrium where the complex ion dissociates into the metal ion (M) and ligands (L), represented as \(ML_x^{(nx-mx)+} \rightleftharpoons M^{n+} + xL^{m-}\), the instability constant is calculated as \(K_{d} = \frac{[M^{n+}]^1[L^{m-}]^x}{[ML_x^{(nx-mx)+}]}\). When \(K_{d}\) has a higher value, it means the complex ion is more prone to dissociate, indicating less stability.

Instability constants are particularly significant in environmental chemistry, where the breakdown of metal complexes can lead to the release of potentially harmful ions. Understanding and manipulating \(K_d\) allows chemists to predict and control the behavior of metal ions in various biological and technological processes.

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Most popular questions from this chapter

Potassium oxide is readily soluble in water, but the resulting solution contains essentially no oxide ion. Explain, using an equation, what happens to the oxide ion.

Write the \(K_{\mathrm{sp}}\) expressions for each of the following compounds: (a) \(\mathrm{Fe}_{3}\left(\mathrm{PO}_{4}\right)_{2},\) (b) \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\), (c) \(\mathrm{PbCrO}_{4}\) (d) \(\mathrm{Al}(\mathrm{OH})_{3}\), (e) \(\mathrm{ZnCO}_{3}\) (f) \(\mathrm{Zn}(\mathrm{OH})_{2}\)

Suppose that \(50.0 \mathrm{~mL}\) of \(0.12 \mathrm{M} \mathrm{AgNO}_{3}\) is added to \(50.0 \mathrm{~mL}\) of \(0.048 \mathrm{M} \mathrm{NaCl}\) solution. (a) What mass of \(\mathrm{AgCl}\) will form? (b) Calculate the final concentrations of all of the ions in the solution that is in contact with the precipitate. (c) What percentage of the \(\mathrm{Ag}^{+}\) ions have precipitated?

Suppose that some dipositive cation, \(M^{2+},\) is able to form a complex ion with a ligand, \(L\), by the following balanced equation: \(M^{2+}+2 L \rightleftharpoons M(\mathrm{~L})_{2}^{2+} .\) The cation also forms a sparingly soluble salt, \(M \mathrm{Cl}_{2}\). In which of the following circumstances would a given quantity of ligand be more able to bring larger quantities of the salt into solution? Explain and justify the calculation involved: (a) \(K_{\text {form }}=1 \times 10^{2}\) and \(K_{\text {sp }}=1 \times 10^{-15}\), (b) \(K_{\text {form }}=1 \times 10^{10}\) and \(K_{\mathrm{sp}}=1 \times 10^{-20}\).

Is \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) a basic salt? Justify your answer. $$ \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3 \mathrm{Ba}^{2+}(a q)+2 \mathrm{PO}_{4}^{3-}(a q) $$

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