Problem 102
Question
Write the equilibria that are associated with the equations for \(K_{\text {form }}\) for each of the following complex ions. Write also the equations for the \(K_{\text {form }}\) of each: (a) \(\mathrm{Hg}\left(\mathrm{NH}_{3}\right)_{4}^{2+},\) (b) \(\mathrm{SnF}_{6}^{2-}\), (c) \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\).
Step-by-Step Solution
Verified Answer
\(K_{\text{form}}\) for (a) is \([\mathrm{Hg}(\mathrm{NH}_3)_4^{2+}]/([\mathrm{Hg}^{2+}][\mathrm{NH}_3]^4)\); for (b) is \([\mathrm{SnF}_6^{2-}]/([\mathrm{Sn}^{4+}][\mathrm{F}^-]^6)\); for (c) is \([\mathrm{Fe(CN)}_6^{3-}]/([\mathrm{Fe}^{3+}][\mathrm{CN}^-]^6)\).
1Step 1 - Write the Equilibrium for (a)
To write the equilibrium for the complex ion \(\mathrm{Hg}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\), start by identifying the metal ion and the ligands. The metal ion is mercury (II), \(\mathrm{Hg}^{2+}\), and the ligands are four ammonia molecules \(\mathrm{NH}_3\). The equilibrium is between the metal ion, the ligands, and the complex ion formed: \[\mathrm{Hg}^{2+} + 4\mathrm{NH}_3 \longleftrightarrow \mathrm{Hg}\left(\mathrm{NH}_3\right)_{4}^{2+}\]
2Step 2 - Write the Formation Constant Equation for (a)
The formation constant \(K_{\text{form}}\) for the complex ion \(\mathrm{Hg}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) is the ratio of the concentration of the complex ion to the product of the concentrations of the metal ion and ligands, raised to the power of their stoichiometric coefficients: \[K_{\text{form}} = \frac{[\mathrm{Hg}\left(\mathrm{NH}_{3}\right)_{4}^{2+}]}{[\mathrm{Hg}^{2+}][\mathrm{NH}_3]^4}\]
3Step 3 - Write the Equilibrium for (b)
For the complex ion \(\mathrm{SnF}_{6}^{2-}\), the metal ion is tin (IV), \(\mathrm{Sn}^{4+}\), and the ligands are six fluoride ions \(\mathrm{F}^-\). The equilibrium is: \[\mathrm{Sn}^{4+} + 6\mathrm{F}^- \longleftrightarrow \mathrm{SnF}_{6}^{2-}\]
4Step 4 - Write the Formation Constant Equation for (b)
The formation constant \(K_{\text{form}}\) for \(\mathrm{SnF}_{6}^{2-}\) is: \[K_{\text{form}} = \frac{[\mathrm{SnF}_{6}^{2-}]}{[\mathrm{Sn}^{4+}][\mathrm{F}^-]^6}\]
5Step 5 - Write the Equilibrium for (c)
For the complex ion \(\mathrm{Fe}({\mathrm{CN}})_{6}^{3-}\), the metal ion is iron (III), \(\mathrm{Fe}^{3+}\), and the ligands are six cyanide ions \(\mathrm{CN}^-\). The equilibrium is: \[\mathrm{Fe}^{3+} + 6\mathrm{CN}^- \longleftrightarrow \mathrm{Fe}({\mathrm{CN}})_{6}^{3-}\]
6Step 6 - Write the Formation Constant Equation for (c)
The formation constant \(K_{\text{form}}\) for \(\mathrm{Fe}({\mathrm{CN}})_{6}^{3-}\) is: \[K_{\text{form}} = \frac{[\mathrm{Fe}({\mathrm{CN}})_{6}^{3-}]}{[\mathrm{Fe}^{3+}][\mathrm{CN}^-]^6}\]
Key Concepts
Formation ConstantChemical EquilibriumCoordination Chemistry
Formation Constant
The formation constant, often denoted as Kform, is a special type of equilibrium constant that measures the stability of a complex ion formed in solution. When a metal ion in solution combines with ligands to form a complex ion, the reaction reaches a state of balance known as chemical equilibrium. The formation constant quantifies the extent of this reaction, with a higher Kform indicating a more stable complex ion.
For example, when explaining the formation constant for the complex ion Hg(NH3)42+, the equation used to calculate Kform shows the concentrations of the products over the reactants. If Kform has a large value, that implies a predominant formation of the complex ion in the equilibrium. Thus, a student can predict that the reaction favors the complex ion formation over the free metal ion and ligands only if the value of Kform is known.
For example, when explaining the formation constant for the complex ion Hg(NH3)42+, the equation used to calculate Kform shows the concentrations of the products over the reactants. If Kform has a large value, that implies a predominant formation of the complex ion in the equilibrium. Thus, a student can predict that the reaction favors the complex ion formation over the free metal ion and ligands only if the value of Kform is known.
Chemical Equilibrium
In chemistry, chemical equilibrium refers to the state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no overall change in the concentrations of the reactants and products over time. It is essential to understand that equilibrium does not mean the reactants and products are present in equal amounts, but rather that their concentrations have stabilized at a certain ratio.
Take, for instance, the formation of SnF62− from Sn4+ and F− ions; initially, as the reactants combine to form the product, the concentration of the product SnF62− increases. When the reverse reaction starts to occur at the same rate as the forward reaction, the system is said to reach equilibrium. The equilibrium constant expression is derived from this state and helps in calculating the formation constants mentioned earlier.
Take, for instance, the formation of SnF62− from Sn4+ and F− ions; initially, as the reactants combine to form the product, the concentration of the product SnF62− increases. When the reverse reaction starts to occur at the same rate as the forward reaction, the system is said to reach equilibrium. The equilibrium constant expression is derived from this state and helps in calculating the formation constants mentioned earlier.
Coordination Chemistry
Coordination chemistry is a field of chemistry that deals with metal complexes, where a central metal atom or ion is surrounded by molecules or anions, known as ligands. These ligands donate electron pairs to the metal, forming coordinate covalent bonds. The number of ligands attached to the metal (its coordination number) and the type of ligands can significantly influence the properties of the complex, including its color, magnetic behavior, and reactivity.
Each of the exercises—such as the formation of Fe(CN)63− from Fe3+ and CN− ions—illustrates key principles of coordination chemistry, including the selectivity of a metal for certain ligands and the resulting geometry of the complex ion. A well-rounded understanding of coordination chemistry is critical for predicting the formation and stability of complex ions, ensuring students grasp how multiplex structures can arise from seemingly simple reactions.
Each of the exercises—such as the formation of Fe(CN)63− from Fe3+ and CN− ions—illustrates key principles of coordination chemistry, including the selectivity of a metal for certain ligands and the resulting geometry of the complex ion. A well-rounded understanding of coordination chemistry is critical for predicting the formation and stability of complex ions, ensuring students grasp how multiplex structures can arise from seemingly simple reactions.
Other exercises in this chapter
Problem 100
Write the chemical equilibria and equilibrium laws that correspond to \(K_{\text {form }}\) for the following complexes: (a) \(\mathrm{Ag}\left(\mathrm{S}_{2} \
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Write equilibria that correspond to \(K_{\text {form }}\) for each of the following complex ions and write the equations for \(K_{\text {form }}:\) (a) \(\mathr
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Write equilibria that correspond to \(K_{\text {inst }}\) for each of the following complex ions and write the equations for \(K_{\text {inst }}:\) (a) \(\mathr
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Write the equilibria that are associated with the equations for \(K_{\text {inst }}\) for each of the following complex ions. Write also the equations for the \
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