Problem 82 - Chemistry The Science in Context

Question

The complexation of mercury( 11 ) ion with cysteine in aqueous solution $$ \mathrm{Hg}^{2+}+\text { cysteinc } \rightleftharpoons \mathrm{Hg} \text { (cysteinc) }^{2+} $$ has a formation constant of \(\log K_{\mathrm{f}}=14.2,\) whereas the formation constant for the \(\mathrm{Hg}^{2+}\) complex with glycine $$ \mathrm{Hg}^{2+}+\text { glycine } \rightleftharpoons \mathrm{Hg}(\text { glycine })^{2+} $$ is \(\log K_{f}=10.3 .\) Calculate the equilibrium constant for the reaction Hg(cysteine) \(^{2+}+\) glycine \(\rightleftharpoons \mathrm{Hg}\) (glycine) \(^{2+}+\) cysteine

Step-by-Step Solution

Answer

Answer: The equilibrium constant for the given reaction is approximately 7943.

Step 1: Write down the given reactions and their formation constants

The first reaction is: $$ \mathrm{Hg}^{2+}+\mathrm{cysteine} \rightleftharpoons \mathrm{Hg}(\mathrm{cysteine})^{2+} $$ with \(\log K_{1}=14.2\) The second reaction is: $$ \mathrm{Hg}^{2+}+\mathrm{glycine} \rightleftharpoons \mathrm{Hg}(\mathrm{glycine})^{2+} $$ with \(\log K_{2}=10.3\)

Step 2: Write down the reaction for which the equilibrium constant is to be calculated

The given reaction is: $$ \mathrm{Hg}(\mathrm{cysteine})^{2+}+\mathrm{glycine} \rightleftharpoons \mathrm{Hg}(\mathrm{glycine})^{2+}+\mathrm{cysteine} $$ Our goal is to find the equilibrium constant, K, for this reaction.

Step 3: Calculate the equilibrium constant

To find the equilibrium constant K, we can first rewrite the given reaction as a combination of the first and second reactions. We can obtain the given reaction by reversing the second reaction and adding it to the first reaction. $$ \mathrm{Hg}(\mathrm{cysteine})^{2+}+\mathrm{glycine} \rightleftharpoons \mathrm{Hg}^{2+}+\mathrm{cysteine}+\mathrm{Hg}^{2+}+\mathrm{glycine} \rightleftharpoons \mathrm{Hg}(\mathrm{glycine})^{2+}+\mathrm{cysteine} $$ Next, we can use the relationship between the formation constants of the first and second reactions and the equilibrium constant of the given reaction: $$ K = \frac{K_{1}}{K_{2}} $$ Now, we can plug in the values of \(K_{1}\) and \(K_{2}\): $$ K = \frac{10^{14.2}}{10^{10.3}} $$ Calculating K, we get: $$ K = 10^{(14.2-10.3)} = 10^{3.9} \approx 7943 $$ The equilibrium constant for the given reaction is approximately 7943.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. This balance means that the concentration of the reactants and products remains constant over time. It's fundamental in understanding how reactions occur and predicting the amounts of substances present when equilibrium is reached.

For the given exercise, equilibrium constants play a pivotal role. They quantify the balance between the reactants and products in a chemical reaction at equilibrium. Specifically, the equilibrium constant (\(K\text{eq}\)) is a numerical value representing the ratio of product concentrations to reactant concentrations, each raised to the power of their respective stoichiometric coefficients, under equilibrium conditions.

Complexation Reactions

Complexation reactions involve the association between two or more molecules to form a complex molecule. In these reactions, a ligand such as cysteine or glycine binds to a central metal ion, in this case, mercury (\(Hg^{2+}\)).

These reactions are essential in fields like biochemistry, where metal ions form coordination complexes with organic molecules. Each complexation reaction has an associated equilibrium constant, known as the formation constant (\(K_f\) or sometimes just \(K\text{eq}\)), which indicates the stability of the complex in solution. The higher the formation constant, the more stable the complex, which means it is less likely to dissociate into its original ions and ligand.

Formation Constants

Formation constants (\(K_f\)) are a specific type of equilibrium constant for complexation reactions. They indicate the strength of the interaction between a metal ion and its ligands to form a complex ion. The logarithmic form of the formation constant (\text{log} \(K_f\)) simplifies comparing the relative stabilities of different metal-ligand complexes.

Values of \text{log} \(K_f\) for metal complexes, like those for mercury-cysteine and mercury-glycine given in the exercise, are vitally important. They guide us in predicting which complexes are more likely to form in solution and thus influence chemical behavior, including reaction mechanisms and pathways.

Logarithmic Calculations

Logarithmic calculations are useful in chemical equilibrium for transforming multiplicative relationships into additive ones. This conversion makes it easier to compare and calculate equilibrium constants. The log of the equilibrium constant, commonly expressed as \text{log} \(K\text{eq}\) or \text{log} \(K_f\text{),}\) is particularly important when dealing with very large or very small numbers, as it provides a more manageable form.

In the solution given, we used logarithmic properties to simplify the calculation of the equilibrium constant for the exchange reaction. Since formation constants were provided in logarithmic form, we subtracted the values directly instead of dividing the constants in their original form, which illustrates the utility of logarithmic calculations in easing complex mathematical operations.