Problem 82 - Chemistry: An Atoms-Focused Approach

Question

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

Step-by-Step Solution

Answer

Question: Calculate the equilibrium constant for the reaction Hg(cysteine)²⁺ + glycine ↔ Hg(glycine)²⁺ + cysteine, given the formation constants \(log K_{f1} = 14.2\) for Hg²⁺ + cysteine ↔ Hg(cysteine)²⁺ and \(log K_{f2} = 10.3\) for Hg²⁺ + glycine ↔ Hg(glycine)²⁺. Answer: The equilibrium constant for the reaction Hg(cysteine)²⁺ + glycine ↔ Hg(glycine)²⁺ + cysteine is \(K_3 = 10^{3.9}\).

Step 1: Write down the given formation constants

We know that: \(\log K_{f1} = 14.2\) (for Hg²⁺ + cysteine ↔ Hg(cysteine)²⁺) \(\log K_{f2} = 10.3\) (for Hg²⁺ + glycine ↔ Hg(glycine)²⁺)

Step 2: Write the reactions

Write the given reactions and the desired reaction we want to find the equilibrium constant for: 1) Hg²⁺ + cysteine ↔ Hg(cysteine)²⁺ 2) Hg²⁺ + glycine ↔ Hg(glycine)²⁺ 3) Hg(cysteine)²⁺ + glycine ↔ Hg(glycine)²⁺ + cysteine

Step 3: Adding the first reaction and the reverse of the second reaction

Add reaction 1 to the reverse of reaction 2. This will give us the desired reaction 3 from the analysis above. (Hg²⁺ + cysteine ↔ Hg(cysteine)²⁺) + (Hg(glycine)²⁺ ↔ Hg²⁺ + glycine) → Hg(cysteine)²⁺ + glycine ↔ Hg(glycine)²⁺ + cysteine Now we have the third reaction we wanted.

Step 4: Calculate the equilibrium constant for reaction 3

Since reaction 3 is the sum of reaction 1 and the reverse of reaction 2, the equilibrium constant for reaction 3 (K₃) is the product of the equilibrium constants for reaction 1 (K₁) and the reverse of reaction 2 (1/K₂). We know K₁ and K₂ from step 1. We can find K₃: \(log(K_3) = log(K_1) - log(K_2)\) (Because \(K_3 = \frac{K_1}{K_2}\)) \(log(K_3) = 14.2 - 10.3\) Now, calculate the value of \(log(K_3)\): \(log(K_3) = 3.9\) Thus, the equilibrium constant for the third reaction is: \(K_3 = 10^{3.9}\)

Key Concepts

Formation Constants

Formation constants, often denoted as \( K_f \), describe the stability of a complex ion in solution. They indicate how easily a metal ion combines with ligands, such as cysteine or glycine, to form a complex. These constants are typically expressed in logarithmic form like \( \log K_f \).

Here's what it tells us:

• A higher \( \log K_f \) value means a more stable complex.
• For mercury(II) with cysteine, \( \log K_{f1} = 14.2 \) suggests a very stable complex.
• In comparison, \( \log K_{f2} = 10.3 \) for mercury(II) with glycine indicates a less stable complex.

Understanding formation constants helps us predict and compare the behavior and reactivity of different metal complexes.

Complexation Reactions

Complexation reactions involve the combination of metal ions with ligands to form a stable complex. These reactions are essential in various chemical processes, including biological systems and industrial applications.

In the given exercise, the reactions can be summarized as follows:

• The mercury(II) ion reacts with cysteine to form \( \text{Hg(cysteine)}^{2+} \).
• Similarly, mercury(II) reacts with glycine to form \( \text{Hg(glycine)}^{2+} \).

The ability of these reactions to proceed depends on factors such as the formation constants and the concentration of reactants. These relationships highlight the intricate balance between reactants and products, governed by their respective \( \log K_f \) values.

Equilibrium Calculations

Equilibrium calculations allow us to determine the direction and extent of a chemical reaction in equilibrium. The equilibrium constant \( K \) quantifies this balance.

To solve for the equilibrium constant of a new reaction, we use the relationship:

• \( K_3 = \frac{K_1}{K_2} \)
• Given \( \log K_1 = 14.2 \) and \( \log K_2 = 10.3 \).

We find \( \log K_3 = 14.2 - 10.3 = 3.9 \).

This means the equilibrium constant \( K_3 \) is \( 10^{3.9} \), which is a large value, indicating the favorability of converting \( \text{Hg(cysteine)}^{2+} \) and glycine into \( \text{Hg(glycine)}^{2+} \) and cysteine. These calculations provide insights into how likely products are formed under specified conditions.