Question

The equilibrium constant for the following reaction is \(1.0 \times 10^{23}\). EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt \(\mathrm{Na}_{2} \mathrm{H}_{2} \mathrm{EDTA}\), are used to treat heavy metal poisoning. Calculate \(\left[\mathrm{Cr}^{3+}\right]\) at equilibrium in a solution originally \(0.0010 M\) in \(\mathrm{Cr}^{3+}\) and \(0.050 \mathrm{M}\) in \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) and buffered at \(\mathrm{pH}=6.00 .\)

Short answer

At equilibrium, the concentration of \(\mathrm{Cr^{3+}}\) ions in the solution is approximately \(0.0010\,\mathrm{M}\).

Step-by-step solution

Reaction Equation

\[\mathrm{Cr^{3+} + H_{2}EDTA^{2-} \rightleftharpoons Cr(EDTA)}\] We have the equilibrium constant Kc = 1.0 × 10²³ for this reaction. Now, let's write the expression for Kc:

Equilibrium Constant Expression

\[K_{c} = \frac{\left[\mathrm{Cr}\left(\mathrm{EDTA}\right)\right]}{\left[\mathrm{Cr^{3+}}\right]\left[\mathrm{H_{2}EDTA^{2-}}\right]}\] Next, we will set up an ICE (Initial, Change, Equilibrium) table for the reaction to find equilibrium concentrations:

ICE Table and Equilibrium Concentrations

ICE Table: Initial: \[ [\mathrm{Cr^{3+}}] = 0.0010\,\mathrm{M }\] \[ [\mathrm{H_{2}EDTA^{2-}}] = 0.050\,\mathrm{M} \] \[ [\mathrm{Cr(EDTA)}] = 0\,\mathrm{M} \] Change: \[ [\mathrm{Cr^{3+}}] = -x\,\mathrm{M} \] \[ [\mathrm{H_{2}EDTA^{2-}}] = -x\,\mathrm{M} \] \[ [\mathrm{Cr(EDTA)}] = +x\,\mathrm{M} \] Equilibrium: \[ [\mathrm{Cr^{3+}}] = (0.0010 - x)\,\mathrm{M} \] \[ [\mathrm{H_{2}EDTA^{2-}}] = (0.050 - x)\,\mathrm{M} \] \[ [\mathrm{Cr(EDTA)}] = x\,\mathrm{M} \] Now, we can substitute the equilibrium concentrations into the expression for Kc:

Substitution of Equilibrium Concentrations

\[\begin{aligned} 1.0 \times 10^{23} &= \frac{x}{(0.0010 - x)(0.050 - x)} \\ \end{aligned}\] Now, solve the equation for x:

Solving for x

Let's assume that the changes (x) are too small compared to the initial concentrations, meaning that the denominator remains essentially constant: \((0.0010)(0.050)\). This simplifies the expression as: \[1.0 \times 10^{23} = \frac{x}{(0.0010)(0.050)}\] Solving for x: \[x = 1.0 \times 10^{23} \times (0.0010)(0.050)\] \[x \approx 5.0 \times 10^{-22}\,\mathrm{M}\] Now that we have the value of x, we can find the equilibrium concentration of Cr^3+:

Calculating Equilibrium Concentration of Cr^3+

\[[\mathrm{Cr^{3+}}]_{equilibrium} = 0.0010 - x\] \[[\mathrm{Cr^{3+}}]_{equilibrium} = 0.0010 - 5.0 \times 10^{-22}\] \[[\mathrm{Cr^{3+}}]_{equilibrium} \approx 0.0010\,\mathrm{M}\] So, at equilibrium, the concentration of Cr^3+ ions in the solution is approximately 0.0010 M.

Key concepts

Equilibrium Constant

An equilibrium constant, denoted as \( K_c \), provides the ratio of concentrations of products to reactants for a reversible chemical reaction at equilibrium. Its value indicates the extent of the reaction; a large \( K_c \) value suggests that the reaction favors product formation. In this specific problem, the given \( K_c \) is extremely large at \( 1.0 \times 10^{23} \), implying that equilibrium heavily favors the formation of the complex \( \text{Cr(EDTA)} \). Small changes in concentrations can be neglected in this context because the amount of reactants consumed is minimal compared to the concentrations of products formed.
Understanding the equilibrium constant is crucial for predicting reaction behavior, especially in complex formations such as in coordination chemistry with EDTA.

ICE Table

The ICE table is a useful tool in chemistry for determining equilibrium concentrations of reactants and products. "ICE" stands for Initial, Change, and Equilibrium, representing the stages of a chemical reaction.

• **Initial**: Refers to the starting concentrations before the reaction begins.
• **Change**: Changes in concentrations that occur as the system moves towards equilibrium.
• **Equilibrium**: The concentrations after the system has reached equilibrium.

In the context of this exercise, the ICE table is used to calculate how much \( \text{Cr}^{3+} \), \( \text{H}_2\text{EDTA}^{2-} \), and \( \text{Cr(EDTA)} \) change as the complex forms. This structured approach helps in visualizing the shift in concentrations and is key in solving equilibrium problems accurately.

EDTA Complexation

EDTA (Ethylenediaminetetraacetic acid) is a powerful chelating agent often used in chemistry to sequester metal ions. It forms very stable complexes with metal ions like \( \text{Cr}^{3+} \), helping in applications such as water treatment and medicine. The reaction explored involves the complexation of \( \text{Cr}^{3+} \) by \( \text{H}_2\text{EDTA}^{2-} \), forming \( \text{Cr(EDTA)} \).
This type of complexation reaction is highly favorable, as evidenced by the large equilibrium constant and contributes to drastically reducing the free metal ion concentration in a solution. Understanding the nature of EDTA complexation reactions is essential for fields requiring precise control over metal ion activity.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of reactants and products when a reaction has reached equilibrium. In this exercise, the equilibrium concentrations of \( \text{Cr}^{3+} \), \( \text{H}_2\text{EDTA}^{2-} \), and \( \text{Cr(EDTA)} \) are determined using the known initial concentrations and the changes from the reaction progress tracked using an ICE table.

For this particular reaction, even though \( x \) represents the change in concentration, it is exceedingly small compared to initial amounts, confirming that the equilibrium concentration of \( \text{Cr}^{3+} \) remains practically unchanged from its initial value. This outcome illustrates the principle that when an equilibrium constant is significantly high, it assures minimal concentrations of uncomplexed metal ions.