Comprehensive Analysis of Ba²⁺ and EDTA Complexation

Understanding Equilibrium Concentrations and Dilution Effects in Aqueous Solutions

Key Takeaways

• High Equilibrium Constant Indicates Complete Reaction: The large value of the equilibrium constant (K = 7.7 × 10⁷) ensures that nearly all Ba²⁺ ions react with EDTA⁴⁻ to form the complex Ba(EDTA)²⁻.
• Limiting Reagent Determines Product Formation: In the initial mixture, Ba²⁺ is the limiting reagent, dictating the maximum concentration of Ba(EDTA)²⁻ formed.
• Dilution Shifts Equilibrium Slightly: Upon dilution, the equilibrium shifts to slightly favor the dissociation of Ba(EDTA)²⁻, resulting in a marginal increase in free Ba²⁺ ions.

Introduction

The complexation of metal ions with ligands is a fundamental concept in coordination chemistry, playing a pivotal role in various biochemical and industrial processes. In this analysis, we delve into the reaction between barium ions (Ba²⁺) and the chelating agent EDTA⁴⁻ to form the complex ion Ba(EDTA)²⁻. By examining the equilibrium concentrations and the effects of dilution, we aim to elucidate the behavior of this system under different conditions.

Part (a): Determining the Concentration of Ba(EDTA)²⁻ in 100.0 mL Solution

1. Initial Concentrations After Mixing

We begin by calculating the initial concentrations of EDTA⁴⁻ and Ba²⁺ ions in the 100.0 mL solution formed by mixing equal volumes of their respective solutions.

Initial Concentrations

2. Identifying the Limiting Reagent

The reaction stoichiometry is 1:1 between Ba²⁺ and EDTA⁴⁻:

$$\text{Ba}^{2+} + \text{EDTA}^{4-} \leftrightarrow \text{Ba(EDTA)}^{2-}$$

Given the moles calculated:

• Ba²⁺: 0.0100 mol
• EDTA⁴⁻: 0.0150 mol

Ba²⁺ is the limiting reagent as it has fewer moles available to react with EDTA⁴⁻.

3. Calculating the Concentration of Ba(EDTA)²⁻

Since nearly all Ba²⁺ ions react with EDTA⁴⁻ (due to the large K value), the concentration of Ba(EDTA)²⁻ formed is equivalent to the initial concentration of Ba²⁺.

Therefore:

$$[\text{Ba(EDTA)}^{2-}] = 0.100 \, \text{M}$$

4. Equilibrium Considerations

The equilibrium expression for the reaction is:

$$K = \frac{[\text{Ba(EDTA)}^{2-}]}{[\text{Ba}^{2+}][\text{EDTA}^{4-}]}$$

Substituting the known values:

$$7.7 \times 10^7 = \frac{0.100}{[\text{Ba}^{2+}][0.150]}$$

Solving for [Ba²⁺]:

$$[\text{Ba}^{2+}] = \frac{0.100}{(7.7 \times 10^7 \times 0.150)}$$

$$[\text{Ba}^{2+}] \approx 8.66 \times 10^{-10} \, \text{M}$$

Since K is significantly large, [Ba²⁺] is negligible, affirming that almost all Ba²⁺ is complexed.

Part (a) Answer

The concentration of Ba(EDTA)²⁻ in the 100.0 mL solution is approximately 0.10 M. This is because the limiting reagent, Ba²⁺, is almost entirely consumed due to the high equilibrium constant, resulting in a stoichiometric formation of Ba(EDTA)²⁻.

Part (b): Effect of Dilution on Ba²⁺ Moles

1. Dilution to 1.00 L

The initial 100.0 mL solution is diluted to 1.00 L, a tenfold increase in volume. This dilution affects the concentrations of all species present.

New Concentrations After Dilution

2. Re-establishing Equilibrium

The dilution perturbs the equilibrium, necessitating a shift to re-establish the equilibrium constant. Applying Le Châtelier's Principle:

Original Equilibrium:
$$\text{Ba}^{2+} + \text{EDTA}^{4-} \leftrightarrow \text{Ba(EDTA)}^{2-}$$

New Equilibrium After Dilution:
$$\text{Ba}^{2+} + \text{EDTA}^{4-} \leftrightarrow \text{Ba(EDTA)}^{2-}$$

The decrease in concentrations shifts the equilibrium to favor the formation of more reactants if the reaction is exothermic or to products if it is endothermic. However, given the substantial value of K, the shift is minimal, but it can still affect the free Ba²⁺ concentration.

3. Quantitative Analysis

Let x be the amount of Ba(EDTA)²⁻ that dissociates upon dilution:

Substituting these into the equilibrium expression:

$$K = \frac{[\text{Ba(EDTA)}^{2-}]}{[\text{Ba}^{2+}][\text{EDTA}^{4-}]} = \frac{0.010 - x}{x (0.015 + x)}$$

Given the large value of K, approximation can be made by neglecting x in the denominator:

$$7.7 \times 10^7 \approx \frac{0.010}{x \times 0.015}$$

Solving for x:

$$x \approx \frac{0.010}{7.7 \times 10^7 \times 0.015} \approx 8.55 \times 10^{-10} \, \text{M}$$

The concentration of free Ba²⁺ after dilution is approximately 8.55 × 10⁻¹⁰ M.

4. Comparing Moles of Ba²⁺ Before and After Dilution

To determine whether the number of moles of Ba²⁺ increases, decreases, or remains the same upon dilution, we compare the moles before dilution (which were nearly zero due to complete complexation) to those after dilution.

Before Dilution:
$$\text{Moles of Ba}^{2+} \approx 0 \, \text{mol}$$

After Dilution:
$$\text{Moles of Ba}^{2+} = 8.55 \times 10^{-10} \, \text{M} \times 1.00 \, \text{L} = 8.55 \times 10^{-10} \, \text{mol}$$

The number of moles of free Ba²⁺ increases from nearly zero to a minuscule but measurable amount upon dilution.

Part (b) Answer

After diluting the solution to a total volume of 1.00 L, the equilibrium shifts slightly to dissociate a very small amount of Ba(EDTA)²⁻. This results in an increase in the number of moles of free Ba²⁺ present in the solution compared to the original concentrated solution prior to dilution. Specifically, while initially nearly all Ba²⁺ was complexed, the dilution causes a minor shift that releases approximately 8.55 × 10⁻¹⁰ moles of Ba²⁺ into the solution.

Conclusion

The complexation of Ba²⁺ with EDTA⁴⁻ is significantly driven by the large equilibrium constant, ensuring that nearly all free Ba²⁺ ions are sequestered in the form of Ba(EDTA)²⁻. In a 100.0 mL solution, this results in a high concentration of the complex ion. Upon dilution to 1.00 L, although the vast majority of Ba²⁺ remains complexed, a slight shift in equilibrium leads to the formation of an exceedingly small concentration of free Ba²⁺ ions. This exemplifies the sensitivity of equilibrium systems to changes in concentration and volume, as articulated by Le Châtelier's Principle.

References

chegg.com

Chegg Homework Help

secure-media.collegeboard.org

College Board AP Chemistry

pearson.com

Pearson Analytical Chemistry