Problem 92 - Chemistry

Question

The equilibrium constant for the following reaction is \(1.0 \times 10^{23} :\) $$\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q) \rightleftharpoons \mathrm{CrEDTA}^{-}(a q)+2 \mathrm{H}^{+}(a q)$$ 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 \(0.050 M\) in \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) and buffered at \(\mathrm{pH}=6.00.\)

Step-by-Step Solution

Answer

The equilibrium concentration of Cr³⁺ in the given reaction is approximately 0.0010 M.

Step 1: Write down the equilibrium constant expression

For the given reaction: K = \(\frac{[CrEDTA^-][H^+]^2}{[Cr^{3+}][H_2EDTA^{2-}]}\)

Step 2: Set up an ICE table to determine concentrations

We can set up an ICE table as follows: Cr^3+ + H₂EDTA²⁻ ↔ CrEDTA⁻ + 2H⁺ Initial: 0.0010 M 0.050 M 0 M Given Change: -x -x +x +2x Equilibrium: 0.0010-x 0.050-x x 2x

Step 3: Calculate H⁺ concentration from given pH

Since pH = 6.00: -log[H^⁺] = 6.00 We can solve for the H⁺ concentration: H⁺ = \(10^{-6}\) M As it is a buffered solution, [H⁺] doesn't change much during the reaction. Thus, [H⁺] ≈ 2x ≈ \(10^{-6}\) M.

Step 4: Substitute concentrations into the equilibrium expression

Substitute these concentrations into the equilibrium expression: \(1.0 \times 10^{23}\) = \(\frac{x(10^{-6})^2}{(0.0010-x)(0.050-x)}\)

Step 5: Solve for x (concentration of CrEDTA⁻ and Cr³⁺ at equilibrium)

Assuming x is very small compared to 0.0010 and 0.050, we can drop x from the denominator: \(1.0 \times 10^{23}\) = \(\frac{x(10^{-6})^2}{(0.0010)(0.050)}\) Solve for x: x ≈ \(5.0 \times 10^{-20}\) M

Step 6: Calculate equilibrium concentration of Cr³⁺

At equilibrium, [Cr³⁺] = 0.0010 - x: [Cr³⁺] ≈ 0.0010 M - \(5.0 \times 10^{-20}\) M ≈ 0.0010 M

Key Concepts

Complexation Reaction

A complexation reaction involves the formation of a complex compound through the combination of molecules or ions. In this exercise, EDTA acts as a chelating agent, combining with metal ions to form stable complexes. The reaction:

• \( \text{Cr}^{3+} + \text{H}_2\text{EDTA}^{2-} \rightleftharpoons \text{CrEDTA}^- + 2 \text{H}^+ \)
• This type of reaction is critical in various applications, including metal extraction, purification, and titration analysis.

Understanding complexation reactions helps in manipulating conditions to favor the desired product. In our example, EDTA captures chromium ions, effectively reducing the free metal ion concentration.

Stability Constant

The stability constant, often denoted as \( K \), reflects the strength of the complex formed in a complexation reaction. It is defined by the equilibrium expression for the reaction. In the given problem, the stability constant is very high \((1.0 \times 10^{23})\), indicating highly favored formation of the CrEDTA complex:

• \( K = \frac{[\text{CrEDTA}^-][\text{H}^+]^2}{[\text{Cr}^{3+}][\text{H}_2\text{EDTA}^{2-}]} \)
• A larger \( K \) value means that the reaction significantly favors the products over the reactants.

Such a high stability constant is crucial in applications where one wants to remove or detect specific metal ions, as it ensures that almost all ions are captured by the chelating agent.

Heavy Metal Poisoning Treatment

Heavy metal poisoning can be addressed using chelating agents like EDTA, which form stable complexes with toxic metal ions, aiding in their removal from the body. The principle behind this treatment lies in:

• Using EDTA to bind heavy metals, thus preventing them from interacting with essential biological molecules.
• The chelated complex is often more soluble, facilitating excretion through the kidneys.

This approach helps detoxify the body efficiently, making EDTA widely used in clinical settings for treating heavy metal exposure.

Buffered Solution

A buffered solution resists changes in its pH when small amounts of acid or base are added. In this exercise, we see that even as hydrogen ions are produced, the pH remains stable at 6.00. This stability comes from:

• A weak acid-base conjugate pair, such as acetic acid and acetate, maintaining the pH.
• In buffered systems, the concentration of ions like \( [\text{H}^+] \) is usually constant; small changes do not affect pH drastically.

Buffered solutions are vital in biological and chemical systems where specific pH levels are necessary for reactions to proceed effectively.