Question

Ethylenediaminetetraacetate \(\left(\mathrm{EDTA}^{4-}\right)\) is used as a complexing agent in chemical analysis with the structure shown in Figure 21.7. Solutions of EDTA \(^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The complex ion essentially eliminates the heavy metal ions from reacting with biochemical systems. The reaction of EDTA \(^{4-}\) with \(\mathrm{Pb}^{2+}\) is \(\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) \quad K=1.1 \times 10^{18}\) Consider a solution with \(0.010 \mathrm{~mol} \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to \(1.0 \mathrm{~L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing \(0.050\) \(M \mathrm{Na}_{4} \mathrm{EDTA} .\) Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution? \(\left(K_{\text {?? }}\right.\) for \(\left.\mathrm{Pb}(\mathrm{OH})_{2}=1.2 \times 10^{-15} .\right)\)

Short answer

The concentration of free Pb²⁺ ions in the solution is \(4.6 \times 10^{-23}\) M, and the concentration of OH⁻ ions is \(0.1 \) M. The reaction quotient Q calculated as \(4.6 \times 10^{-25}\) is less than the solubility product Ksp = \(1.2 \times 10^{-15}\) for Pb(OH)₂. Therefore, Pb(OH)₂ does not precipitate from the solution.

Step-by-step solution

Calculate the concentration of free Pb²⁺ ions.

First, we need to find the concentration of Pb²⁺ ions in the solution. To do this, we'll use the equilibrium constant K for the reaction of Pb²⁺ with EDTA⁴⁻. The reaction: \(\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q)\), K = \(1.1 \times 10^{18}\) Let \(x\) be the concentration of free Pb²⁺ ions. Then, we have: \[K = \frac{[\mathrm{PbEDTA}^{2-}]}{[\mathrm{Pb}^{2+}][\mathrm{EDTA}^{4-}]} = \frac{[Pb^{2+}] - x}{x([EDTA^{4-}] - x)}\] We can consider \(x\) to be very small because K is very large. In this case, let's neglect \(x\) in both [Pb²⁺] and [EDTA⁴⁻] terms: \[K = \frac{([Pb^{2+}])([EDTA^{4-}])}{x}\] Now, calculate the value of x: \[x = \frac{([Pb^{2+}])([EDTA^{4-}])}{K} = \frac{(0.010)(0.050)}{1.1 \times 10^{18}}\) \[x = 4.6 \times 10^{-23}\]

Calculate the concentration of OH⁻ ions.

We are given the pH of the solution, which is 13. We will now calculate the concentration of OH⁻ ions using the pOH: pOH = 14 - pH = 1 Therefore, the concentration of OH⁻ ions in the solution is: [OH⁻] = \(10^{-pOH} = 10^{-1} = 0.1\)

Determine if Pb(OH)₂ precipitates.

To determine whether Pb(OH)₂ will precipitate, we will calculate the reaction quotient (Q) and compare it with the solubility product (Ksp). The reaction is: \(\mathrm{Pb}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q) + 2\mathrm{OH}^{-}(a q)\) The reaction quotient Q: \(Q = [\mathrm{Pb}^{2+}][\mathrm{OH}^{-}]^{2}\) The given Ksp for Pb(OH)₂ is \(1.2 \times 10^{-15}\). If Q < Ksp, then the Pb(OH)₂ does not precipitate. If Q > Ksp, then the Pb(OH)₂ precipitates. Let's calculate Q: \(Q = (4.6 \times 10^{-23})(0.1)^{2}\) \(Q = 4.6 \times 10^{-25}\) Since Q < Ksp, the Pb(OH)₂ does not precipitate from the solution.

Key concepts

Complex Ion Formation

Understanding complex ion formation is essential when examining reactions like the one between EDTA and heavy metal ions. A complex ion consists of a central metal ion bonded to one or more molecules or ions, called ligands. In this case, EDTA acts as a ligand and forms a stable complex with lead ions (Pb²⁺), creating the complex ion PbEDTA²⁻.

Complex ions are significant in various fields including medical treatment, where they can be used to deactivate heavy metal ions by preventing them from interfering with biological systems. The complex ion’s stability is governed by the strength of the interactions between the metal ion and the ligands. EDTA, with its multiple electron-donating groups, is particularly effective at binding to metal ions, thus making it an excellent chelating agent.

Equilibrium Constant (K)

The equilibrium constant (K) quantifies the balance between products and reactants in a reversible chemical reaction at equilibrium. It is determined by the concentrations of the species involved once the reaction has stabilized and no further changes in concentrations are observed. The higher the value of K, the more the equilibrium favors the formation of products.

In the provided exercise, the reaction between EDTA and Pb²⁺ has an exceptionally large equilibrium constant (\(K = 1.1 \times 10^{18}\)), indicating that the formation of the PbEDTA²⁻ complex is highly favored. This implies that virtually all the lead ions will be bound to EDTA in the solution, resulting in an extremely low concentration of free Pb²⁺ ions, which is critical in assessing whether lead hydroxide will precipitate in the solution.

Solubility Product (Ksp)

The solubility product (Ksp) is an equilibrium constant that applies to the dissolution of a sparingly soluble compound. Specifically, it represents the product of the ion concentrations in a saturated solution, each raised to the power of its stoichiometric coefficient. This concept helps predict whether a precipitate will form when solutions are mixed.

In our exercise scenario, the solubility product of \(Pb(OH)_{2}\) is given as \(1.2 \times 10^{-15}\). If the reaction quotient (Q), also calculated through the product of the relevant ion concentrations, is less than the Ksp, the solution is undersaturated, and no precipitate will form. The calculation shows that Q for the Pb²⁺ and OH⁻ ions is far below the Ksp, assuring that no PB(OH)₂ precipitate will form, effectively demonstrating the power of EDTA chelation in preventing lead hydroxide precipitation.