Question

Calculate the concentration of \(\mathrm{Pb}^{2+}\) in each of the following. a. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=1.2 \times 10^{-15}\) b. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13.00\) c. Ethylenediaminetetraacetate \(\left(\mathrm{EDTA}^{4-}\right)\) is used as a complexing agent in chemical analysis and has the following structure: 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 reaction of EDTA \(^{4-}\) with \(\mathrm{Pb}^{2+}\) is \(\begin{aligned} \mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) & \\ K &=1.1 \times 10^{18} \end{aligned}\) Consider a solution with \(0.010\) mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 L of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing \(0.050 M\) NasEDTA. Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?

Short answer

In the three given scenarios, the concentration of \(\mathrm{Pb}^{2+}\) ions is found to be: a. \(1.7 \times 10^{-6} M\) in a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) with \(K_{\mathrm{sp}}=1.2 \times 10^{-15}\) b. \(1.2 \times 10^{-13} M\) in a saturated solution buffered at \(\mathrm{pH}=13.00\) c. In the solution with \(0.010\) mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 L of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing \(0.050 M\) Na\(\mathrm{EDTA}^{4-}\), \(\mathrm{Pb}(\mathrm{OH})_{2}\) will not precipitate.

Step-by-step solution

Write the dissolution equation for \(\mathrm{Pb}(\mathrm{OH})_{2}\)

The dissolution equation for \(\mathrm{Pb}(\mathrm{OH})_{2}\) is: \[\mathrm{Pb}(\mathrm{OH})_{2} (s) \rightleftharpoons \mathrm{Pb}^{2+} (aq) + 2 \mathrm{OH}^- (aq)\]

Write the \(K_{\mathrm{sp}}\) expression

The \(K_{\mathrm{sp}}\) expression for the dissolution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) is: \[K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}] [(\mathrm{OH}^-)^2]\]

Calculate the concentration of \(\mathrm{Pb}^{2+}\) and \(\mathrm{OH}^-\)

Let \([\mathrm{Pb}^{2+}] = x\) and \([\mathrm{OH}^-] = 2x\). Now, substitute the \(K_{\mathrm{sp}}\) value and the concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{OH}^-\) ions in the \(K_{\mathrm{sp}}\) expression: \[1.2 \times 10^{-15} = x (2x)^2\] Solving for x, we get, \[\boxed{[\mathrm{Pb}^{2+}] = x = 1.7 \times 10^{-6} M}\] For part b:

Calculate \([\mathrm{OH}^-]\) from the given pH

Since the solution is buffered at \(\mathrm{pH}=13.00\), we can find the concentration of \(\mathrm{OH}^-\) ions using the formula for \(\mathrm{pOH}\): \[\mathrm{pOH} = 14.00 - \mathrm{pH} = 14.00 - 13.00 = 1\] Now, using the relation between \(\mathrm{pOH}\) and the concentration of \(\mathrm{OH}^-\): \[[\mathrm{OH}^-] = 10^{-\mathrm{pOH}} = 10^{-1} = 0.1\, M\]

Calculate the concentration of \(\mathrm{Pb}^{2+}\)

We can now substitute this \([\mathrm{OH}^-]\) value in the \(K_{\mathrm{sp}}\) expression and solve for \([\mathrm{Pb}^{2+}]\): \[1.2 \times 10^{-15} = [\mathrm{Pb}^{2+}] (0.1)^2\] Solving for the concentration of \(\mathrm{Pb}^{2+}\), we get, \[\boxed{[\mathrm{Pb}^{2+}] = 1.2 \times 10^{-13}\, M}\] For part c:

Write the equation for the \(\mathrm{Pb}^{2+}\) and \(\mathrm{EDTA}^{4-}\) complex formation

The equation representing the complex formation between \(\mathrm{Pb}^{2+}\) and \(\mathrm{EDTA}^{4-}\) is: \[\mathrm{Pb}^{2+}(aq) + \mathrm{EDTA}^{4-}(aq) \rightleftharpoons \mathrm{PbEDTA}^{2-}(aq)\]

Calculate the \(K\) expression for the complex and find the concentration of \(\mathrm{Pb}^{2+}\)

Using the \(K\) value of \(1.1 \times 10^{18}\) for the complex formation, we can find the concentration of \(\mathrm{Pb}^{2+}\) in the solution. Keeping in mind that the given conditions are: \(0.010\) mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 L of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing \(0.050 M\) Na\(\mathrm{EDTA}^{4-}\), we can perform an equilibrium calculation to find whether \(\mathrm{Pb}(\mathrm{OH})_{2}\) will precipitate from this solution or not. Since we have already calculated the concentration of \(\mathrm{Pb}^{2+}\) in part b, we can compare it with the concentration of \(\mathrm{Pb}^{2+}\) required for \(\mathrm{Pb}(\mathrm{OH})_{2}\) to precipitate. If it is higher than the calculated concentration, then \(\mathrm{Pb}(\mathrm{OH})_{2}\) will precipitate, otherwise, it will not. In this case, since the calculated concentration of \(\mathrm{Pb}^{2+}\) in part b is \(1.2 \times 10^{-13}\, M\), which is greater than the concentration of \(\mathrm{Pb}^{2+}\) required for \(\mathrm{Pb}(\mathrm{OH})_{2}\) to precipitate (\(1.7 \times 10^{-6} M\)), we can conclude that: \[\boxed{\text{No, }\mathrm{Pb}(\mathrm{OH})_{2} \text{ will not precipitate from this solution.}}\]

Key concepts

Chemical Equilibrium

The concept of chemical equilibrium is vital in understanding various reactions, including the solubility of substances. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentrations of reactants and products remain constant over time. This principle applies to solubility where a solid, such as Pb(OH)2, is in a dynamic equilibrium with its dissolved ions in solution.

For the exercise provided where we consider the dissolution of lead(II) hydroxide, we write an equation representing this equilibrium. Then, we apply the solubility product constant (Ksp), which quantifies this equilibrium. Understanding this balance helps us predict whether a precipitate will form in different scenarios, which is especially important in questions about solubility and complex ion formation.

Concentration Calculation

Concentration calculation is an essential skill in chemistry that allows us to understand the extent of a reaction and the amounts of substances in solution. In the context of the exercise, it involves using the value of the solubility product constant and the stoichiometry of the dissociation equation to calculate the concentration of lead ions ([Pb2+]) under various conditions.

For instance, when dealing with the solubility of Pb(OH)2 at a given pH, it's important to calculate the concentration of hydroxide ions ([OH-]) first by using the pOH. Then, we can use the equilibrium expression to solve for the concentration of lead ions. This step-by-step process simplifies complex problems and helps predict the behaviour of ions in solution. Always ensure the units are consistent and remember that concentration is typically expressed in moles per litre (M).

Complex Ion Formation

In chemistries such as complex ion formation, equilibrium principles are applied to predict the formation of new species in solution. A complex ion consists of a central metal ion bonded to one or more molecules or ions, called ligands. The formation of a complex ion can significantly increase the solubility of certain metals in water.

In the exercise, EDTA acts as a ligand to form a complex with lead ions. The extremely high equilibrium constant (K) for the formation of the PbEDTA2- complex indicates that this reaction is highly favored, essentially pulling lead ions out of any precipitate that might have formed, keeping them in solution. This is a prime example of the Le Châtelier's Principle, illustrating how the addition of a reactant (in this case, the EDTA ligand) will shift the equilibrium to favor the formation of products (the complex ion).