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 (EDTA \(^{4-}\) ) 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 \(1.0 \mathrm{L}\) 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?

Short answer

a. The concentration of \(\mathrm{Pb}^{2+}\) in a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) is found by solving the equation \(x(2x)^2 = 1.2 \times 10^{-15}\), which gives \([\mathrm{Pb}^{2+}] = 1.31 \times 10^{-6}\,\mathrm{M}\). b. In the buffered solution at pH 13, the concentration of \(\mathrm{OH}^-\) is 0.1 M, and the concentration of \(\mathrm{Pb}^{2+}\) is found using the \(K_{\mathrm{sp}}\) expression: \([\mathrm{Pb}^{2+}] = \frac{1.2 \times 10^{-15}}{(0.1)^2} = 1.2 \times 10^{-13}\,\mathrm{M}\). c. To find if \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitates, calculate the equilibrium concentration of \(\mathrm{Pb}^{2+}\) using the equation \(K = \frac{y}{(0.010 - y)(0.050 - y)}\), then compare \([\mathrm{Pb}^{2+}][\mathrm{OH}^-]^2\) to \(K_{\mathrm{sp}}\). With the provided concentrations, no precipitation occurs in this scenario.

Step-by-step solution

a. Saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\)

Given the formula: \(\mathrm{Pb}(\mathrm{OH})_{2} \rightleftharpoons \mathrm{Pb}^{2+} + 2\,\mathrm{OH}^{-}\) We need to find the concentration of \(\mathrm{Pb}^{2+}\) in the saturated solution. Let x be the concentration of \(\mathrm{Pb}^{2+}\) in the solution. Since it dissociates into two \(\mathrm{OH}^-\) ions, the concentration of \(\mathrm{OH}^-\) will be 2x. Now we can use the \(K_{\mathrm{sp}}\) expression and plug in the values: \(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{OH}^-]^2 = (x)(2x)^2\) As we know, \(K_{\mathrm{sp}} = 1.2 \times 10^{-15}\), we can solve for x. \(x(2x)^2 = 1.2 \times 10^{-15}\)

b. Saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at pH = 13

We're given the pH of the solution, which is 13. From pH, we can find the concentration of \(\mathrm{OH}^-\) present in the solution. Using the formula: \(pOH = 14 - pH\) \(pOH = 14 - 13 = 1\) So we can calculate the concentration of \(\mathrm{OH}^-\) ions as: \([\mathrm{OH}^-] = 10^{-pOH} = 10^{-1} = 0.1\,M\) Now, we can use the \(K_{\mathrm{sp}}\) expression and plug in the values: \(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{OH}^-]^2 = [\mathrm{Pb}^{2+}](0.1)^2\) We can solve for \([\mathrm{Pb}^{2+}]\): \([\mathrm{Pb}^{2+}] = \frac{K_{sp}}{[\mathrm{OH}^-]^2} = \frac{1.2 \times 10^{-15}}{(0.1)^2}\)

c. Solution containing \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\), \(\mathrm{Na}_{4}\mathrm{EDTA}\), and buffered at pH = 13

Given the equation for the reaction: \(\mathrm{Pb}^{2+}(aq) + \mathrm{EDTA}^{4-}(aq) \rightleftharpoons \mathrm{PbEDTA}^{2-}(aq)\) The given moles and concentrations: \([\mathrm{Pb}^{2+}] = 0.010 \,M\) \([\mathrm{EDTA}^{4-}] = 0.050 \,M\) As we know, \(K = 1.1 \times 10^{18}\) Let y be the equilibrium concentration of \(\mathrm{PbEDTA}^{2-}\), then we can write the reaction table with these equilibrium concentrations: \(\mathrm{Pb}^{2+}: 0.010 - y\) \(\mathrm{EDTA}^{4-}: 0.050 - y\) \(\mathrm{PbEDTA}^{2-}: y\) Now we can use the given K value and plug in the values: \(K = \frac{[\mathrm{PbEDTA}^{2-}]}{[\mathrm{Pb}^{2+}][\mathrm{EDTA}^{4-}]} = \frac{y}{(0.010 - y)(0.050 - y)}\) We can solve for y to find the concentration of \(\mathrm{Pb}^{2+}\). Then check if \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitates: If \([\mathrm{Pb}^{2+}][\mathrm{OH}^-]^2 > K_{sp}\), then \(\mathrm{Pb}(\mathrm{OH})_{2}\) will precipitate.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is essential for analyzing reactions in solutions, such as those involving solubility. In a chemical system, equilibrium is the state where the rates of the forward and reverse reactions are equal, resulting in no net change in the amounts of reactants and products. This state can be characterized by an equilibrium constant, denoted as K, which remains constant at a given temperature.

The equilibrium constant for a reaction is the ratio of the concentrations of the products raised to the power of their coefficients to the concentrations of the reactants raised to the power of their coefficients. For solubility products, we use the symbol K_sp, which represents the maximum product of the ion concentrations that can exist in solution without precipitating. K_sp is a special type of equilibrium constant that applies to saturated solutions of sparingly soluble salts. It is important to remember that equilibrium constants are not influenced by the amounts of reactants or products, only by the temperature.

In step 1 of the solution, we utilize the concept of chemical equilibrium to calculate the concentrations of ions in a saturated solution of Pb(OH)₂. By setting up an equilibrium table and using the solubility product constant (K_sp), we can solve for the concentration of the lead ion (Pb²⁺).

Complex Ion Formation

Complex ion formation is a process where a metal ion binds with one or more ligands to form a complex ion. A ligand is a molecule or ion that can form a coordination bond with a metal ion by donating a pair of electrons. Complex ions can have significant impacts on the solubility of a compound in a solution.

In step 3, the complex ion formation between lead ions, Pb²⁺, and the ligand EDTA⁴⁻ is illustrated. The extraordinarily large equilibrium constant, K, indicates that the formation of the complex ion PbEDTA²⁻ is highly favored. This strong tendency to form a complex ion can potentially increase the solubility of lead ions in solution, altering the expected behavior of precipitation based on the solubility product constant alone.

The formation of a complex ion can substantially change the concentration of free metal ions in a solution, often making them more soluble than they would be without the presence of a complexing agent. This phenomenon is important in various applications, including medical treatment for heavy metal poisoning, water softening, and analytical chemistry.

pH and Solubility

The solubility of a substance can be influenced by the pH of the solution in which it is dissolved. This is particularly true for substances like metal hydroxides, whose solubility depends on the concentration of hydroxide ions (OH⁻) in the solution. When the pH of the solution is changed, the solubility of these compounds can increase or decrease accordingly.

In steps 1 and 2, the role of pH comes into play while analyzing the solubility of Pb(OH)₂. In step b, the pH of the buffer solution is 13, which implies a high concentration of hydroxide ions due to the relationship between pH and pOH, as well as the self-ionization of water. This high concentration of hydroxide ions affects the solubility product and, consequently, the equilibrium concentrations of soluble lead ions.

Furthermore, when pH is adjusted using a buffer or any other means, it can shift the equilibrium towards the formation or dissociation of soluble or insoluble compounds. Predicting the effect of pH on solubility is essential for correctly anticipating the outcome in reactions, whether a precipitate will form or dissolve in the context of a saturated solution.