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 \(\left(\mathrm{ED} \mathrm{TA}^{4-}\right)\) is used as a complexing agent in chemical analysis and has the following structure: Solutions of \(\mathrm{EDTA}^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of \(\mathrm{EDTA}^{4-}\) with \(\mathrm{Pb}^{2+}\) is 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 \(0.050 \mathrm{M} \mathrm{Na}_{4} \mathrm{EDTA} .\) Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?

Short answer

The concentration of \(\mathrm{Pb}^{2+}\) is approximately \(1.14 \times 10^{-6} \mathrm{M}\) in a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\), and \(1.2 \times 10^{-13} \mathrm{M}\) in a solution buffered at pH 13. In the given solution with EDTA, \(\mathrm{Pb}(\mathrm{OH})_{2}\) will precipitate because the reaction quotient Q \( (1 \times 10^{-4})\) is much greater than Ksp \((1.2 \times 10^{-15})\).

Step-by-step solution

Write the solubility product expression

For a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\), the dissolution equation and solubility product (Ksp) expression are given by: \[\mathrm{Pb}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{OH}^-(aq)\] \[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{OH}^-]^2\]

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

Let the solubility of \(\mathrm{Pb}(\mathrm{OH})_{2}\) be S. Then, the concentration of \(\mathrm{Pb}^{2+}\) will be S and the concentration of \(\mathrm{OH}^-\) will be 2S. Now substituting the values in the Ksp expression: \[K_{sp} = [\mathrm{Pb}^{2+}] \times (2 \times S)^2\] \[1.2 \times 10^{-15} = S \times (2S)^2\]

Calculate S

Solve the equation to get the value for S: \[S =\sqrt[3]{\frac{1.2 \times 10^{-15}}{8}} = \sqrt[3]{1.5 \times 10^{-16}}\] \[S \approx 1.14 \times 10^{-6}\] So, the concentration of \(\mathrm{Pb}^{2+}\) in a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) is approximately \(1.14 \times 10^{-6} \mathrm{M}\). **For part b.**

Find the concentration of \(\mathrm{OH}^-\) ions using the given pH

The given pH is 13. To find the concentration of hydroxide ions, we can use the formula: \[\mathrm{pOH} = 14 - \mathrm{pH}\] \[\mathrm{pOH} = 14 - 13 = 1\] Now, the concentration of \(\mathrm{OH}^-\) ions can be calculated as: \[[\mathrm{OH}^-] = 10^{-\mathrm{pOH}} = 10^{-1} = 0.1 \mathrm{M}\]

Use the Ksp expression to find the concentration of \(\mathrm{Pb}^{2+}\)

Using the Ksp expression, we can find the concentration of \(\mathrm{Pb}^{2+}\): \[K_{sp} = [\mathrm{Pb}^{2+}] \times [\mathrm{OH}^-]^2\] \[1.2 \times 10^{-15} = [\mathrm{Pb}^{2+}] \times (0.1)^2\] Now, solve for the concentration of \(\mathrm{Pb}^{2+}\): \[[\mathrm{Pb}^{2+}] = \frac{1.2 \times 10^{-15}}{(0.1)^2} = 1.2 \times 10^{-13}\] So, the concentration of \(\mathrm{Pb}^{2+}\) in the solution buffered at pH 13 is \(1.2 \times 10^{-13} \mathrm{M}\). **For part c.**

Evaluate if precipitation occurs

To evaluate if precipitation occurs, we must compare the reaction quotient (Q) with the Ksp value. The total concentration of \(\mathrm{Pb}^{2+}\) and \(\mathrm{OH}^-\) ions can be deduced from the given problem. Since the total Molarity of Pb2+ is 0.010 mol/ 1.0 L = 0.010 M and the concentration of OH- ions is given from pH = 13 (0.1 M), let's now compute the Q value. \[Q = [\mathrm{Pb}^{2+}][\mathrm{OH}^-]^2 = (0.010)(0.1)^2 = 1 \times 10^{-4}\] Now, compare Q and Ksp values: If Q > Ksp, precipitation occurs. If Q < Ksp, no precipitation.

Compare Q and Ksp

We have: \[Q = 1 \times 10^{-4}\] \[K_{sp} = 1.2 \times 10^{-15}\] Since Q (1 × 10⁻⁴) is much greater than Ksp (1.2 × 10⁻¹⁵), \(\mathrm{Pb}(\mathrm{OH})_{2}\) will precipitate from the solution.

Key concepts

Understanding the Ksp Expression

The solubility product constant, commonly abbreviated as Ksp, plays a crucial role in predicting whether a solid will dissolve in a solution or a precipitate will form. It's a unique value for sparingly soluble salts and is calculated at a given temperature. For a general salt compound, which dissolves into its constituent ions, A and B, in the ratio of a:b, the Ksp expression would look like this:

\[Ksp = [A]^a[B]^b\]
In our case, the salt is lead(II) hydroxide, \(\mathrm{Pb}(\mathrm{OH})_{2}\). Upon dissolution, it separates into \(\mathrm{Pb}^{2+}\) and \(\mathrm{OH}^-\) ions. The Ksp expression becomes:
\[Ksp = [\mathrm{Pb}^{2+}][\mathrm{OH}^-]^2\]
This equation shows the relationship between the concentrations of the ions in a saturated solution: when the product of these concentrations reaches a certain point, the solution is at equilibrium, and no more solid can dissolve. If the product exceeds the Ksp value, the solution becomes supersaturated, and precipitation occurs. Understanding this concept helps students predict solubility outcomes and, by calculating the Ksp, quantify the solubility of compounds under different conditions.

Calculating Concentration from the Ksp Value

Often in chemistry, we come across situations where we need to calculate the concentration of ions in a solution. To achieve this from the Ksp value, we can employ stoichiometry and algebra. The first step is to establish the stoichiometry of dissolution, from which we assign the solubility value, S:

For lead(II) hydroxide, the process looks like this:
\[\mathrm{Pb}(\mathrm{OH})_{2}(s) \leftrightharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{OH}^-(aq)\]
If the compound's solubility is S moles per liter, then \([\mathrm{Pb}^{2+}] = S\) and \([\mathrm{OH}^-] = 2S\). Therefore:
\[Ksp = S(2S)^2 = 4S^3\]
Solving this cubic equation for S gives us the molar concentration of \(\mathrm{Pb}^{2+}\) in the saturated solution. This approach applies to any sparingly soluble salt where we use its Ksp value to calculate the molar solubility, which is essential for understanding the limitations of a solute's solubility in a given solvent.

Exploring the Relationship Between pH and pOH

In aqueous solutions, the pH and pOH are measures of the acidity and basicity, respectively. They are inversely related and provide quick insights into the concentration of hydronium \(\mathrm{H}_3\mathrm{O}^+\) and hydroxide \(\mathrm{OH}^-\) ions in solution.

The pH scale commonly ranges from 0 to 14, with 7 as neutral. Acidic solutions have a pH less than 7, while basic solutions have a pH greater than 7. The pH and pOH are connected by the following relationship:
\[\mathrm{pH} + \mathrm{pOH} = 14\]
The relationship is derived from the ion product of water, \(K_w = [\mathrm{H}_3\mathrm{O}^+][\mathrm{OH}^-] = 1.0 \times 10^{-14} \) at room temperature. The pOH can be found by subtracting the pH from 14, from which we can then find the hydroxide ion concentration:
\[[\mathrm{OH}^-] = 10^{-\mathrm{pOH}}\]
As seen in the exercise, understanding this relationship allows us to calculate ion concentrations needed to determine solubility or precipitation in equilibrium reactions. Adequate knowledge of these calculations is vital for those studying aqueous equilibrium, acid-base titrations, and similar processes in chemistry.