Question

Lead(II) chloride, \(\mathrm{PbCl}_{2}(s),\) dissolves in water to the extent of approximately \(3.6 \times 10^{-2} \mathrm{M}\) at \(20^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{PbCl}_{2}(s),\) and calculate its solubility in grams per liter.

Short answer

The \(K_{sp}\) value for PbCl\(_{2}\) is 1.87 x 10\(^{-4}\) and its solubility at \(20^{\circ}C\) is approximately 10 g/L.

Step-by-step solution

Write the balanced chemical equation for the dissolution of PbCl\(_{2}\)

The balanced chemical equation for the dissolution of lead(II) chloride in water is: \[PbCl_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)\]

Set up the reaction table for the dissolution of PbCl\(_{2}\)

Now let's set up a table showing the concentrations of each ion involved in the reaction: Initial concentrations: [Pb\(^{2+}\)] = 0 M [Cl\(^-\)] = 0 M Change in concentrations: Since the solubility of PbCl\(_{2}\) provided is \(3.6 \times 10^{-2} \mathrm{M}\), we can say that for every mole of PbCl\(_{2}\) dissolved there will be +1 mole of Pb\(^{2+}\) and +2 moles of Cl\(^-\). Thus, \(\Delta\)[Pb\(^{2+}\)] = +S and \(\Delta\)[Cl\(^-\)] = +2S. Final concentrations: [Pb\(^{2+}\)] = S = \(3.6 \times 10^{-2} \mathrm{M}\) [Cl\(^-\)] = 2S = \(2(3.6 \times 10^{-2} \mathrm{M})\) = \(7.2 \times 10^{-2}\mathrm{M}\)

Calculate the \(K_{sp}\) value for PbCl\(_{2}\)

We can determine the \(K_{sp}\) value using the following formula: \[K_{sp} = [Pb^{2+}] [Cl^-]^2\] Plugging in the concentrations we found in step 2: \[K_{sp} = (3.6 \times 10^{-2} \mathrm{M}) (7.2 \times 10^{-2} \mathrm{M})^2\] \[K_{sp} = 1.87 \times 10^{-4}\]

Convert the solubility from moles per liter to grams per liter

First, we will calculate the molar mass of PbCl\(_{2}\) using the atomic weights of lead (Pb) and chlorine (Cl): Molar mass of PbCl\(_{2}\) = (207.2 g/mol for Pb) + 2(35.45 g/mol for Cl) = 278.1 g/mol Now we will convert the solubility in moles per liter to grams per liter: Solubility (g/L) = Moles per liter * Molar mass Solubility (g/L) = (3.6 × 10⁻² M) * (278.1 g/mol) Solubility (g/L) ≈ 10 g/L Thus, the solubility in grams per liter for PbCl\(_{2}\) at \(20^{\circ}C\) is approximately 10 g/L.

Key concepts

Lead(II) Chloride Solubility

Understanding the solubility of lead(II) chloride, \(\mathrm{PbCl}_{2}\), is crucial to grasping the concept of how substances dissolve in water. This compound is only slightly soluble, meaning it dissolves to a limited extent. At \(20^{\circ} \mathrm{C}\), the solubility is expressed as \(3.6 \times 10^{-2}\ \mathrm{M}\), which tells us how many moles of \(\mathrm{PbCl}_{2}\) can dissolve in a liter of water to reach saturation. The solubility is determined under specific conditions, with temperature being a major factor. In practical terms:

• For every mole of \(\mathrm{PbCl}_{2}\) that dissolves in water, 1 mole of \(\mathrm{Pb^{2+}}\) ions and 2 moles of \(\mathrm{Cl^-}\) ions are formed.
• This ratio comes from its balanced chemical equation during dissolution:

\[\mathrm{PbCl}_{2}(s) \rightleftharpoons \mathrm{Pb^{2+}}(aq) + 2\mathrm{Cl^-}(aq)\]Understanding this concept helps students calculate the amounts of ions in solution and relate this to the idea of saturation. It connects the physical process of dissolving with the chemical descriptions of ion concentrations.

Chemical Equilibrium

Chemical equilibrium plays a pivotal role when studying the dissolution of salts like lead(II) chloride in water. When \(\mathrm{PbCl}_{2}\) dissolves, it doesn't just disappear into its ions and never look back. Instead, it establishes a dynamic equilibrium between the solid \(\mathrm{PbCl}_{2}\) and the ions in solution. At equilibrium:

• The rate of the solid dissolving into \(\mathrm{Pb^{2+}}\) and \(\mathrm{Cl^-}\) ions matches the rate of these ions precipitating back into solid \(\mathrm{PbCl}_{2}\).
• This dynamic balance implies that the ion concentrations will remain consistent over time so long as conditions remain unchanged.

The solubility product constant, \(K_{sp}\), is a measure of this equilibrium condition. It provides a quantitative way to express how much \(\mathrm{PbCl}_{2}\) can dissolve in water before reaching saturation. The values of \(K_{sp}\) are derived from the molar concentrations of the ions at equilibrium: \[K_{sp} = [\mathrm{Pb^{2+}}] [\mathrm{Cl^-}]^2\]High \(K_{sp}\) values indicate higher solubility, whereas low values point to limited solubility. Learning about chemical equilibrium concepts is essential for comprehending reactions and solutions in chemistry.

Solubility Conversion

Converting solubility from moles per liter to grams per liter is crucial for practical applications, such as preparing solutions in laboratories. For lead(II) chloride, knowing its solubility as \(3.6 \times 10^{-2} \ \mathrm{M}\) provides a foundational step. However, many applications require the solubility expressed in terms of mass per volume. The conversions involve:

• Calculating the molar mass of \(\mathrm{PbCl}_{2}\), which is fundamental to converting between these units.
• The molar mass is derived by adding the atomic masses of its constituent elements:

\[\text{Molar mass} = \text{(Pb)} + 2 \times \text{(Cl)} = 207.2 \, \text{g/mol} + 2 \times 35.45 \, \text{g/mol} = 278.1 \, \text{g/mol}\]Once we have the molar mass, multiply it by the molarity to find solubility in grams per liter:\[\text{Solubility} (\text{g/L}) = \text{Molarity} \times \text{Molar mass} = 3.6 \times 10^{-2} \, \mathrm{M} \times 278.1 \, \mathrm{g/mol} \approx 10 \, \mathrm{g/L}\]This approach translates chemical information into practical, usable data.

Molarity Calculations

Molarity is a key concept in chemistry that quantifies the concentration of a solution by indicating the number of moles of solute per liter of solution. For the solubility of \(\mathrm{PbCl}_{2}\), it means understanding that \(3.6 \times 10^{-2} \ \mathrm{M}\) translates directly to how many moles are dissolved per liter of water. Calculating molarity is straightforward:

• Moles of solute are calculated or measured directly in the experiment.
• It's crucial to keep the volume of the solution in liters to ensure the units match the definition of molarity.

Understanding molarity helps in various chemical calculations, including:

• Predicting the outcomes of reactions by knowing the concentration of reactants.
• Determining the precise amount of reactants and products in a chemical reaction.
• Molarity is used to calculate the \(K_{sp}\) in equilibrium expressions, emphasizing its practical importance in both academic and real-world chemistry.

By grasping molarity, students gain a solid foundation in quantitative analysis and are better equipped for advanced chemistry topics.