Problem 53 - Chemistry The Molecular Science

Question

The solubility of \(\mathrm{PbBr}_{2}\) is \(2.2 \times 10^{-2} \mathrm{~g}\) per \(100.0 \mathrm{~mL}\) at \(25^{\circ} \mathrm{C}\). Calculate the \(K_{\mathrm{sp}}\) of \(\mathrm{PbBr}_{2}\), assuming that the solute dissociates completely into \(\mathrm{Pb}^{2+}\) and \(\mathrm{Br}^{-}\) ions and that these ions do not react with water.

Step-by-Step Solution

Answer

The \( K_{\mathrm{sp}} \) of \( \mathrm{PbBr}_{2} \) is \( 8.6 \times 10^{-10} \).

Step 1: Determine Molar Mass of PbBr2

Calculate the molar mass of \(\mathrm{PbBr}_{2}\) using the atomic masses: \(\mathrm{Pb} = 207.2\, \text{g/mol}\) and \(\mathrm{Br} = 79.9\, \text{g/mol}\).\ The molar mass of \(\mathrm{PbBr}_{2} = 207.2 + 2\times 79.9 = 367.0\, \text{g/mol}\).

Step 2: Convert Solubility from Grams to Moles

The solubility of \(\mathrm{PbBr}_{2}\) is given as \(2.2 \times 10^{-2}\, \text{g/100 mL}\). Convert this to moles per liter (molarity):\ \[ \text{Solubility (mol/L)} = \frac{2.2 \times 10^{-2}\, \text{g}}{367.0\, \text{g/mol}} \times \frac{1000\, \text{mL/L}}{100\, \text{mL/L} } = 6.0 \times 10^{-4}\, \text{mol/L} \].

Step 3: Write Dissociation Equation

The dissociation of \(\mathrm{PbBr}_{2}\) in water can be expressed as: \\[ \mathrm{PbBr}_{2} (s) \rightleftharpoons \mathrm{Pb}^{2+} (aq) + 2\mathrm{Br}^{-} (aq) \] \ Each mole of \(\mathrm{PbBr}_{2}\) dissociates to form 1 mole of \(\mathrm{Pb}^{2+}\) and 2 moles of \(\mathrm{Br}^{-}\).

Step 4: Calculate Ion Concentrations

From the dissociation equation, the concentration of \(\mathrm{Pb}^{2+}\) equals the molarity of the \(\mathrm{PbBr}_{2}\) solution (\(6.0 \times 10^{-4} \ \text{mol/L}\)), and the concentration of \(\mathrm{Br}^{-}\) is twice that amount: \ \[ \text{[Pb}^{2+}\text{]} = 6.0 \times 10^{-4}\, \text{mol/L} \] \\[ \text{[Br}^{-}\text{]} = 2 \times 6.0 \times 10^{-4}\, \text{mol/L} = 1.2 \times 10^{-3}\, \text{mol/L} \].

Step 5: Calculate Ksp

The solubility product \(K_{\mathrm{sp}}\) is calculated using the equilibrium concentrations of the ions:\ \[ K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}][\mathrm{Br}^{-}]^2 \]. \ Substitute the concentrations: \ \[ K_{\mathrm{sp}} = (6.0 \times 10^{-4})(1.2 \times 10^{-3})^2 = 8.6 \times 10^{-10} \].

Key Concepts

Ionic Dissociation

Ionic dissociation is a fundamental concept in chemistry where an ionic compound breaks apart into its individual ions. Consider the compound \(\mathrm{PbBr}_{2}\), which when dissolved in water dissociates completely into its constituent ions. In this reaction, each mole of \(\mathrm{PbBr}_{2}\) dissociates into one mole of \(\mathrm{Pb}^{2+}\) ions and two moles of \(\mathrm{Br}^{-}\) ions. This process can be represented by the chemical equation:
\[\mathrm{PbBr}_{2} (s) \rightleftharpoons \mathrm{Pb}^{2+} (aq) + 2\mathrm{Br}^{-} (aq)\]
The assumption here is that the dissociation is complete and that these ions do not interact with the water molecules but rather remain freely separate as \(\mathrm{Pb}^{2+}\) and \(\mathrm{Br}^{-}\) ions. This is critical for accurate calculation of equilibrium concentrations and subsequently, the solubility product constant \(K_{\mathrm{sp}}\). Understanding how ionic compounds behave in solutions allows us to predict the solubility and precipitation reactions that are crucial in various chemical processes.

Molar Mass Calculation

Calculating molar mass is a crucial step in determining the solubility and equilibrium concentrations of a compound in solution. Here, we calculate the molar mass of \(\mathrm{PbBr}_{2}\) by using the atomic masses of its constituent elements:

• Lead (\(\mathrm{Pb}\)): 207.2 \(\mathrm{g/mol}\)
• Bromine (\(\mathrm{Br}\)): 79.9 \(\mathrm{g/mol}\)

To find the molar mass of \(\mathrm{PbBr}_{2}\), combine these atomic masses using the formula:
\[207.2 + 2 \times 79.9 = 367.0 \; \mathrm{g/mol}\]
This calculated molar mass is essential because it allows conversion of the given solubility from grams per milliliter to moles per liter (molarity). This is critical in subsequent steps when calculating the ion concentrations and ultimately, the solubility product constant \(K_{\mathrm{sp}}\). Proper molar mass calculation ensures the accuracy and reliability of these further calculations.

Equilibrium Concentrations

Once the molar mass is known, we can determine the equilibrium concentrations of ions in solution, which are necessary for calculating \(K_{\mathrm{sp}}\). Given that the solubility of \(\mathrm{PbBr}_{2}\) is \(2.2 \times 10^{-2}\ \text{g/100 mL}\), we convert it to molarity:
\[\text{Solubility (mol/L)} = \frac{2.2 \times 10^{-2}\, \text{g}}{367.0\, \text{g/mol}} \times \frac{1000\, \text{mL/L}}{100\, \text{mL/L} } = 6.0 \times 10^{-4}\, \text{mol/L}\]
From the dissociation equation, the molarities of the ions can be determined. The concentration of \(\mathrm{Pb}^{2+}\) equals the molarity of \(\mathrm{PbBr}_{2}\), while the concentration of \(\mathrm{Br}^{-}\) is twice that:

• \([\mathrm{Pb}^{2+}] = 6.0 \times 10^{-4}\, \text{mol/L}\)
• \([\mathrm{Br}^{-}] = 2 \times 6.0 \times 10^{-4}\, \text{mol/L} = 1.2 \times 10^{-3}\, \text{mol/L}\)

Accurate determination of equilibrium concentrations is vital for calculating the solubility product constant \(K_{\mathrm{sp}}\). Understand that these concentrations represent how ions are balanced in the solution at equilibrium, reflecting the maximum concentration before precipitation occurs.

Chemical Equilibrium

Chemical equilibrium is a dynamic state where the rate of the forward reaction equals the rate of the reverse reaction. For the dissolution of \(\mathrm{PbBr}_{2}\), equilibrium is reached when the solute's rate of dissolution equals the rate of crystallization. The equilibrium concentrations of the ions determined before can now be used to find the solubility product constant \(K_{\mathrm{sp}}\), defined as:
\[ K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}][\mathrm{Br}^{-}]^2 \]
Plug in the ion concentrations:
\[ K_{\mathrm{sp}} = (6.0 \times 10^{-4})(1.2 \times 10^{-3})^2 = 8.6 \times 10^{-10} \]
This value represents the product of the ions' concentrations at equilibrium, a constant for a given temperature. The \(K_{\mathrm{sp}}\) helps predict whether a precipitation reaction will occur when two solutions are mixed. Understanding chemical equilibrium is essential as it explains how reactions proceed and reach a state of balance, providing insight into the behavior of ions in solutions.