Problem 47 - Chemistry and Chemical Reactivity

Question

You add 0.979 g of \(\mathrm{Pb}(\mathrm{OH})_{2}\) to \(1.00 \mathrm{L}\) of pure water at \(25^{\circ} \mathrm{C}\). The \(\mathrm{pH}\) is \(9.15 .\) Estimate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{Pb}(\mathrm{OH})_{2}.\)

Step-by-Step Solution

Answer

The estimated value of \( K_{sp} \) for \( \mathrm{Pb(OH)_2} \) is \( 1.40 \times 10^{-15} \).

Step 1: Write Dissolution Equation

When \( \mathrm{Pb(OH)}_2 \) dissolves in water, it dissociates into its ions: \( \mathrm{Pb(OH)_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2OH^- (aq)} \). We need to find the solubility product constant \( K_{sp} \).

Step 2: Determine Hydroxide Ion Concentration

Given \( \text{pH} = 9.15 \), we can find \( \text{pOH} \) using the relation \( \text{pOH} = 14 - \text{pH} = 4.85 \). The hydroxide ion concentration \( [OH^-] \) can be calculated as \( [OH^-] = 10^{-\text{pOH}} = 10^{-4.85} \approx 1.41 \times 10^{-5} \text{ M} \).

Step 3: Relate Ion Concentration to Solubility

From the dissolution equation, the concentration of \( \mathrm{Pb^{2+}} \) is \( s \) and \( [OH^-] = 2s \). Since \( [OH^-] = 1.41 \times 10^{-5} \text{ M} \), we find \( s = \frac{1.41 \times 10^{-5}}{2} = 0.705 \times 10^{-5} \text{ M} \approx 7.05 \times 10^{-6} \text{ M} \).

Step 4: Calculate \( K_{sp} \)

The \( K_{sp} \) expression for \( \mathrm{Pb(OH)_2} \) is written as \( K_{sp} = [\mathrm{Pb^{2+}}][OH^-]^2 \). Substitute the values: \( K_{sp} = (7.05 \times 10^{-6})(1.41 \times 10^{-5})^2 \). Calculate \( K_{sp} = (7.05 \times 10^{-6})(1.9881 \times 10^{-10}) \approx 1.40 \times 10^{-15} \).

Key Concepts

Dissolution Equation

The dissolution equation is a representation of how a solid compound breaks down into its ions in a solution. In our example, the dissolution of lead hydroxide, \( \mathrm{Pb(OH)_2} \), into water can be expressed as:\[ \mathrm{Pb(OH)_2(s) \rightleftharpoons Pb^{2+}(aq) + 2OH^- (aq)} \]This equation shows that one mole of solid \( \mathrm{Pb(OH)_2} \) dissociates completely to release one mole of \( \mathrm{Pb^{2+}} \) ions and two moles of \( \mathrm{OH^-} \) ions into the solution.

• The arrow \( \rightleftharpoons \) indicates a reversible process, signifying dynamic equilibrium between dissolving and precipitation.
• The equation helps define the relationship between the solid compound and its ions in solution, key for calculating the solubility product constant \( K_{sp} \).

Understanding this equation is vital as it illustrates the chemical equilibrium established during the dissolution of sparingly soluble salts like \( \mathrm{Pb(OH)_2} \).
This helps in predicting how the salt will behave in various conditions, which is critical in fields like chemistry and environmental science.

Hydroxide Ion Concentration

The hydroxide ion concentration \([\mathrm{OH^-}]\) in a solution can be determined by using the \( \text{pH} \) and \( \text{pOH} \) relationship.
As given in the exercise, the \( \text{pH} \) of the solution is 9.15. To find \([\mathrm{OH^-}]\), you must first calculate \( \text{pOH} \) using the equation:\[ \text{pOH} = 14 - \text{pH} \]Substituting the known \( \text{pH} \) value:\[ \text{pOH} = 14 - 9.15 = 4.85 \]
The concentration \([\mathrm{OH^-}]\) is then determined using the formula:\[ [\mathrm{OH^-}] = 10^{-\text{pOH}} = 10^{-4.85} \approx 1.41 \times 10^{-5} \text{ M} \]

• This value indicates the presence of \( \mathrm{OH^-} \) ions, crucial for calculating \( K_{sp} \) and understanding the basic nature of the solution.
• Knowing the \([\mathrm{OH^-}]\) helps describe the system's behavior in terms of chemical equilibrium and the solubility product.

This approach is vital to analyze how changes in pH can affect the solubility of compounds like \( \mathrm{Pb(OH)_2} \) in water.

Solubility

Solubility refers to the extent to which a substance (solute) dissolves in a solvent to form a homogeneous mixture at a given temperature. In this exercise, the focus is on the solubility of \( \mathrm{Pb(OH)_2} \) in water.
Applying the dissolution equation's stoichiometry, we know:\[ [\mathrm{OH^-}] = 2s \] Given \([\mathrm{OH^-}] = 1.41 \times 10^{-5} \text{ M} \), it follows that:\[ s = \frac{1.41 \times 10^{-5}}{2} \approx 7.05 \times 10^{-6} \text{ M} \]

• The symbol \( s \) represents the molarity of \( \mathrm{Pb^{2+}} \) ions in the saturated solution.
• This calculation implies that at equilibrium, a very small amount of \( \mathrm{Pb(OH)_2} \) is dissolved, indicating low solubility.

Understanding solubility is crucial for determining how effectively \( \mathrm{Pb(OH)_2} \) or similar compounds will dissolve in given conditions, impacting fields such as water treatment and waste management.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions of a dissolution process are equal, leading to no net change over time. For \( \mathrm{Pb(OH)_2} \), equilibrium is represented as:\[ \mathrm{Pb(OH)_2(s) \rightleftharpoons Pb^{2+}(aq) + 2OH^- (aq)} \]At equilibrium, the concentration values used to determine \( K_{sp} \) help illustrate this state.The expression for the solubility product constant \( K_{sp} \) is:\[ K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{OH^-}]^2 \] Substituting the determined concentrations:\[ K_{sp} = (7.05 \times 10^{-6})(1.41 \times 10^{-5})^2 \approx 1.40 \times 10^{-15} \]

• \( K_{sp} \) value characterizes the solubility equilibrium state.
• It predicts the compound's solubility under different conditions and helps identify precipitate formation.

This concept is fundamental in understanding solution dynamics, impacting many chemical processes and industries where equilibrium consideration is indispensable.