Problem 43 - Chemistry and Chemical Reactivity

Question

You add \(0.979 \mathrm{g}\) 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 \( 1.40 \times 10^{-16} \).

Step 1: Write the Dissociation Equation

First, consider the dissociation of lead(II) hydroxide, \( \mathrm{Pb(OH)_2} \), in water. The reaction is as follows: \[ \mathrm{Pb(OH)_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2OH^{-} (aq)} \]

Step 2: Calculate Hydroxide Ion Concentration

Given the pH is 9.15, first calculate the pOH: \[ \text{pOH} = 14 - \text{pH} = 14 - 9.15 = 4.85 \]. Then calculate the hydroxide ion concentration \( [\mathrm{OH}^-] \) using \( [\mathrm{OH}^-] = 10^{-\text{pOH}} \): \[ [\mathrm{OH}^-] = 10^{-4.85} = 1.41 \times 10^{-5} \ \mathrm{M} \].

Step 3: Set Up the Expression for \( K_{sp} \)

The equilibrium expression for the solubility product, \( K_{sp} \), is given by: \[ K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{OH}^-]^2 \]. Let \( s \) represent the solubility of \( \mathrm{Pb(OH)_2} \) in mol/L, which equals the concentration \([\mathrm{Pb^{2+}}] \), and \( 2s \) for \([\mathrm{OH}^{-}]\) for one mole of \( \mathrm{Pb^{2+}} \) giving two moles of \([\mathrm{OH}^{-}]\).

Step 4: Solve for \( s \) Using Hydroxide Concentration

Since \( [\mathrm{OH}^-] = 2s \), we know \( 1.41 \times 10^{-5} \ \mathrm{M} \) = \( 2s \), so \[ s = \frac{1.41 \times 10^{-5}}{2} = 7.05 \times 10^{-6} \ \mathrm{M} \]. This is the solubility of \( \mathrm{Pb(OH)_2} \) in mol/L.

Step 5: Calculate \( K_{sp} \) Using Solubility

Substitute the values back into the \( K_{sp} \) expression: \[ K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{OH}^-]^2 = s \times (2s)^2 = s \times 4s^2 = 4s^3 \]. Simplifying with \( s = 7.05 \times 10^{-6} \), we get: \[ K_{sp} = 4 \times (7.05 \times 10^{-6})^3 = 1.40 \times 10^{-16} \].

Key Concepts

Dissociation Equation

Understanding the dissociation equation is key in chemistry, especially when dealing with solubility and equilibrium concepts. When a substance like lead(II) hydroxide dissolves in water, it dissociates into its constituent ions. This process is represented by a dissociation equation. For lead(II) hydroxide, the dissociation equation is given as:\[ \mathrm{Pb(OH)_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2OH^{-} (aq)} \]This equation tells us that one mole of solid lead(II) hydroxide, \( \mathrm{Pb(OH)_2} \), splits into one mole of \( \mathrm{Pb^{2+}} \) ions and two moles of \( \mathrm{OH^{-}} \) ions when it dissolves in water.

• \( \mathrm{Pb(OH)_2} \) indicates the solid reactant.
• \( \mathrm{Pb^{2+}} \) is the cation formed in solution.
• \( \mathrm{2OH^{-}} \) represents the anions formed.

This equilibrium nature of this reaction demonstrates that the dissociation is not complete and can shift based on conditions like concentration and temperature.

Hydroxide Ion Concentration

To determine the solubility product constant, understanding the concentration of hydroxide ions, \([\mathrm{OH}^-]\), is crucial. The hydroxide ion concentration can be calculated using the pH of the solution. Given that the pH of the solution is 9.15:

• First, calculate the pOH, since \( \text{pOH} = 14 - \text{pH} \).
• Therefore, \( \text{pOH} = 14 - 9.15 = 4.85 \).
• Now, use the formula to find the concentration: \( [\mathrm{OH}^-] = 10^{-\text{pOH}} \).
• Substituting the pOH value gives \( [\mathrm{OH}^-] = 10^{-4.85} \approx 1.41 \times 10^{-5} \; \mathrm{M} \).

This value is crucial in further calculations, as it directly affects the solubility product, \( K_{sp} \). The concentration of hydroxide ions is doubled due to the stoichiometry of the dissociation equation, adding an additional layer of complexity to the calculations.

Equilibrium Concentration

In equilibrium calculations, one aims to find the concentration of ions at the point where the reaction rate of dissociation equals the rate of formation. This is vital for computing the solubility product constant, \( K_{sp} \). In this scenario, the concentration of each ion at equilibrium is linked to the solubility of the original compound.Given that the hydroxide ion concentration is \( 1.41 \times 10^{-5} \; \mathrm{M} \) and recalling from the dissociation equation that one mole of \( \mathrm{Pb(OH)_2} \) gives two moles of \( \mathrm{OH}^- \):

• The relationship \( [\mathrm{OH}^-] = 2s \) applies, where \( s \) is the solubility in \( \mathrm{mol/L} \).
• Solve for \( s \): \( s = \frac{1.41 \times 10^{-5}}{2} = 7.05 \times 10^{-6} \; \mathrm{M} \).
• This \( s \) value is equivalent to the equilibrium concentration of \( \mathrm{Pb^{2+}} \) ions.

Now, insert these values into the \( K_{sp} \) expression to find its value: \( K_{sp} = s \times (2s)^2 = 4s^3 \). Solving gives \( K_{sp} = 1.40 \times 10^{-16} \). Thus, knowing the equilibrium concentrations enables a deeper understanding of the solubility dynamics.