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

Step 1: Write the Dissolution Equation

Lead hydroxide, \( \mathrm{Pb(OH)_2} \), dissolves as follows: \[ \mathrm{Pb(OH)_2(s) \rightleftharpoons Pb^{2+}(aq) + 2OH^-(aq)} \] We need this equation to understand the relationship between the solid and ions.

Step 2: Solve for Hydroxide Ion Concentration

Given that \( \mathrm{pH} = 9.15 \), calculate \( \mathrm{pOH} \) using the relationship: \[ \mathrm{pOH} = 14.00 - pH = 14.00 - 9.15 = 4.85 \] Next, determine the hydroxide ion concentration by using: \[ [\mathrm{OH^-}] = 10^{-\mathrm{pOH}} = 10^{-4.85} \approx 1.41 \times 10^{-5} \mathrm{M} \]

Step 3: Determine Lead Ion Concentration

From the dissolution equation, every mole of \( \mathrm{Pb(OH)_2} \) that dissolves produces one mole of \( \mathrm{Pb^{2+}} \) and two moles of \( \mathrm{OH^-} \). Thus, \[ [\mathrm{Pb^{2+}}] = \frac{1}{2}[\mathrm{OH^-}] = \frac{1}{2}(1.41 \times 10^{-5}) = 7.05 \times 10^{-6} \mathrm{M} \]

Step 4: Calculate the Solubility Product

The solubility product \( K_{\mathrm{sp}} \) for \( \mathrm{Pb(OH)_2} \) can be expressed as: \[ K_{\mathrm{sp}} = [\mathrm{Pb^{2+}}][\mathrm{OH^-}]^2 \] Substitute the ion concentrations: \[ K_{\mathrm{sp}} = (7.05 \times 10^{-6})(1.41 \times 10^{-5})^2 \approx 1.40 \times 10^{-15} \]

Step 5: Verify Assumptions

To verify, we ensure that the ionic product matches the solid present and that no deviations are expected in a saturated solution. Given the calculated \( K_{\mathrm{sp}} \), it shows equilibrium is established between dissolved ions and the remaining solid.

Key Concepts

pH

The term "pH" stands for the potential of hydrogen. It is a measure of how acidic or basic a solution is. The pH scale ranges from 0 to 14, where a pH of 7 is considered neutral, values below 7 indicate acidity, and values above 7 indicate alkalinity. In this exercise, the solution's pH of 9.15 indicates that the solution is slightly basic.
Understanding the pH value allows us to infer the concentration of hydrogen and hydroxide ions in the solution. There is a relationship between pH and \(\text{pOH}\) (potential of hydroxide), given by the equation:

• \(\text{pH} + \text{pOH} = 14 \)

In this situation, we used the pH value to find \(\text{pOH}\), which then led us to the hydroxide ion concentration in the solution.
By calculating \(\text{pOH}\), you can determine the balance of ions in the water, which further provides insights into the chemical equilibrium in solution.

solubility equilibrium

Solubility equilibrium refers to the balance between a solute and its components in solution. For a solid ionic compound like lead hydroxide \(\mathrm{Pb(OH)_2}\), the dissolution forms lead ions \(\mathrm{Pb^{2+}}\) and hydroxide ions \(\mathrm{OH^-}\) in solution. This equilibrium is dynamic, meaning the dissolution and precipitation occur at equal rates, maintaining constant ion concentrations.
In our exercise, solubility equilibrium plays a crucial role as we determine how much of the \(\mathrm{Pb(OH)_2}\) dissolves in water to reach this state. By writing the dissolution equation:

• \(\mathrm{Pb(OH)_2(s) \rightleftharpoons Pb^{2+}(aq) + 2OH^-(aq)}\)

we understand the stoichiometry of the reaction. Every mole of \(\mathrm{Pb(OH)_2}\) produces one mole of lead ions and two moles of hydroxide ions, which directly relates to the balance achieved at solubility equilibrium.

chemical equilibrium

Chemical equilibrium in this context shows the state where the rate of the forward reaction (dissolution of \(\mathrm{Pb(OH)_2}\)) equals the rate of the reverse reaction (precipitation). This balance means the concentrations of ions don't change over time once equilibrium is reached.
The expression for the equilibrium constant, specifically the solubility product \((K_{\mathrm{sp}})\), is necessary to describe this balance mathematically. For \(\mathrm{Pb(OH)_2}\), the equilibrium constant is defined by:

• \(K_{\mathrm{sp}} = [\mathrm{Pb^{2+}}][\mathrm{OH^-}]^2\)

This constant provides crucial information about how much solute will dissolve in the solution, allowing for predictions about solubility under different conditions. It helps in understanding how chemical equilibrium maintains the proportions of dissolved ions within the system.

ionic product

The ionic product signifies the product of the concentrations of the ions in a saturated solution, each raised to the power of their coefficients in the balanced equation. For lead hydroxide, \(\mathrm{Pb(OH)_2}\), the ionic product is an essential tool in assessing the degree of solubility and stability of the dissolved components.
In our calculation, the ionic product is given by:

• \([\mathrm{Pb^{2+}}][\mathrm{OH^-}]^2\)

By substituting the concentrations of ions into this expression, we evaluated \(K_{\mathrm{sp}}\) as an indicator of the maximum solubility the substance could achieve in water. The calculated value of \(K_{\mathrm{sp}}\) at a given temperature reflects the extent of dissolution before reaching the equilibrium.Understanding the ionic product aids in visualizing how these ions interact in solutions, and what conditions will lead to precipitation or further dissolution.