Problem 44 - Chemistry and Chemical Reactivity

Question

You place \(1.234 \mathrm{g}\) of solid \(\mathrm{Ca}(\mathrm{OH})_{2}\) in \(1.00 \mathrm{L}\) of pure water at \(25^{\circ} \mathrm{C} .\) The \(\mathrm{pH}\) of the solution is found to be \(12.68 .\) Estimate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{Ca}(\mathrm{OH})_{2}.\)

Step-by-Step Solution

Answer

The estimated \( K_{sp} \) of \( \text{Ca(OH)}_2 \) is approximately \( 5.5 \times 10^{-5} \).

Step 1: Understand the Dissolution Reaction

First, we need to write out the dissolution equation for calcium hydroxide in water. The chemical equation is:\[ \text{Ca(OH)}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{OH}^-(aq) \]This shows that one mole of \( \text{Ca(OH)}_2 \) dissolves to form one mole of \( \text{Ca}^{2+} \) ions and two moles of \( \text{OH}^- \) ions.

Step 2: Use pH to Find Hydroxide Concentration

The solution pH is given as 12.68. Use the pH to find the pOH.\[ \text{pOH} = 14.00 - \text{pH} = 14.00 - 12.68 = 1.32 \]Now, calculate the hydroxide ion concentration:\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-1.32} \approx 0.0479 \text{ M} \]

Step 3: Determine Molarity of Calcium Ions

In the dissolution equation, for every mole of \( \text{Ca(OH)}_2 \) that dissolves, 2 moles of \( \text{OH}^- \) are formed. So, the concentration of calcium ions, \( [\text{Ca}^{2+}] \), is half the concentration of \( \text{OH}^- \):\[ [\text{Ca}^{2+}] = \frac{1}{2} \times 0.0479 \approx 0.02395 \] Thus, the molarity of \( \text{Ca}^{2+} \) ions is approximately 0.02395 M.

Step 4: Calculate the Ksp

The expression for the solubility product constant, \(K_{sp}\), is given by:\[ K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 \]Substitute the values obtained:\[ K_{sp} = (0.02395)(0.0479)^2 \approx 5.5 \times 10^{-5} \]

Step 5: Ensure Units and Assumptions Are Clear

We assume that all \( \text{Ca(OH)}_2 \) initially dissolves and the volume change of the solution due to the solute dissolution is negligible. Calculations are based purely on stoichiometry and the dissolution of \( \text{Ca(OH)}_2 \) up to its saturation point.

Key Concepts

Dissolution Reaction

When calcium hydroxide (\( \text{Ca(OH)}_2 \)) is added to water, it undergoes a dissolution reaction. This means that the solid (\( \text{Ca(OH)}_2 \)) breaks down into ions in the solution:

• One mole of calcium ions (\( \text{Ca}^{2+} \))
• Two moles of hydroxide ions (\( \text{OH}^- \))

The balanced chemical equation for this process is:\[\text{Ca(OH)}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2 \text{OH}^-(aq)\]This reaction is important for understanding the solubility and behavior of \( \text{Ca(OH)}_2 \) in water. Knowing the stoichiometry (the ratio of reacting particles) helps us calculate the concentration of ions formed in the solution, essential for further steps in determining solubility dynamics like the solubility product constant.

pH Calculation

The pH scale measures the acidity or basicity of a solution. In this case, the solution has a pH of 12.68, indicating it's basic. To find the hydroxide ion concentration, we refer to the relationship between pH and pOH:

• The pOH is calculated as: \( \text{pOH} = 14.00 - \text{pH} = 14.00 - 12.68 = 1.32\)
• This relationship helps us understand the balance between hydrogen ions (\( \text{H}^+ \)) and hydroxide ions (\( \text{OH}^- \)) in the solution.

By calculating the pOH, we can find the concentration of hydroxide ions using the formula:\[[\text{OH}^-] = 10^{-\text{pOH}}\]This approach is crucial for linking the solution's pH to its chemical components, paving the way for further analysis.

Hydroxide Ion Concentration

Determining the hydroxide ion concentration (\( [\text{OH}^-] \))is a critical step. From the pOH calculated earlier, use the expression:\[ [\text{OH}^-] = 10^{-1.32} \approx 0.0479 \text{ M}\]This calculation reveals the molarity of hydroxide ions in the solution.

• The concentration obtained is directly influenced by the dissolution of \( \text{Ca(OH)}_2 \)
• Understanding this concentration allows us to assess how much of the solid has dissolved and the resulting ion balance.

This piece of information is instrumental in discovering other concentration values in the solution and ultimately solving for the solubility product constant, \( K_{sp} \) .

Molarity Determination

Once we have the concentration of hydroxide ions, we can figure out the molarity of calcium ions (\( [\text{Ca}^{2+}] \)). According to the dissolution reaction equation:

• For every mole of \( \text{Ca(OH)}_2 \)dissolved, one mole of calcium ions is formed alongside two moles of hydroxide ions.
• Therefore, \( [\text{Ca}^{2+}] = \frac{1}{2} \times [\text{OH}^-] \approx \frac{1}{2} \times 0.0479 \approx 0.02395 \text{ M}\)

This molarity is significant for calculating the solubility product constant, \( K_{sp} \) .The conversion from [(OH^-)] to[(Ca^{2+})] using stoichiometry ensures the consistency and accuracy of the results, reflecting the dissolution's impact on ion concentrations.