Problem 122 - Chemistry The Molecular Science

Question

An experiment found that \(0.0050 \mathrm{~mol} \mathrm{Ca}(\mathrm{OH})_{2}\) dissolved to form \(0.100 \mathrm{~L}\) of a saturated aqueous solution. (a) Calculate the pH of the solution. (b) Calculate the \(K_{\mathrm{sp}}\) of \(\mathrm{Ca}(\mathrm{OH})_{2} .\) Explain why the calculated \(K_{\mathrm{sp}}\) differs from the value of \(5.5 \times 10^{-6}\) given in Appendix \(\mathrm{H}\).

Step-by-Step Solution

Answer

(a) pH = 13.00 (b) Calculated \(K_{sp}\) is \(5 \times 10^{-4}\), differing due to experimental conditions.

Step 1: Determine concentration of hydroxide ions

First, calculate the concentration of hydroxide ions in the solution. The reaction \( ext{Ca(OH)}_2 ightarrow ext{Ca}^{2+} + 2 ext{OH}^-\) implies each mole of \( ext{Ca(OH)}_2\) provides 2 moles of \( ext{OH}^-\). Thus, the concentration of \( ext{OH}^-\) is calculated as follows:\[\text{Concentration of OH}^- = 2 \times \frac{0.0050 \text{ mol}}{0.100 \text{ L}} = 0.100 \text{ M}\]

Step 2: Calculate the pOH

Use the concentration of \(\text{OH}^-\) ions to find the \(pOH\):\[pOH = -\log [\text{OH}^-] = -\log(0.100) = 1.00\]

Step 3: Calculate the pH of the solution

Convert the \(pOH\) to \(pH\) using the relationship:\[pH + pOH = 14\]So,\[pH = 14 - 1.00 = 13.00\]

Step 4: Calculate molar solubility and determine Ksp

Since 0.0050 mol of \(\text{Ca(OH)}_2\) dissolves in 0.100 L, the molar solubility \(s\) is 0.050 M. The solubility product \(K_{sp}\) of \(\text{Ca(OH)}_2\) is the product of the concentrations of the ions each raised to the power of their respective coefficients:\[K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 = (s)(2s)^2 = 4s^3\]Substitute \(s = 0.050\):\[K_{sp} = 4(0.050)^3 = 5 \times 10^{-4}\]

Step 5: Compare calculated Ksp with given value

The calculated \(K_{sp}\) of \(5 \times 10^{-4}\) is higher than the given value of \(5.5 \times 10^{-6}\) because the conditions during the experiment might differ from standard conditions or assumptions in the Appendix, potentially due to temperature, ion interactions, impurities, or measurement error.

Key Concepts

pH Calculation

Understanding pH calculation involves using the concentration of hydroxide ions. When a solution involves calcium hydroxide (\(\text{Ca(OH)}_2\)), which dissociates into calcium ions (\(\text{Ca}^{2+}\)) and hydroxide ions (\(\text{OH}^-\)), the dissociation reaction is: \[ \text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^- \] Each mole of \(\text{Ca(OH)}_2\) produces two moles of \(\text{OH}^-\) ions. Start by calculating the \(\text{OH}^-\) ion concentration:

• Given: 0.0050 mol \(\text{Ca(OH)}_2\) in 0.100 L solution, giving \(0.100 \, \text{M} \) of \(\text{OH}^-\) ions.

Next, calculate the \(\text{pOH}\) value using the formula: \[ \text{pOH} = -\log[\text{OH}^-] = 1.00 \] To find the \(\text{pH}\), use the universal relation \(\text{pH} + \text{pOH} = 14\). Plugging in the values gives:

• \(\text{pH} = 14 - 1.00 = 13.00\)

Ksp Determination

The solubility product constant, \(K_{sp}\), indicates how much of a compound can dissolve in water. For calcium hydroxide, \(K_{sp}\) is determined through the concentrations of dissolved ions. The formula for this equilibrium is: \[ K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 \] For calculation:

• Molar solubility \(s = 0.050 \, \text{M}\) for \(\text{Ca(OH)}_2\).
• [\text{Ca}^{2+}] = s
• [\text{OH}^-] = 2s

Solving further: \[ K_{sp} = (s)(2s)^2 = 4s^3 \] Substitute \(s = 0.050\): \[ K_{sp} = 4(0.050)^3 = 5 \times 10^{-4} \] This explains why the calculated \(K_{sp}\) might differ from a listed value due to variations in experimental or environmental conditions.

Molar Solubility

Molar solubility refers to the number of moles of a solute that can dissolve in one liter of solution before the solution becomes saturated. Here, it helps find how much \(\text{Ca(OH)}_2\) can dissolve:

• The problem states 0.0050 mol of \(\text{Ca(OH)}_2\) in 0.100 L, leading to a molar solubility \(s = 0.050 \, \text{M} \).

With molar solubility, one can derive the concentrations of the ions in a saturated solution to calculate \(K_{sp}\).

• For \(\text{Ca(OH)}_2\), \(s\) gives the concentration of \(\text{Ca}^{2+}\) ions and \(2 \times s\) gives the concentration of \(\text{OH}^-\) ions.
• It helps in understanding how solubility and equilibrium concepts are intertwined.

This plays a critical role in predicting how substances dissolve, especially in varying conditions like temperature and pressure.