Problem 48 - 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 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 \(K_{sp}\) for \(\mathrm{Ca(OH)}_2\) is approximately \(5.44 \times 10^{-5}\).

Step 1: Determine pOH from pH

Given the pH of the solution is 12.68, we can find the pOH using the relation:\[\text{pOH} = 14 - \text{pH} = 14 - 12.68 = 1.32.\]

Step 2: Calculate Hydroxide Ion Concentration

The pOH gives us the concentration of hydroxide ions. We use the formula: \[[\text{OH}^-] = 10^{-\text{pOH}} = 10^{-1.32} \approx 0.0477 \, \text{M}.\]

Step 3: Write the Dissociation Equation

The dissociation of \(\text{Ca(OH)}_2\) in water is represented by:\[\text{Ca(OH)}_2 (s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{OH}^-(aq).\]

Step 4: Determine Calcium Ion Concentration

From the stoichiometry of the dissociation equation, for every mole of \(\text{Ca}^{2+}\) ions, there are 2 moles of \(\text{OH}^-\) ions:\[[\text{Ca}^{2+}] = \frac{[\text{OH}^-]}{2} = \frac{0.0477}{2} = 0.02385 \, \text{M}.\]

Step 5: Calculate the Solubility Product Constant \(K_{sp}\)

The solubility product constant \(K_{sp}\) for \(\text{Ca(OH)}_2\) is given by:\[K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 = (0.02385)(0.0477)^2.\]Calculating this gives:\[K_{sp} \approx 5.44 \times 10^{-5}.\]

Key Concepts

pH and pOH relationship

The concepts of pH and pOH are vital when discussing the acidity and basicity of solutions. In any aqueous solution, at 25°C, the sum of the pH and pOH is always 14. This relationship is derived from the ion-product constant of water (kw=1×10^−14). The pH measures the hydrogen ion concentration, while the pOH measures the hydroxide ion concentration.

• If you know the pH, you can easily find the pOH using the formula: \[ ext{pOH} = 14 - ext{pH} \]


• Similarly, if you know the pOH, you can find the pH by rearranging to: \[ ext{pH} = 14 - ext{pOH} \]

The understanding of this relationship allows us to convert between these scales easily, as demonstrated when determining the pOH of the calcium hydroxide solution given in the exercise.

Hydroxide ion concentration calculation

Once you have the pOH of a solution, finding the hydroxide ion concentration ([ ext{OH}^-]) becomes straightforward. The concentration can be found using the following calculation: \[[\text{OH}^-] = 10^{-\text{pOH}}\]This formula comes from the definition of pOH, which measures the power of ten corresponding to the hydroxide concentration in moles per liter (M).
For example, with a solution pOH of 1.32, substituting into the equation gives: \[[\text{OH}^-] = 10^{-1.32} \approx 0.0477 \text { M}\]
This means the solution has a hydroxide ion concentration of approximately 0.0477 M. Understanding this process is crucial as it forms a foundation for many subsequent calculations in chemistry.

Dissociation equation of calcium hydroxide

The dissociation of calcium hydroxide (Ca(OH)_2) in aqueous solutions is a critical concept, especially when dealing with solubility and equilibrium. When calcium hydroxide dissolves in water, it separates into calcium ions and hydroxide ions.
This dissociation is expressed by the equation:\[Ca(OH)_2 (s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq) \]The double arrow indicates equilibrium between the solid and its dissolved ions.
This stoichiometry tells us that one mole of calcium hydroxide will produce one mole of calcium ions and two moles of hydroxide ions. Thus, understanding this equation is crucial for predicting concentrations of ions in solution.

Calcium ion concentration calculation

After understanding the dissociation of calcium hydroxide, calculating the calcium ion concentration ([ ext{Ca}^{2+}]) becomes possible using stoichiometry principles. Given that each formula unit of (Ca(OH)_2) produces one calcium ion and two hydroxide ions upon dissolving:

• The concentration of calcium ions is half the concentration of hydroxide ions.

For instance:
From the earlier calculated hydroxide concentration ([ ext{OH}^-] = 0.0477 \text{ M} )\, the calcium ion concentration would be:\[[\text{Ca}^{2+}] = \frac{[\text{OH}^-]}{2} = \frac{0.0477}{2} \approx 0.02385 \text{ M}\]Comprehending this calculation is essential in determining the equilibrium concentration of ions in solution and plays a central role in calculating the solubility product constant.