Problem 42 - Chemistry and Chemical Reactivity

Question

Calcium hydroxide, \(\mathrm{Ca}(\mathrm{OH})_{2},\) dissolves in water to the extent of \(1.3 \mathrm{g}\) per liter. What is the value of \(K_{\mathrm{sp}}\) for \(\mathrm{Ca}(\mathrm{OH})_{2} ?\) $$\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s}) \rightleftarrows \mathrm{Ca}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})$$

Step-by-Step Solution

Answer

The \( K_{sp} \) for \( \mathrm{Ca(OH)}_2 \) is \( 2.1 \times 10^{-5} \).

Step 1: Convert grams to moles

Determine the molar mass of \( \mathrm{Ca(OH)}_2 \). It consists of 1 calcium atom, 2 oxygen atoms, and 2 hydrogen atoms. The molar mass is: \(40.08 + 2(16.00) + 2(1.01) = 74.10 \text{ g/mol} \). Convert the solubility from grams per liter to moles per liter: \( 1.3 \text{ g/L} \times \frac{1 \text{ mol}}{74.10 \text{ g}} = 0.0175 \text{ mol/L} \). This is the molar solubility of \( \mathrm{Ca(OH)}_2 \).

Step 2: Determine ion concentrations at equilibrium

For every mole of \( \mathrm{Ca(OH)}_2 \) that dissolves, 1 mole of \( \mathrm{Ca}^{2+} \) ions and 2 moles of \( \mathrm{OH}^- \) ions are produced. The concentration of \( \mathrm{Ca}^{2+} \) ions is equal to the molar solubility: \( 0.0175 \text{ mol/L} \). The concentration of \( \mathrm{OH}^- \) ions is twice the molar solubility: \( 2 \times 0.0175 \text{ mol/L} = 0.035 \text{ mol/L} \).

Step 3: Write the expression for the solubility product constant, \( K_{sp} \)

The solubility product constant expression for \( \mathrm{Ca(OH)}_2 \) is given by: \( K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{OH}^-]^2 \).

Step 4: Substitute the ion concentrations into the \( K_{sp} \) expression

Substitute the determined concentrations into the \( K_{sp} \) expression from the earlier step: \( K_{sp} = (0.0175)(0.035)^2 \).

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

Calculate the product: \( K_{sp} = 0.0175 \times 0.035^2 = 2.1 \times 10^{-5} \).

Step 6: Provide the final value for \( K_{sp} \)

The solubility product constant, \( K_{sp} \), for \( \mathrm{Ca(OH)}_2 \) is \( 2.1 \times 10^{-5} \).

Key Concepts

Molar Solubility

Molar solubility refers to the number of moles of a solute that can dissolve in one liter of solution before the system reaches an equilibrium state. It’s a way of expressing how much of a substance can be dissolved in a given amount of solvent, typically water. In the given exercise involving calcium hydroxide, we start by converting the solubility from grams per liter to moles per liter. This is achieved by determining the molar mass of the solute and using it to convert grams to moles. Once this conversion is completed, we find that the molar solubility of calcium hydroxide in water is 0.0175 moles per liter. This value indicates how much calcium hydroxide dissolves in water to reach chemical equilibrium.

Ion Concentration

When calcium hydroxide dissolves in water, it dissociates into ions: calcium ions, \(\text{Ca}^{2+}\), and hydroxide ions, \(\text{OH}^{-}\). This dissociation is key to understanding ion concentration in solutions at equilibrium. For every mole of calcium hydroxide that dissolves, one mole of \(\text{Ca}^{2+}\) ions and two moles of \(\text{OH}^{-}\) ions are released into the solution. Thus, the concentration of \(\text{Ca}^{2+}\) ions directly equals the molar solubility, which is 0.0175 mol/L.The concentration of \(\text{OH}^{-}\) ions, however, is twice this value, given that two hydroxide ions are produced for each molecule of calcium hydroxide. Therefore, the concentration of \(\text{OH}^{-}\) ions is 0.035 mol/L.

Chemical Equilibrium

Chemical equilibrium occurs in a saturated solution where a dynamic balance exists between the dissolution and precipitation of a solid compound. It is significant in determining the solubility product constant, \(K_{sp}\). In the case of calcium hydroxide, the \(K_{sp}\) expression involves the equilibrium concentrations of its dissociated ions: \([\text{Ca}^{2+}]\) and \([\text{OH}^{-}]^2\). The equilibrium constant \(K_{sp}\) is calculated using these concentrations to express the solution's saturation level.For the exercise at hand, once the ion concentrations are known, they are substituted into the \(K_{sp}\) expression: \(K_{sp} = [\text{Ca}^{2+}] [\text{OH}^{-}]^2\). After performing the arithmetic with the given concentrations, we find that \(K_{sp}\) for calcium hydroxide is \(2.1 \times 10^{-5}\). This value reflects the specific conditions under which calcium hydroxide reaches chemical equilibrium in the solution.