Problem 46 - Chemistry and Chemical Reactivity

Question

Calcium hydroxide, \(\mathrm{Ca}(\mathrm{OH})_{2},\) dissolves in water to the extent of 1.78 g per liter. What is the value of \(K_{\mathrm{sp}}\) for \(\mathrm{Ca}(\mathrm{OH})_{2} ?\) $$\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Ca}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})$$

Step-by-Step Solution

Answer

The \(K_{sp}\) for \(\mathrm{Ca(OH)_2}\) is approximately 0.0000553.

Step 1: Molecular Weight Calculation

Start by calculating the molar mass of calcium hydroxide, \( \mathrm{Ca(OH)_2} \). The atomic weight of calcium (Ca) is 40.08 g/mol, oxygen (O) is 16.00 g/mol, and hydrogen (H) is 1.01 g/mol. Therefore, the molar mass is calculated as follows: \( 40.08 + 2 \times (16.00 + 1.01) = 40.08 + 34.02 = 74.10 \text{ g/mol} \).

Step 2: Convert Solubility to Molarity

Convert the solubility from grams per liter to molarity (moles per liter) using the molar mass calculated previously. Given solubility is 1.78 g/L, so: \[ \text{Molarity} = \frac{1.78 \text{ g/L}}{74.10 \text{ g/mol}} \approx 0.0240 \text{ mol/L}. \]

Step 3: Determine Ion Concentrations

In solution, \( \mathrm{Ca(OH)_2} \) dissociates into one \( \mathrm{Ca^{2+}} \) ion and two \( \mathrm{OH^-} \) ions. If the concentration of \( \mathrm{Ca(OH)_2} \) is \( 0.0240 \text{ mol/L} \), then the concentration of \( \mathrm{Ca^{2+}} \) ions is \( 0.0240 \text{ mol/L} \) and the concentration of \( \mathrm{OH^-} \) ions is \( 2 \times 0.0240 \text{ mol/L} = 0.0480 \text{ mol/L} \).

Step 4: Calculate \(K_{sp}\) Value

The solubility product \(K_{sp}\) is given by the expression \( K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{OH^-}]^2 \). Substitute the concentrations from Step 3: \( K_{sp} = (0.0240)(0.0480)^2 = 0.0000553 \).

Key Concepts

Calcium Hydroxide Solubility

Calcium hydroxide, abbreviated as \( \mathrm{Ca(OH)_2} \), is commonly referred to as "slaked lime" and is known for its ability to dissolve in water, albeit sparingly. The solubility of a substance is a measure of how much of it can be dissolved in a solvent, in this case, water, to form a saturated solution. For \( \mathrm{Ca(OH)_2} \), this solubility is given as 1.78 grams per liter. It's important to understand that solubility varies with temperature, but for a typical textbook problem, assume a constant condition unless otherwise stated.

In practical calculations, solubility is often converted from grams per liter into molarity, which is the number of moles of solute per liter of solution. Molarity provides a direct link to chemical reactions, as it allows us to use stoichiometry to determine concentrations of ions produced during dissolution. By calculating the molar mass of \( \mathrm{Ca(OH)_2} \), we find it to be 74.10 g/mol, which facilitates this conversion. Consequently, 1.78 g/L translates to approximately 0.0240 moles per liter. This conversion is pivotal for determining the equilibrium concentrations of ions in the solution when \( \mathrm{Ca(OH)_2} \) dissolves.

Dissolution Equilibrium

When \( \mathrm{Ca(OH)_2} \) is added to water, it dissociates into its constituent ions, \( \mathrm{Ca^{2+}} \) and \( \mathrm{OH^-} \). The process of reaching a state where the rates of dissolution and precipitation of \( \mathrm{Ca(OH)_2} \) are equal is known as dissolution equilibrium. The chemical reaction can be represented as: \[ \mathrm{Ca(OH)_2(s)} \rightleftharpoons \mathrm{Ca^{2+}(aq)} + 2\mathrm{OH^{-}(aq)} \] At equilibrium, the concentration of \( \mathrm{Ca^{2+}} \) in the solution equals the solubility we previously determined as 0.0240 mol/L. However, because each unit of \( \mathrm{Ca(OH)_2} \) produces two \( \mathrm{OH^-} \) ions, the hydroxide ion concentration is twice this, or 0.0480 mol/L.

This balance between dissolution and precipitation is crucial for calculating the solubility product constant, \( K_{sp} \). Understanding these dynamics help in predicting the solubility under different conditions, such as changes in pH or the presence of other ions.

Chemical Calculations

The solubility product constant \( K_{sp} \) is a unique value for ionic compounds dissolved in water. It signifies the level of solubility of the compound in the solution. For \( \mathrm{Ca(OH)_2} \), the expression for the solubility product is derived from the equilibrium concentrations of its ions:

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

By substituting the ion concentrations obtained from solubility data, we calculate \( K_{sp} \) as follows: \( K_{sp} = (0.0240)(0.0480)^2 = 0.0000553 \).

The use of a squared term underscores the importance of stoichiometry—each molecule of \( \mathrm{Ca(OH)_2} \) produces two hydroxide ions, which is why the hydroxide concentration is squared in the \( K_{sp} \) expression.

• Stability of solutions: A higher \( K_{sp} \) value indicates higher solubility, which means the solution can hold more dissolved ions before precipitating.
• Predicting reactions: Knowing \( K_{sp} \) helps anticipate product formation in reactions involving ionic solids.

Understanding \( K_{sp} \) equips you with the ability to solve complex equilibrium problems in chemistry that involve determining solute ion concentrations in various scenarios.