Problem 62 - Chemistry and Chemical Reactivity

Question

Calculate the solubility, in moles per liter, of calcium hydroxide, \(\mathrm{Ca}(\mathrm{OH})_{2},\) in a solution buffered to a pH of 12.60.

Step-by-Step Solution

Answer

The solubility is approximately \( 3.47 \times 10^{-3} \text{ M} \).

Step 1: Understand the Problem

We need to find the solubility of calcium hydroxide \( \text{Ca(OH)}_2 \) in a solution with a pH of 12.60. This involves finding how much \( \text{Ca(OH)}_2 \) can dissolve under these conditions.

Step 2: Calculate Hydroxide Ion Concentration

Since pH is given as 12.60, we first find the pOH, which is \( \text{pOH} = 14 - \text{pH} = 14 - 12.60 = 1.40 \). Then, calculate \( \text{[OH]}^- = 10^{-\text{pOH}} = 10^{-1.40} \approx 3.98 \times 10^{-2} \text{ M}\).

Step 3: Set up the Equilibrium Expression

The dissolution of calcium hydroxide is given by the equation: \( \text{Ca(OH)}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{OH}^- (aq) \). We represent the solubility as \( s \). Therefore, \( [\text{Ca}^{2+}] = s \) and \( [\text{OH}^{-}] = 2s \).

Step 4: Apply the Solubility Product Constant (Ksp)

The solubility product (\( K_{sp} \)) for \( \text{Ca(OH)}_2 \) is known to be approximately \( 5.5 \times 10^{-6} \). However, in a buffered solution, we have \( [\text{OH}^-] = 3.98 \times 10^{-2} \). Substitute this into the expression: \( K_{sp} = [\text{Ca}^{2+}] [\text{OH}^{-}]^2 \).

Step 5: Solve for Calcium Ion Concentration

Substitute \( K_{sp} = 5.5 \times 10^{-6} \) and \( [\text{OH}^-] = 3.98 \times 10^{-2} \) into the equation: \[ [\text{Ca}^{2+}] = \frac{K_{sp}}{[\text{OH}^{-}]^2} = \frac{5.5 \times 10^{-6}}{(3.98 \times 10^{-2})^2} \approx 3.47 \times 10^{-3} \text{ M} \].

Step 6: Conclusion

The solubility of \( \text{Ca(OH)}_2 \) in a solution buffered to a pH of 12.60 is approximately \( 3.47 \times 10^{-3} \text{ M} \).

Key Concepts

pH and pOH

The concepts of pH and pOH are essential in understanding the acidity and basicity of a solution. The pH scale measures how acidic or basic a liquid is, ranging from 0 to 14, with 7 being neutral. A pH below 7 indicates acidity, while a pH above 7 indicates a basic (or alkaline) solution. In this exercise, the pH is given as 12.60, which signifies a highly basic solution.

To find the pOH, which gives us a measure similar to pH but focuses on the concentration of hydroxide ions ([OH]^−), we use the relationship: \[ ext{pH} + ext{pOH} = 14 \].Given a pH of 12.60, the pOH is calculated as 1.40 using the formula: \[ ext{pOH} = 14 - ext{pH} \].
This pOH value helps us move to the next step, knowing that the concentration of hydroxide ions is relatively high due to the basic nature of the solution.

Solubility Product Constant (Ksp)

The solubility product constant, represented as \( K_{sp} \),is a crucial value that helps us determine the solubility of a sparingly soluble ionic compound. It provides insight into the equilibrium concentrations of the ions in a saturated solution.For calcium hydroxide, \( ext{Ca(OH)}_2 \),its \( K_{sp} \)is approximately \( 5.5 \times 10^{-6} \).This means that when the solid is in equilibrium with its ions in solution, the product of the concentrations of these ions equals \( K_{sp} \).

• Causes the dissolution equation for \( ext{Ca(OH)}_2 \):\[ ext{Ca(OH)}_2 (s) ightleftharpoons ext{Ca}^{2+} (aq) + 2 ext{OH}^- (aq) \]
• At equilibrium:\[ K_{sp} = [ ext{Ca}^{2+}][ ext{OH}^-]^2 \]

Understanding \( K_{sp} \)enables us to predict how much of a substance can dissolve and to ascertain the composition of the saturated solution.

Equilibrium Expressions

Equilibrium expressions are mathematical ways to describe the state of a system in dynamic equilibrium. For the dissolution of a solid like calcium hydroxide \( ext{Ca(OH)}_2 \),equilibrium exists when the rate at which the solid dissolves matches the rate at which the ions recombine to form the solid.
This can be expressed with the chemical equation:\[ ext{Ca(OH)}_2 (s) ightleftharpoons ext{Ca}^{2+} (aq) + 2 ext{OH}^- (aq) \].
The equilibrium expression associated with the solubility process is:\[ K_{sp} = [ ext{Ca}^{2+}][ ext{OH}^-]^2 \].Each concentration in the expression is raised to the power of the coefficient from the balanced equation, indicating how each species' concentration affects equilibrium.

• In this case, the hydroxide ion concentration square reflects the stoichiometry of two \( ext{OH}^- \)ions produced per molecule of calcium hydroxide dissolved.
• These expressions allow us to manipulate the solubility balance by adding or removing ions.

Hydroxide Ion Concentration

Hydroxide ion concentration is a key aspect of understanding how basic a solution is. It greatly affects the solubility of certain compounds, such as calcium hydroxide.Given that the pH of the solution is 12.60, which was determined earlier, the hydroxide ion concentration is found through the pOH:\[ ext{pOH} = 14 - ext{pH} = 1.40 \].
Calculating the \( [ ext{OH}^-] \)involves:\[ [ ext{OH}^-] = 10^{- ext{pOH}} = 10^{-1.40} \ ext{M} \approx 3.98 \times 10^{-2} \].This concentration directly feeds into the equilibrium expression to determine the solubility:

• It affects the expression \( K_{sp} = [ ext{Ca}^{2+}][ ext{OH}^-]^2 \)by squaring the \( [ ext{OH}^-] \)value.
• Higher concentrations of hydroxide inhibit further dissolution of \( ext{Ca(OH)}_2 \),thus determining the overall solubility.

This step is vital as it explains the amount of compound that can be dissolved, suiting the pH conditions set in the problem.