Question

What are \(\left[\mathrm{Ca}^{2+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{~s}) ?\) The \(K_{\mathrm{sp}}\) of \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{~s})\) is \(5.0 \times 10^{-6}\)

Short answer

In the saturated solution, \([\mathrm{Ca}^{2+}] \approx 1.08 \times 10^{-2}\, \text{M}\) and \([\mathrm{OH}^-] \approx 2.16 \times 10^{-2}\, \text{M}\).

Step-by-step solution

Understanding Solubility Product

The solubility product constant, \(K_{sp}\), represents the equilibrium between a solid and its ions in a solution. For \(\mathrm{Ca(OH)_2}\), the expression for \(K_{sp}\) is given by: \[ K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{OH}^-]^2 \] where \([\mathrm{Ca}^{2+}]\) is the concentration of calcium ions and \([\mathrm{OH}^-]\) is the concentration of hydroxide ions in a saturated solution.

Setting Up the Expression

In a saturated solution of \(\mathrm{Ca(OH)_2} \), it dissolves as: \[\mathrm{Ca(OH)_2(s)} \rightleftharpoons \mathrm{Ca^{2+}(aq)} + 2\mathrm{OH^-}(aq)\] Let's denote the molar solubility of \(\mathrm{Ca(OH)_2}\) as \(s\). Therefore, \([\mathrm{Ca}^{2+}] = s\) and \([\mathrm{OH}^-] = 2s\) in the solution.

Substituting into the \(K_{sp}\) Expression

Substitute the concentrations \([\mathrm{Ca}^{2+}] = s\) and \([\mathrm{OH}^-] = 2s\) into the \(K_{sp}\) expression: \[ K_{sp} = s \cdot (2s)^2 = 4s^3 \] We know \(K_{sp} = 5.0 \times 10^{-6}\), so we have: \[ 4s^3 = 5.0 \times 10^{-6} \]

Solving for \(s\)

To find \(s\), solve the equation \(4s^3 = 5.0 \times 10^{-6}\). Divide both sides by 4: \[ s^3 = \frac{5.0 \times 10^{-6}}{4} = 1.25 \times 10^{-6} \] Then, take the cube root of both sides: \[ s = \sqrt[3]{1.25 \times 10^{-6}} \]

Calculate \(s\) and Substituting Back

By calculating the cube root, we find: \[ s \approx 1.08 \times 10^{-2} \] Now substitute \(s\) back to find the concentrations: \([\mathrm{Ca}^{2+}] = s = 1.08 \times 10^{-2} \ \text{M} \) and \([\mathrm{OH}^-] = 2s = 2.16 \times 10^{-2} \ \text{M} \).

Key concepts

Calcium Hydroxide

Calcium hydroxide, often known as slaked lime, is a chemical compound with the formula \( \mathrm{Ca(OH)_2} \). It is a white, crystalline substance commonly used in various applications like cement, agricultural lime, and even food preparation. In water, calcium hydroxide can dissolve to form a slightly alkaline solution, releasing calcium ions \( \mathrm{Ca^{2+}} \) and hydroxide ions \( \mathrm{OH^-} \). This reaction reaches a state of dynamic equilibrium where the solid and its ions coexist.
Understanding the behavior of calcium hydroxide in solutions is crucial in fields like chemistry and environmental science. Its solubility is described by a solubility product constant \( K_{sp} \), which quantifies how much \( \mathrm{Ca(OH)_2} \) can dissolve in water until the solution becomes saturated.

Saturated Solution

A saturated solution is a state where a solution contains the maximum concentration of a solute, at a given temperature, that can dissolve in a solvent before any additional solute remains undissolved. In the case of calcium hydroxide, when it is dissolved in water to form a saturated solution, it disperses as calcium ions \( \mathrm{Ca^{2+}} \) and hydroxide ions \( \mathrm{OH^-} \) in a specific ratio.
In a saturated solution of \( \mathrm{Ca(OH)_2} \):

• Each unit of \( \mathrm{Ca(OH)_2} \) that dissolves, forms one calcium ion \( \mathrm{Ca^{2+}} \) and two hydroxide ions \( \mathrm{OH^-} \).
• The concentrations of these ions at saturation are determined by the \( K_{sp} \).

Equilibrium is reached when the rate of dissolution of \( \mathrm{Ca(OH)_2} \) equals the rate at which the ions recombine to form the solid. At this point, the product of the ion concentrations equals the \( K_{sp} \). This balance maintains the saturated state.

Ion Concentration

Ion concentration refers to the amount of a specific ion present in a solution, usually measured in moles per liter (Molarity, M). For calcium hydroxide in its saturated form, the ion concentrations depend on its solubility characteristics, represented by the solubility product constant \( K_{sp} \).
The dissociation of \( \mathrm{Ca(OH)_2} \) produces one calcium ion \( \mathrm{Ca^{2+}} \) and two hydroxide ions \( \mathrm{OH^-} \) for every unit that dissolves:

• Calcium ion concentration \( [\mathrm{Ca^{2+}}] = s \)
• Hydroxide ion concentration \( [\mathrm{OH^-}] = 2s \)

Where \( s \) is the molar solubility of \( \mathrm{Ca(OH)_2} \). By using the equation \( K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{OH^-}]^2 \), and substituting the expressions for ion concentration \( (s) \) and \( (2s) \), one can solve for \( s \).
This allows us to accurately predict the concentrations of ions in the saturated solution, which is essential in chemistries dealing with equilibria and saturation dynamics.