Question

A saturated solution of calcium hydroxide, \(\mathrm{Ca}(\mathrm{OH})_{2},\) has a \(\mathrm{pH}\) of \(12.35 .\) Find the hydroxide ion concentration and calculate the equilibrium constant.

Short answer

Hydroxide ion concentration: \(2.24 \times 10^{-2}\,\text{M},\) Ksp: \(5.62 \times 10^{-6}\).

Step-by-step solution

Convert pH to pOH

We start by utilizing the relationship between pH and pOH: \[ \text{pH} + \text{pOH} = 14 \]Given \[ \text{pH} = 12.35 \]\[ \text{pOH} = 14 - 12.35 = 1.65 \] This tells us the value of pOH.

Find Hydroxide Ion Concentration

To find the concentration of hydroxide ions, we use the formula:\[ [\text{OH}^-] = 10^{-\text{pOH}} \]Substitute the pOH value:\[ [\text{OH}^-] = 10^{-1.65} \approx 2.24 \times 10^{-2} \text{ M} \]This provides the hydroxide ion concentration in the solution.

Find Calcium Ion Concentration

Because calcium hydroxide dissociates according to the equation:\[ \text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^- \] The ratio of \( \text{Ca}^{2+} \) to \( \text{OH}^- \) is 1:2. Thus, the calcium ion concentration \( [\text{Ca}^{2+}] \) is half of \( [\text{OH}^-] \):\[ [\text{Ca}^{2+}] = \frac{[\text{OH}^-]}{2} \approx \frac{2.24 \times 10^{-2}}{2} = 1.12 \times 10^{-2} \text{ M} \]

Calculate the Equilibrium Constant, Ksp

Equilibrium constant (Ksp) for \( \text{Ca(OH)}_2 \) dissociation is given by:\[ K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 \]Substituting the ion concentrations:\[ K_{sp} = (1.12 \times 10^{-2})(2.24 \times 10^{-2})^2 \]Calculate:\[ K_{sp} = 1.12 \times 10^{-2} \times 5.02 \times 10^{-4} = 5.62 \times 10^{-6} \]

Key concepts

pH and pOH Relationship

Understanding the relationship between pH and pOH is crucial in the study of chemical equilibrium, especially when working with acidic and basic solutions. The formula that connects pH and pOH is:\[ \text{pH} + \text{pOH} = 14 \]This equation arises from water's auto-ionization, where water slightly dissociates into hydrogen ions (\[ \text{H}^+ \]) and hydroxide ions (\[ \text{OH}^- \]) resulting in a product that remains constant at 25°C.

• If a solution has a high pH (greater than 7), it indicates a basic environment.
• Conversely, a low pH (less than 7) indicates acidity.
• The pH and pOH should always add up to 14, which helps you easily calculate one if you know the other.

When dealing with the exercise, you began with a pH of 12.35. Subtracting this from 14 gave us the pOH of 1.65. This quick conversion is fundamental in chemical equilibria calculations, allowing us to dive deeper into ion concentration assessments.

Hydroxide Ion Concentration

Determining the concentration of hydroxide ions is vital in analyzing the extent of a substance's basicity. The concentration of hydroxide ions \([\text{OH}^-]\) can be calculated using the known pOH value. Here's how:\[ [\text{OH}^-] = 10^{-\text{pOH}} \]When you substitute the pOH of 1.65:\[ [\text{OH}^-] = 10^{-1.65} \approx 2.24 \times 10^{-2} \text{ M} \]This expression gives the molarity of the hydroxide ions in the solution. Higher hydroxide ion concentrations typically indicate a stronger base.
Calculating and understanding \([\text{OH}^-]\) is not only integral in fields like analytical chemistry but also in industries such as agriculture and food science, where pH balance is critical. Ensure you practice calculating hydroxide concentrations, as they are a gateway to learning about the solution's behavior and interaction with other compounds.

Equilibrium Constant (Ksp) Calculation

The equilibrium constant, specifically known as the solubility product constant \( K_{sp} \) for slightly soluble salts, is a measure of the solubility of ionic compounds. For calcium hydroxide \( \text{Ca(OH)}_2 \), it dissociates in water as:\[ \text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^- \]The solubility product \( K_{sp} \) is formulated as:\[ K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 \]Given the concentrations from our previous calculations:

• \([\text{OH}^-] \approx 2.24 \times 10^{-2} \text{ M}\)
• \([\text{Ca}^{2+}] = \frac{[\text{OH}^-]}{2} \approx 1.12 \times 10^{-2} \text{ M}\)

By substituting these values into the \( K_{sp} \) equation:\[ K_{sp} = (1.12 \times 10^{-2})(2.24 \times 10^{-2})^2 = 5.62 \times 10^{-6} \]This final calculation of \( K_{sp} \) is pivotal for predicting how much of a substance will dissolve in solution. It allows chemists to anticipate precipitation, apply safety measures, and understand material behaviors in natural and industrial processes.