Question

You place \(1.234 \mathrm{g}\) of solid \(\mathrm{Ca}(\mathrm{OH})_{2}\) in \(1.00 \mathrm{L}\) of pure water at \(25^{\circ} \mathrm{C}\). The \(\mathrm{pH}\) of the solution is found to be 12.68 Estimate the value of \(K_{\mathrm{rp}}\) for \(\mathrm{Ca}(\mathrm{OH})_{2}.\)

Short answer

The estimated value of \(K_{sp}\) for \(\mathrm{Ca(OH)}_2\) is approximately \(5.496 \times 10^{-5}\).

Step-by-step solution

Determine Concentration of OH⁻ Ions

The pH of the solution is 12.68. We can use this value to find the pOH, which is given by: \[\text{pOH} = 14 - \text{pH} = 14 - 12.68 = 1.32\]Next, calculate the concentration of OH⁻ ions using the formula:\[[\text{OH}^-] = 10^{-\text{pOH}} = 10^{-1.32} \approx 4.79 \times 10^{-2} \text{ M}\]

Calculate Moles of Dissolved Ca(OH)₂

The dissolution reaction is \(\text{Ca(OH)}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{OH}^-\). From this reaction, we see that one mole of \(\text{Ca(OH)}_2\) produces two moles of \(\text{OH}^-\) ions. Therefore, the concentration of \(\text{Ca}^{2+}\) ions is half of that of \(\text{OH}^-\) ions:\[[\text{Ca}^{2+}] = \frac{[\text{OH}^-]}{2} = \frac{4.79 \times 10^{-2}}{2} = 2.395 \times 10^{-2} \text{ M}\]

Calculate the Solubility Product Ksp

The solubility product \(K_{sp}\) is given by the equation:\[K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2\]Substitute the concentrations found in previous steps:\[K_{sp} = (2.395 \times 10^{-2})(4.79 \times 10^{-2})^2\]\[K_{sp} = (2.395 \times 10^{-2})(2.295 \times 10^{-3})\]\[K_{sp} \approx 5.496 \times 10^{-5}\]

Conclusion

The calculated value of the solubility product \(K_{sp}\) for \(\text{Ca(OH)}_2\) is approximately \(5.496 \times 10^{-5}\). This value represents the equilibrium constants for the dissolution of \(\text{Ca(OH)}_2\) in water.

Key concepts

Solubility Product (Ksp)

In the realm of chemistry, the solubility product constant, known as \(K_{sp}\), is a crucial concept for understanding the dissolving of sparingly soluble salts in water. It represents the product of the concentrations of the ions present in a saturated solution of a compound, each raised to the power of their respective coefficients in the balanced equation. In the context of calcium hydroxide, the dissolution process can be written as:\[\text{Ca(OH)}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{OH}^- \]Since one molecule of calcium hydroxide produces one calcium ion and two hydroxide ions, the \(K_{sp}\) expression becomes:\[K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 \]This equilibrium constant tells us how much calcium hydroxide can dissolve in water under equilibrium conditions. An important aspect to remember is that a low \(K_{sp}\) value indicates very little solubility, meaning only a minimal amount of solid dissolves.

pH and pOH Calculations

When dealing with solutions, knowing how to compute pH and pOH is essential as they indicate the acidity or basicity of a solution. The pH scale measures how acidic or basic a solution is, ranging from 0 (very acidic) to 14 (very basic), with 7 being neutral. To find pOH, you subtract the pH value from 14, as shown in this formula:\[\text{pOH} = 14 - \text{pH} \]This relationship provides a straightforward way to determine the concentration of hydroxide ions \([\text{OH}^-]\) in a solution. Once you have the pOH, you can find \([\text{OH}^-]\) by utilizing this expression:\[[\text{OH}^-] = 10^{-\text{pOH}} \]This calculation is crucial for determining the ion concentrations necessary to calculate the solubility product \(K_{sp}\) for compounds like calcium hydroxide.

Calcium Hydroxide Dissolution

Calcium hydroxide, commonly known as slaked lime, undergoes dissolution in water, forming calcium ions \(\text{Ca}^{2+}\) and hydroxide ions \(\text{OH}^-\). The overall dissolution equation can be illustrated as follows:\[\text{Ca(OH)}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{OH}^- \]This reaction shows a 1:2 ratio between calcium hydroxide and hydroxide ions. Therefore, if you know the concentration of \([\text{OH}^-]\), you can easily find \([\text{Ca}^{2+}]\) by dividing \([\text{OH}^-]\) by 2. In practical applications, calcium hydroxide is not very soluble, as reflected by its relatively low \(K_{sp}\). Understanding how this compound behaves in water is important for applications ranging from water treatment to cement production.