Question

\(pH\) of a saturated solution of \(Ca(OH)_2\) is \(9\). The solubility product \((K_{sp})\) of \( Ca(OH)_2\) is: 

• A. \(0.5 \times 10^{-15} \)
• B. \(0.25 \times 10^{-10} \)
• C. \(0.125 \times 10^{-15} \)
• D. \(0.5 \times 10^{-10} \)

Answer 1

The Correct Option is A

To determine the solubility product \((K_{sp})\) of \(Ca(OH)_2\) given that the \(pH\) of its saturated solution is 9, we need to follow a few logical steps. Let's break down the solution:

1. First, recall that the \(pH\) of a solution is related to its \(pOH\) by the equation: \( pH + pOH = 14 \). Given that \(pH = 9\), we can calculate \(pOH\) as follows: \( pOH = 14 - 9 = 5 \).
2. Next, using the relation between \(pOH\) and the concentration of hydroxide ions \([\text{OH}^-]\): \( [\text{OH}^-] = 10^{-pOH} \), substitute the calculated \(pOH\): \( [\text{OH}^-] = 10^{-5} \).
3. Considering the dissolution equation of \(Ca(OH)_2\): \( Ca(OH)_2 \rightleftharpoons Ca^{2+} + 2OH^- \), the stoichiometry shows that for every mole of \(Ca(OH)_2\) that dissolves, one mole of \(Ca^{2+}\) and two moles of \(\text{OH}^-\) are produced. Let the solubility of \(Ca(OH)_2\) be \(s\), so: \( [Ca^{2+}] = s \) and \( [OH^-] = 2s \).
4. Since we know \([\text{OH}^-] = 10^{-5}\), we equate this to \(2s\): \( 2s = 10^{-5} \). Solving for \(s\), we get: \( s = \frac{10^{-5}}{2} = 0.5 \times 10^{-5} \).
5. The solubility product \((K_{sp})\) is given by: \( K_{sp} = [Ca^{2+}][OH^-]^2 \). Substitute \( [Ca^{2+}] = s = 0.5 \times 10^{-5} \) and \([OH^-] = 10^{-5}\): \( K_{sp} = (0.5 \times 10^{-5}) \times (10^{-5})^2 \).
6. Calculate \(K_{sp}\): \( K_{sp} = 0.5 \times 10^{-15} \).

Thus, the correct answer is \(\boxed{0.5 \times 10^{-15}}\).

This step-by-step approach not only provides the correct answer but also clarifies the relationship between the pH, solubility, and solubility product.