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 solve this problem, we need to determine the solubility product \(K_{sp}\) of calcium hydroxide, \(Ca(OH)_2\), given the pH of its saturated solution is 9.

1. Calculate the pOH: Since \(pH\) of the solution is 9, we can find the \(pOH\) using the relation:

   \(pH + pOH = 14\)

   Therefore,

   \(pOH = 14 - 9 = 5\)

2. Find the concentration of OH^- ions: The hydroxide ion concentration can be calculated using the relation:

   \[ [OH^-] = 10^{-pOH} = 10^{-5} \, \text{mol/L} \]

3. Determine the solubility (s) of \(Ca(OH)_2\): The dissociation of calcium hydroxide in water is represented by the equation below:

   \( Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^- \)

   From this equation, it follows that if s is the solubility of \(Ca(OH)_2\), the concentration of \(OH^-\) ions will be \(2s\). Therefore, we have:

   \( 2s = 10^{-5} \)

   Solving for s gives:

   \( s = \frac{10^{-5}}{2} = 0.5 \times 10^{-5} \, \text{mol/L} \)

4. Calculate the solubility product \(K_{sp}\) of \(Ca(OH)_2\): The expression for the solubility product is:

   \( K_{sp} = [Ca^{2+}][OH^-]^2 \)

   Since the concentration of \(Ca^{2+}\) is equal to the solubility, \(s\), and the concentration of \(OH^-\) is \(2s\), we have:

   \( K_{sp} = s \times (2s)^2 = s \times 4s^2 = 4s^3 \)

   Substitute \(s = 0.5 \times 10^{-5} \) into the formula:

   \[ K_{sp} = 4 \times (0.5 \times 10^{-5})^3 \]

   = 4 \times (0.125 \times 10^{-15}) = 0.5 \times 10^{-15} \]

Conclusion: The correct option for the solubility product \(K_{sp}\) of \(Ca(OH)_2\) is \(0.5 \times 10^{-15} \), which matches the first option provided.