The solubility of `Ca(OH)_2` in water is : [Given : The solubility product of `Ca(OH)_2` in water `= 5.5 xx 10^(-6)` ]

Options

A. `1.77 xx 10^(-6)`
B. `1.11 xx 10^(-6)`
C. `1.11 xx 10^(-2)`
D. `1.77 xx 10^(-2)`

Answer

To solve the problem of finding the solubility of \( Ca(OH)_2 \) in water given its solubility product \( K_{sp} = 5.5 \times 10^{-6} \), we can follow these steps:

### Step 1: Write the dissociation equation
Calcium hydroxide, \( Ca(OH)_2 \), dissociates in water as follows:
\[ Ca(OH)_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2 OH^{-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \( Ca(OH)_2 \) be \( S \) mol/L. When \( Ca(OH)_2 \) dissolves, it produces:
- \( S \) mol/L of \( Ca^{2+} \)
- \( 2S \) mol/L of \( OH^{-} \)

### Step 3: Write the expression for the solubility product
The solubility product \( K_{sp} \) is given by the formula:
\[ K_{sp} = [Ca^{2+}][OH^{-}]^2 \]
Substituting the concentrations from the dissociation:
\[ K_{sp} = (S)(2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the given \( K_{sp} \) value
We know that \( K_{sp} = 5.5 \times 10^{-6} \), so we can set up the equation:
\[ 4S^3 = 5.5 \times 10^{-6} \]

### Step 5: Solve for \( S^3 \)
To find \( S^3 \), we rearrange the equation:
\[ S^3 = \frac{5.5 \times 10^{-6}}{4} \]
Calculating this gives:
\[ S^3 = 1.375 \times 10^{-6} \]

### Step 6: Calculate \( S \)
Now, we take the cube root to find \( S \):
\[ S = \sqrt[3]{1.375 \times 10^{-6}} \]
Calculating the cube root:
\[ S \approx 1.11 \times 10^{-2} \, \text{mol/L} \]

### Final Answer
The solubility of \( Ca(OH)_2 \) in water is approximately:
\[ \boxed{1.11 \times 10^{-2} \, \text{mol/L}} \]

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