The solubility of `Ca(OH)_(2)` is 'S' moles `lit^(-1)`, the solubility product is

Options

A. `4 S^(3)`
B. `4 S^(2)`
C. `S^(3)`
D. `S^(2)`

Answer

To find the solubility product (Ksp) of calcium hydroxide, \( Ca(OH)_2 \), we can follow these steps:

### Step 1: Write the dissociation equation
When calcium hydroxide dissolves in water, it dissociates according to the following equation:
\[ Ca(OH)_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2OH^{-} (aq) \]

### Step 2: Define solubility
Let the solubility of \( Ca(OH)_2 \) be \( S \) moles per liter. This means that at equilibrium:
- The concentration of \( Ca^{2+} \) ions will be \( S \) moles per liter.
- The concentration of \( OH^{-} \) ions will be \( 2S \) moles per liter (since 2 moles of \( OH^{-} \) are produced for every mole of \( Ca(OH)_2 \)).

### Step 3: Write the expression for Ksp
The solubility product \( Ksp \) is given by the product of the concentrations of the ions, each raised to the power of their coefficients in the balanced equation:
\[ Ksp = [Ca^{2+}][OH^{-}]^2 \]

### Step 4: Substitute the concentrations
Substituting the concentrations we found in Step 2 into the Ksp expression:
\[ Ksp = [S][2S]^2 \]

### Step 5: Simplify the expression
Now simplify the expression:
\[ Ksp = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Final Answer
Thus, the solubility product \( Ksp \) of \( Ca(OH)_2 \) is:
\[ Ksp = 4S^3 \]

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