If the solubility of `Al(OH)_3` is S moles/litre, the solubility product is

Options

A. `S^3`
B. `27S^4`
C. `S^2`
D. `4S^3`

Answer

To determine the solubility product (Ksp) of aluminum hydroxide, Al(OH)₃, given its solubility S moles per liter, we can follow these steps:

### Step 1: Write the dissociation equation
Aluminum hydroxide dissociates in water as follows:

\[ \text{Al(OH)}_3 (s) \rightleftharpoons \text{Al}^{3+} (aq) + 3 \text{OH}^- (aq) \]

### Step 2: Define the solubility
Let the solubility of Al(OH)₃ be S moles per liter. This means that at equilibrium:

- The concentration of Al³⁺ ions will be S moles/liter.
- The concentration of OH⁻ ions will be 3S moles/liter (since each formula unit of Al(OH)₃ produces 3 hydroxide ions).

### Step 3: Write the expression for the solubility product (Ksp)
The solubility product Ksp is given by the formula:

\[ K_{sp} = [\text{Al}^{3+}][\text{OH}^-]^3 \]

### Step 4: Substitute the concentrations into the Ksp expression
Substituting the concentrations we found:

\[ K_{sp} = [S][3S]^3 \]

### Step 5: Simplify the expression
Calculating the expression:

\[ K_{sp} = S \cdot (3S)^3 \]
\[ K_{sp} = S \cdot 27S^3 \]
\[ K_{sp} = 27S^4 \]

### Final Answer
Thus, the solubility product \( K_{sp} \) of Al(OH)₃ is:

\[ K_{sp} = 27S^4 \]

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