Solubility of `Al(OH)_(3) = S, K_(sp)` will be

Options

A. `108 S^(3)`
B. `27 S^(3)`
C. `4 S^(2)`
D. `27 S^(4)`

Explanation

To find the solubility product constant (\(K_{sp}\)) of aluminum hydroxide (\(Al(OH)_3\)) in terms of its solubility (\(S\)), we can follow these steps:

### Step 1: Write the Dissociation Reaction
The dissociation of aluminum hydroxide in water can be represented as:
\[ Al(OH)_3 (s) \rightleftharpoons Al^{3+} (aq) + 3 OH^{-} (aq) \]

### Step 2: Define the Solubility (\(S\))
Let the solubility of \(Al(OH)_3\) be \(S\) mol/L. This means that when \(Al(OH)_3\) dissolves, it produces:
- \(S\) moles of \(Al^{3+}\)
- \(3S\) moles of \(OH^{-}\)

### Step 3: Write the Expression for \(K_{sp}\)
The solubility product constant (\(K_{sp}\)) is given by the expression:
\[ K_{sp} = [Al^{3+}][OH^{-}]^3 \]
Substituting the concentrations in terms of \(S\):
\[ K_{sp} = [S][3S]^3 \]

### Step 4: Simplify the Expression
Now, simplify the expression:
\[ K_{sp} = S \cdot (3S)^3 = S \cdot 27S^3 = 27S^4 \]

### Step 5: Final Expression for \(K_{sp}\)
Thus, the solubility product constant for aluminum hydroxide in terms of its solubility is:
\[ K_{sp} = 27S^4 \]

### Conclusion
The final expression for the solubility product constant \(K_{sp}\) of \(Al(OH)_3\) is:
\[ K_{sp} = 27S^4 \]