If the solubility of `Al_2(SO_4)_3` is S , then its solubility product is

Options

A. `27 S^3`
B. `54 S^4`
C. `108 S^5`
D. `64S^3`

Answer

To find the solubility product (Ksp) of \( Al_2(SO_4)_3 \) given its solubility \( S \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of aluminum sulfate in water can be represented as:
\[ Al_2(SO_4)_3 (s) \rightleftharpoons 2Al^{3+} (aq) + 3SO_4^{2-} (aq) \]

### Step 2: Determine the molar concentrations at equilibrium
If the solubility of \( Al_2(SO_4)_3 \) is \( S \), then at equilibrium:
- The concentration of \( Al^{3+} \) ions will be \( 2S \) (since 2 moles of \( Al^{3+} \) are produced for every mole of \( Al_2(SO_4)_3 \)).
- The concentration of \( SO_4^{2-} \) ions will be \( 3S \) (since 3 moles of \( SO_4^{2-} \) are produced for every mole of \( Al_2(SO_4)_3 \)).

### Step 3: Write the expression for Ksp
The solubility product \( Ksp \) is given by the formula:
\[ Ksp = [Al^{3+}]^m [SO_4^{2-}]^n \]
where \( m \) and \( n \) are the stoichiometric coefficients from the balanced equation.

For our case:
- \( m = 2 \) (for \( Al^{3+} \))
- \( n = 3 \) (for \( SO_4^{2-} \))

Substituting the equilibrium concentrations into the Ksp expression, we get:
\[ Ksp = (2S)^2 \cdot (3S)^3 \]

### Step 4: Calculate Ksp
Now, we can simplify the expression:
\[ Ksp = (2^2 \cdot S^2) \cdot (3^3 \cdot S^3) \]
\[ Ksp = 4S^2 \cdot 27S^3 \]
\[ Ksp = 108S^5 \]

### Final Answer
Thus, the solubility product \( Ksp \) of \( Al_2(SO_4)_3 \) is:
\[ Ksp = 108S^5 \]

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