The solubility of `Sb_(2)S_(3)` in water is `1.0 xx 10^(-8)` mol/litre at 298 K. What will be its solubility product:

Options

A. `108 xx 10^(-25)`
B. `1.0 xx 10^(-25)`
C. `144 xx 10^(-25)`
D. `126 xx 1^(-28)`

Answer

To find the solubility product (Ksp) of `Sb2S3`, we will follow these steps:

### Step 1: Write the dissociation equation
The dissociation of `Sb2S3` in water can be represented as:
\[ Sb_2S_3 (s) \rightleftharpoons 2 Sb^{3+} (aq) + 3 S^{2-} (aq) \]

### Step 2: Define solubility (S)
Let the solubility of `Sb2S3` be denoted as \( S \). According to the problem, the solubility is given as:
\[ S = 1.0 \times 10^{-8} \, \text{mol/L} \]

### Step 3: Determine the concentrations of ions at equilibrium
From the dissociation equation:
- For every 1 mole of `Sb2S3` that dissolves, it produces 2 moles of `Sb^{3+}` ions and 3 moles of `S^{2-}` ions.
- Therefore, at equilibrium:
- Concentration of `Sb^{3+}` ions = \( 2S \)
- Concentration of `S^{2-}` ions = \( 3S \)

### Step 4: Substitute the solubility value into the ion concentrations
Substituting \( S = 1.0 \times 10^{-8} \):
- Concentration of `Sb^{3+}` ions = \( 2 \times (1.0 \times 10^{-8}) = 2.0 \times 10^{-8} \, \text{mol/L} \)
- Concentration of `S^{2-}` ions = \( 3 \times (1.0 \times 10^{-8}) = 3.0 \times 10^{-8} \, \text{mol/L} \)

### Step 5: Write the expression for Ksp
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [Sb^{3+}]^2 \times [S^{2-}]^3 \]

### Step 6: Substitute the concentrations into the Ksp expression
Substituting the values we calculated:
\[ K_{sp} = (2.0 \times 10^{-8})^2 \times (3.0 \times 10^{-8})^3 \]

### Step 7: Calculate Ksp
Now calculate \( K_{sp} \):
1. Calculate \( (2.0 \times 10^{-8})^2 = 4.0 \times 10^{-16} \)
2. Calculate \( (3.0 \times 10^{-8})^3 = 27.0 \times 10^{-24} = 2.7 \times 10^{-23} \)

Now multiply these two results:
\[ K_{sp} = 4.0 \times 10^{-16} \times 2.7 \times 10^{-23} \]
\[ K_{sp} = 10.8 \times 10^{-39} \]
\[ K_{sp} = 1.08 \times 10^{-38} \]

### Final Answer
The solubility product \( K_{sp} \) of `Sb2S3` is:
\[ K_{sp} = 1.08 \times 10^{-38} \]

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