The solubility of `Sb_(2)S_(3)` in water is `1.0 xx 10^(-5) ` mol/letre at 298K. 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 10^(-24)`

Answer

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

### Step 1: Write the Dissociation Equation
The dissociation of `Sb2S3` in water can be represented as:
\[ \text{Sb}_2\text{S}_3 (s) \rightleftharpoons 2 \text{Sb}^{3+} (aq) + 3 \text{S}^{2-} (aq) \]

### Step 2: Define Molar Solubility
Let the molar solubility of `Sb2S3` be \( s = 1.0 \times 10^{-5} \) mol/L. This means that at equilibrium:
- The concentration of `Sb^{3+}` ions will be \( 2s \).
- The concentration of `S^{2-}` ions will be \( 3s \).

### Step 3: Calculate Ion Concentrations
Substituting \( s \) into the concentrations:
- Concentration of `Sb^{3+}`:
\[ [\text{Sb}^{3+}] = 2s = 2 \times (1.0 \times 10^{-5}) = 2.0 \times 10^{-5} \, \text{mol/L} \]
- Concentration of `S^{2-}`:
\[ [\text{S}^{2-}] = 3s = 3 \times (1.0 \times 10^{-5}) = 3.0 \times 10^{-5} \, \text{mol/L} \]

### Step 4: Write the Expression for Ksp
The solubility product (Ksp) expression for the dissociation reaction is:
\[ Ksp = [\text{Sb}^{3+}]^2 [\text{S}^{2-}]^3 \]

### Step 5: Substitute the Ion Concentrations into the Ksp Expression
Substituting the values we calculated:
\[ Ksp = (2.0 \times 10^{-5})^2 \times (3.0 \times 10^{-5})^3 \]

### Step 6: Calculate Ksp
Calculating each part:
- \( (2.0 \times 10^{-5})^2 = 4.0 \times 10^{-10} \)
- \( (3.0 \times 10^{-5})^3 = 27.0 \times 10^{-15} = 2.7 \times 10^{-14} \)

Now, multiply these results:
\[ Ksp = 4.0 \times 10^{-10} \times 2.7 \times 10^{-14} = 1.08 \times 10^{-23} \]

### Final Answer
Thus, the solubility product \( Ksp \) of `Sb2S3` is:
\[ Ksp = 1.08 \times 10^{-23} \]

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