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 \( Sb_2S_3 \), we can follow these steps:

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

### Step 2: Define the solubility
Let the solubility of \( Sb_2S_3 \) be \( S \) mol/L. According to the stoichiometry of the dissociation:
- For every 1 mole of \( Sb_2S_3 \) that dissolves, it produces 2 moles of \( Sb^{3+} \) and 3 moles of \( S^{2-} \).
- Therefore, the concentrations of the ions at equilibrium will be:
- \( [Sb^{3+}] = 2S \)
- \( [S^{2-}] = 3S \)

### Step 3: Substitute the given solubility
From the problem, we know that the solubility \( S = 1.0 \times 10^{-5} \) mol/L.

### Step 4: Calculate the concentrations of the ions
Substituting \( S \) into the expressions for the ion concentrations:
- \( [Sb^{3+}] = 2(1.0 \times 10^{-5}) = 2.0 \times 10^{-5} \) mol/L
- \( [S^{2-}] = 3(1.0 \times 10^{-5}) = 3.0 \times 10^{-5} \) mol/L

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

### Step 6: 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 7: 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 \( Sb_2S_3 \) is:
\[ Ksp = 1.08 \times 10^{-23} \]

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