The solubility product of `As_(2)S_(3)` is `2.8 xx 10^(-72)`. What is the solubility of `As_(2)S_(3)`

Options

A. `1.09 xx 10^(-15)` mole/litre
B. `1.72 xx 10^(-15)` mole/litre
C. `2.3 xx 10^(-16)` mole/litre
D. `1.65 xx 10^(-36)` mole/litre

Explanation

To find the solubility of \( As_2S_3 \) given its solubility product \( K_{sp} = 2.8 \times 10^{-72} \), we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of \( As_2S_3 \) in water can be represented as:
\[ As_2S_3 (s) \rightleftharpoons 2 As^{3+} (aq) + 3 S^{2-} (aq) \]

### Step 2: Define the Solubility
Let the solubility of \( As_2S_3 \) be \( S \) mol/L. From the dissociation equation, we can see that:
- For every 1 mole of \( As_2S_3 \) that dissolves, it produces 2 moles of \( As^{3+} \) and 3 moles of \( S^{2-} \).
- Therefore, at equilibrium:
- The concentration of \( As^{3+} \) will be \( 2S \)
- The concentration of \( S^{2-} \) will be \( 3S \)

### Step 3: Write the Expression for \( K_{sp} \)
The solubility product \( K_{sp} \) can be expressed as:
\[ K_{sp} = [As^{3+}]^2 [S^{2-}]^3 \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (2S)^2 (3S)^3 \]

### Step 4: Simplify the Expression
Now, simplify the expression:
\[ K_{sp} = 4S^2 \cdot 27S^3 = 108S^5 \]

### Step 5: Set Up the Equation
Now, we can set up the equation using the given \( K_{sp} \):
\[ 108S^5 = 2.8 \times 10^{-72} \]

### Step 6: Solve for \( S^5 \)
Rearranging the equation gives:
\[ S^5 = \frac{2.8 \times 10^{-72}}{108} \]

### Step 7: Calculate \( S^5 \)
Calculating \( S^5 \):
\[ S^5 = \frac{2.8}{108} \times 10^{-72} = 0.0259259 \times 10^{-72} = 2.59259 \times 10^{-74} \]

### Step 8: Calculate \( S \)
Now, take the fifth root to find \( S \):
\[ S = (2.59259 \times 10^{-74})^{1/5} \]

Calculating \( S \):
\[ S \approx 1.09 \times 10^{-15} \text{ mol/L} \]

### Final Answer
The solubility of \( As_2S_3 \) is approximately:
\[ S \approx 1.09 \times 10^{-15} \text{ mol/L} \]

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