The solubility of `BaSO_4` in water is 0.00233 g per litre at `30^@C.` The solubility of `BaSO_4` in 0.1 M `(NH_4)_2SO_4` solution at the same temperature is

Options

A. `10^(-5)` mole/litre
B. `10^(-6)` mole/litre
C. `10^(-8)` mole/litre
D. `10^(-9)` mole/litre

Answer

To find the solubility of barium sulfate (BaSO₄) in a 0.1 M (NH₄)₂SO₄ solution, we can follow these steps:

### Step 1: Calculate the solubility of BaSO₄ in pure water
Given that the solubility of BaSO₄ in water is 0.00233 g/L, we first convert this to moles per liter (M).

**Calculation:**
- Molar mass of BaSO₄:
- Ba = 137.33 g/mol
- S = 32.07 g/mol
- O₄ = 4 × 16.00 g/mol = 64.00 g/mol
- Total = 137.33 + 32.07 + 64.00 = 233.40 g/mol

- Convert grams to moles:
\[ S = \frac{0.00233 \text{ g/L}}{233.40 \text{ g/mol}} = 1.00 \times 10^{-5} \text{ moles/L} \]

### Step 2: Write the dissociation equation for BaSO₄
Barium sulfate dissociates in water as follows:
\[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

### Step 3: Write the expression for Ksp
The solubility product constant (Ksp) for BaSO₄ can be expressed as:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = S \cdot S = S^2 \]

### Step 4: Calculate Ksp using the solubility in water
Substituting the value of S:
\[ K_{sp} = (1.00 \times 10^{-5})^2 = 1.00 \times 10^{-10} \text{ mol}^2/\text{L}^2 \]

### Step 5: Determine the effect of (NH₄)₂SO₄ on the solubility of BaSO₄
In a 0.1 M (NH₄)₂SO₄ solution, the sulfate ions (SO₄²⁻) are already present. This will affect the solubility of BaSO₄ due to the common ion effect.

### Step 6: Write the new Ksp expression in the presence of the common ion
Let the solubility of BaSO₄ in the (NH₄)₂SO₄ solution be S'. The concentration of sulfate ions from (NH₄)₂SO₄ is 0.1 M. Therefore, the Ksp expression becomes:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = S' \cdot 0.1 \]

### Step 7: Set the Ksp equal to the new expression
Using the Ksp value we calculated:
\[ 1.00 \times 10^{-10} = S' \cdot 0.1 \]

### Step 8: Solve for S'
\[ S' = \frac{1.00 \times 10^{-10}}{0.1} = 1.00 \times 10^{-9} \text{ moles/L} \]

### Step 9: Convert back to grams per liter if needed
To convert moles back to grams:
\[ \text{grams} = S' \times \text{molar mass of BaSO}_4 = 1.00 \times 10^{-9} \text{ moles/L} \times 233.40 \text{ g/mol} = 2.33 \times 10^{-7} \text{ g/L} \]

### Final Answer
The solubility of BaSO₄ in 0.1 M (NH₄)₂SO₄ solution is \( 1.00 \times 10^{-9} \) moles/L or \( 2.33 \times 10^{-7} \) g/L.

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