The solubility product of `BaSO_(4)" is " 1.5 xx 10^(-9)`. Find the solubility of `BaSO_(4)` in
`0.1 M BaCl_(2)`

Answer

To find the solubility of barium sulfate (BaSO₄) in a 0.1 M BaCl₂ solution, we can follow these steps:

### Step 1: 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 2: Define the solubility product (Ksp).
The solubility product expression for BaSO₄ is given by:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]
Given that \( K_{sp} = 1.5 \times 10^{-9} \).

### Step 3: Set up the concentration terms.
Let \( S \) be the solubility of BaSO₄ in moles per liter. In a 0.1 M BaCl₂ solution, the concentration of Ba²⁺ ions from BaCl₂ is 0.1 M. Therefore, the total concentration of Ba²⁺ ions will be:
\[ [\text{Ba}^{2+}] = 0.1 + S \]
Since \( S \) is expected to be very small compared to 0.1, we can approximate this as:
\[ [\text{Ba}^{2+}] \approx 0.1 \]

The concentration of sulfate ions will be:
\[ [\text{SO}_4^{2-}] = S \]

### Step 4: Substitute into the Ksp expression.
Substituting these concentrations into the Ksp expression gives:
\[ K_{sp} = (0.1)(S) \]

### Step 5: Solve for S.
Now substituting the value of Ksp:
\[ 1.5 \times 10^{-9} = (0.1)(S) \]
\[ S = \frac{1.5 \times 10^{-9}}{0.1} \]
\[ S = 1.5 \times 10^{-8} \, \text{M} \]

### Step 6: Convert to grams per liter.
To convert the solubility from moles per liter to grams per liter, we need the molar mass of BaSO₄. The molar mass is calculated as follows:
- Molar mass of Ba = 137 g/mol
- Molar mass of S = 32 g/mol
- Molar mass of O = 16 g/mol × 4 = 64 g/mol

Total molar mass of BaSO₄:
\[ 137 + 32 + 64 = 233 \, \text{g/mol} \]

Now, we can calculate the solubility in grams per liter:
\[ \text{Solubility in g/L} = S \times \text{Molar mass} \]
\[ \text{Solubility in g/L} = (1.5 \times 10^{-8} \, \text{mol/L}) \times (233 \, \text{g/mol}) \]
\[ \text{Solubility in g/L} = 3.495 \times 10^{-6} \, \text{g/L} \]

### Final Answer:
The solubility of BaSO₄ in 0.1 M BaCl₂ is approximately \( 3.5 \times 10^{-6} \, \text{g/L} \).

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