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 BaSO₄ in a 0.1 M BaCl₂ solution, we can follow these steps:

### Step 1: Write the dissociation equations
The dissociation of BaSO₄ can be represented as:
\[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

The dissociation of BaCl₂ can be represented as:
\[ \text{BaCl}_2 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + 2 \text{Cl}^- (aq) \]

### Step 2: Define solubility
Let the solubility of BaSO₄ in the solution be \( S \) mol/L. This means that:
- The concentration of Ba²⁺ from BaSO₄ will be \( S \).
- The concentration of SO₄²⁻ will also be \( S \).

### Step 3: Calculate the concentration of Ba²⁺ from BaCl₂
Since BaCl₂ is 0.1 M, it will dissociate to give:
- Concentration of Ba²⁺ from BaCl₂ = 0.1 M
- Concentration of Cl⁻ from BaCl₂ = 2 × 0.1 M = 0.2 M

### Step 4: Total concentration of Ba²⁺
The total concentration of Ba²⁺ in the solution will be the sum of the contributions from BaSO₄ and BaCl₂:
\[ [\text{Ba}^{2+}]_{\text{total}} = S + 0.1 \]

### Step 5: Write the expression for the solubility product (Ksp)
The solubility product \( K_{sp} \) for BaSO₄ is given by:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]
Given that \( K_{sp} = 1.5 \times 10^{-9} \), we can substitute the concentrations:
\[ 1.5 \times 10^{-9} = (S + 0.1)(S) \]

### Step 6: Make an assumption
Since BaSO₄ is a sparingly soluble salt, we can assume that \( S \) is very small compared to 0.1 M. Therefore, we can approximate:
\[ S + 0.1 \approx 0.1 \]

### Step 7: Substitute and solve for S
Now substituting this approximation into the Ksp expression:
\[ 1.5 \times 10^{-9} = (0.1)(S) \]
\[ S = \frac{1.5 \times 10^{-9}}{0.1} \]
\[ S = 1.5 \times 10^{-8} \, \text{M} \]

### Conclusion
The solubility of BaSO₄ in 0.1 M BaCl₂ is:
\[ S = 1.5 \times 10^{-8} \, \text{M} \]

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