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The question asks for the solubility product (\(K_{sp}\)) of BaSO\(_4\) at 25°C based on its solubility in water. Here's the breakdown:

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### Given:
1. Maximum solubility: \(0.25 \, \text{mg}\) of BaSO\(_4\) in \(100 \, \text{mL}\) water.
2. Molar mass of BaSO\(_4\): \(233 \, \text{g/mol}\).

### Required:
Determine the solubility product, \(K_{sp}\), for BaSO\(_4\).

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### Step 1: Convert solubility into moles per liter.
The solubility in mg/L:
\[
0.25 \, \text{mg} \times 10 = 2.5 \, \text{mg/L}.
\]

Convert mg to grams:
\[
2.5 \, \text{mg/L} = 0.0025 \, \text{g/L}.
\]

Convert grams to moles:
\[
\text{Moles of BaSO}_4 = \frac{0.0025}{233} \approx 1.07 \times 10^{-5} \, \text{mol/L}.
\]

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### Step 2: Write the dissociation equation for BaSO\(_4\):
\[
\text{BaSO}_4 \leftrightharpoons \text{Ba}^{2+} + \text{SO}_4^{2-}.
\]

From the solubility:
\[
[\text{Ba}^{2+}] = [\text{SO}_4^{2-}] = 1.07 \times 10^{-5} \, \text{mol/L}.
\]

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### Step 3: Calculate \(K_{sp}\).
\[
K_{sp} = [\text{Ba}^{2+}] \cdot [\text{SO}_4^{2-}] = (1.07 \times 10^{-5})^2 \approx 1.15 \times 10^{-10}.
\]

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### Final Answer:
The correct option is **D. \(1.15 \times 10^{-10}\)**.

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Would you like further explanation or details on any step?  

### 5 Related Questions:
1. What is the significance of the solubility product \(K_{sp}\) in chemistry?
2. How does temperature affect the solubility of BaSO\(_4\)?
3. How can the solubility of a compound be experimentally determined?
4. What other factors (besides temperature) influence \(K_{sp}\)?
5. Can \(K_{sp}\) be used to predict precipitation in a solution?

### Tip:
Always ensure units are consistent when solving chemistry problems to avoid calculation errors.

At 25°C, a maximum of 0.25 mg of BaSO4 dissolves in 100 mL of water. Determine the solubility product (Ksp) of BaSO4. Given: M(BaSO4) = 233 g/mol.

Calculate the solubility product (Ksp) of BaSO4 at 25°C given its solubility of 0.25 mg in 100 mL of water. This solution involves stoichiometry, chemical equilibrium, and solubility product principles. Suitable for advanced chemistry students in Grades 11-12.

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