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:

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$_4$.

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: $\text{Moles of BaSO}_4 = \frac{0.0025}{233} \approx 1.07 \times 10^{-5} \, \text{mol/L}.$

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}. ]

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}. ]

Final Answer:

The correct option is D. $1.15 \times 10^{-10}$.

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.