At 25 degrees celsius only 0.0390 mol of generric salt AB3 is soluable in 1.00L of water. what is the ksp of the salt at 25 degrees celsius. use standard notation for answer.  AB3(s)<->A3+(aq)+3B-(aq)

To determine the solubility product constant \( K_{sp} \) for the salt \( \text{AB}_3 \) at 25°C, let's analyze the given dataTo determine the solubility product constant \( K_{sp} \) for the salt \( \text{AB}_3 \) at 25°C, let's analyze the given 

1. **Given solubility**: Only 0.0390 mol of \( \text{AB}_3 \) is soluble in 1.00 L of water at 25°C.
   - Therefore, the molar solubility \( s \) of \( \text{AB}_3 \) is 0.0390 M.

2. **Dissociation of \( \text{AB}_3 \)**:
   \[
   \text{AB}_3(s) \leftrightarrow \text{A}^{3+}(aq) + 3\text{B}^-(aq)
   \]

   For each mole of \( \text{AB}_3 \) that dissolves, it produces 1 mole of \( \text{A}^{3+} \) ions and 3 moles of \( \text{B}^- \) ions.

3. **Ion concentrations in terms of solubility**:
   - The concentration of \( \text{A}^{3+} \) ions will be \( s = 0.0390 \) M.
   - The concentration of \( \text{B}^- \) ions will be \( 3s = 3 \times 0.0390 = 0.117 \) M.

4. **Write the \( K_{sp} \) expression**:
   \[
   K_{sp} = [\text{A}^{3+}][\text{B}^-]^3
   \]

5. **Substitute the values**:
   \[
   K_{sp} = (0.0390)(0.117)^3
   \]

6. **Calculate \( K_{sp} \)**:
   \[
   K_{sp} = (0.0390)(0.00160329) \approx 6.25 \times 10^{-5}
   \]

**Answer**: The solubility product constant is:
\[
K_{sp} = 6.25 \times 10^{-5}
\]

Would you like further details, or have any questions?

### Related Questions
1. What would happen to \( K_{sp} \) if the temperature were increased?
2. How does the \( K_{sp} \) expression change if \( \text{AB}_3 \) dissociates into different ion ratios?
3. How would the solubility change if common ions are added to the solution?
4. What units are used for \( K_{sp} \), and why is it unitless in this case?
5. How is \( K_{sp} \) useful in predicting precipitation reactions?

### Tip
Remember, \( K_{sp} \) is constant at a given temperature, and changes only with temperature adjustments.

At 25 degrees celsius only 0.0390 mol of generic salt AB3 is soluble in 1.00L of water. What is the Ksp of the salt at 25 degrees celsius? Use standard notation for the answer. AB3(s) <-> A3+(aq) + 3B-(aq)

Learn how to calculate the solubility product constant (Ksp) for the salt AB3 at 25°C, given that 0.0390 mol is soluble in 1.00 L of water. This step-by-step solution covers stoichiometry, ion concentration calculations, and equilibrium concepts. Suitable for high school and early college students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation