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.