The solubility in water of a sparingly soluble salt AB2 is 1.0×10-5 mol L-1. Its solubility product number will be

To determine the solubility product (Ksp) of the sparingly soluble salt AB2, we first need to understand how the dissociation of the salt in water relates to its solubility. When AB2 dissolves, it separates into its constituent ions. For a salt like AB2, which dissociates into one cation (A) and two anions (B), the dissociation can be represented as follows:

Dissociation Equation

The dissociation of AB2 in water can be written as:

AB2 (s) ⇌ A²⁺ (aq) + 2B^- (aq)

Understanding Solubility

The solubility of AB2 is given as 1.0 × 10^-5 mol L^-1. This means that in a saturated solution, the concentration of AB2 that has dissolved is 1.0 × 10^-5 mol L^-1.

Ion Concentrations

From the dissociation equation, we can see that for every 1 mole of AB2 that dissolves, it produces:

• 1 mole of A²⁺
• 2 moles of B^-

Thus, if the solubility of AB2 is 1.0 × 10^-5 mol L^-1, the concentrations of the ions in solution will be:

• [A²⁺] = 1.0 × 10^-5 mol L^-1
• [B^-] = 2 × (1.0 × 10^-5) = 2.0 × 10^-5 mol L^-1

Calculating the Solubility Product (Ksp)

The solubility product constant (Ksp) is calculated using the concentrations of the ions at equilibrium. The expression for Ksp for the dissociation of AB2 is:

Ksp = [A²⁺][B^-]²

Substituting the concentrations we found:

Ksp = (1.0 × 10^-5)(2.0 × 10^-5)²

Calculating this gives:

Ksp = (1.0 × 10^-5) × (4.0 × 10^-10)

Ksp = 4.0 × 10^-15

Final Result

Therefore, the solubility product (Ksp) of the sparingly soluble salt AB2 is 4.0 × 10^-15. This value indicates how much of the salt can dissolve in water before reaching saturation, and it provides insight into the salt's solubility characteristics.