Introduction to Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is an equilibrium constant used to describe the solubility of sparingly soluble ionic compounds. It represents the maximum concentration of the ions in a saturated solution at a particular temperature. Understanding how to calculate molar solubility from Ksp is essential in fields such as chemistry, environmental science, and pharmacology.

In this article, we will explore three diverse examples that illustrate how to calculate molar solubility from Ksp.

Example 1: Molar Solubility of Calcium Fluoride (CaF2)

Calcium fluoride is a common ionic compound used in dental treatments and as a fluoride supplement. Its solubility product constant at 25°C is approximately 3.9 × 10^-11. We can calculate the molar solubility of CaF2 using its Ksp.

Given the dissociation of CaF2 in water:

CaF2 (s) ⇌ Ca²⁺ (aq) + 2F⁻ (aq)

Let the molar solubility of CaF2 be represented by ‘s’. Therefore, at equilibrium:

• [Ca²⁺] = s
• [F⁻] = 2s

Using the Ksp expression:

Ksp = [Ca²⁺][F⁻]² = (s)(2s)² = 4s³

Now, substituting the known Ksp value:

3.9 × 10^-11 = 4s³

Solving for ‘s’:

s³ = (3.9 × 10^-11) / 4
s³ = 9.75 × 10^-12
s = (9.75 × 10^-12)^(1/3) ≈ 2.15 × 10^-4 M

Thus, the molar solubility of CaF2 is approximately 2.15 × 10^-4 M.

Notes:

• This example demonstrates how to derive concentrations from the Ksp equation for a 1:2 ionic dissociation.

Example 2: Molar Solubility of Silver Chloride (AgCl)

Silver chloride is often used in photography and as a bactericide. Its Ksp at 25°C is known to be 1.77 × 10^-10. We will calculate the molar solubility of AgCl.

The dissociation equation for AgCl is:

AgCl (s) ⇌ Ag⁺ (aq) + Cl⁻ (aq)

Let the molar solubility be ‘s’, then at equilibrium:

• [Ag⁺] = s
• [Cl⁻] = s

The Ksp expression is:

Ksp = [Ag⁺][Cl⁻] = s * s = s²

Substituting the Ksp value:

1.77 × 10^-10 = s²

To find ‘s’:

s = √(1.77 × 10^-10) ≈ 1.33 × 10^-5 M

Thus, the molar solubility of AgCl is approximately 1.33 × 10^-5 M.

Notes:

• This case illustrates a 1:1 ionic dissociation, making the calculations straightforward.

Example 3: Molar Solubility of Barium Sulfate (BaSO4)

Barium sulfate is a compound utilized in medical imaging and as a pigment. Its Ksp value at 25°C is 1.0 × 10^-10. We will calculate the molar solubility of BaSO4.

The dissociation reaction for BaSO4 is:

BaSO4 (s) ⇌ Ba²⁺ (aq) + SO4²⁻ (aq)

Let the molar solubility be ‘s’. Thus:

• [Ba²⁺] = s
• [SO4²⁻] = s

The Ksp expression becomes:

Ksp = [Ba²⁺][SO4²⁻] = s * s = s²

Substituting the Ksp value:

1.0 × 10^-10 = s²

Solving for ‘s’:

s = √(1.0 × 10^-10) ≈ 1.0 × 10^-5 M

Therefore, the molar solubility of BaSO4 is approximately 1.0 × 10^-5 M.

Notes:

• This example is another 1:1 ionic dissociation, showcasing the consistency of the Ksp calculations across different compounds.