Solubility Product Constant (Ksp) and Solubility Equilibria

Solubility and Solubility Product Constant (Ksp)

Solubility is a chemical property that describes how much of a solute can dissolve in a solvent to form a saturated solution. The Solubility Product Constant (Ksp) is an equilibrium constant that quantifies the solubility of ionic solids in water.

• Solubility can also be referred to as concentration or molarity (M).

• A higher Ksp value indicates a more soluble solid; a lower Ksp value indicates a less soluble solid.

Key Point: Upward Ksp = more soluble solid; downward Ksp = less soluble solid.

Example: Comparing Solubility Using Ksp

Given the following Ksp values, which substance is the most soluble?

Substance	Ksp
AgCl	1.5 × 10-5
MgCO3	1.5 × 10-8
CaSO4	7.1 × 10-5
CuS	8.5 × 10-45

Answer: CaSO4 is the most soluble because it has the highest Ksp value.

Equilibrium Expressions for Solubility

When an ionic solid dissolves in water, it establishes an equilibrium between the solid and its dissolved ions. The equilibrium can be represented as:

$A_2B_3 (s) \rightleftharpoons 2A^{3+} (aq) + 3B^{2-} (aq)$

• Dissolution is the process of the solid dissolving into ions.

• The equilibrium expression for this reaction is written in terms of the concentrations of the products (ions), ignoring solids and liquids.

General Ksp Expression:

$K_{sp} = [A^{3+}]^2 [B^{2-}]^3$

Example: Equilibrium Expression for Calcium Nitrate

Write the equilibrium expression for calcium nitrate, Ca(NO3)2:

1. Write the dissociation equation:

2. $Ca(NO_3)_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2NO_3^- (aq)$

3. Write the Ksp expression:

4. $K_{sp} = [Ca^{2+}][NO_3^-]^2$

Calculating Ksp from Solubility

If the solubility (in molar concentration) of an ionic solid is known, the Ksp can be calculated.

1. Write the equilibrium equation for the solid dissolving into its aqueous ions.

2. Write the Ksp expression based on the equilibrium equation.

3. Make concentrations of the ions equal to their stoichiometric coefficients multiplied by the solubility variable (x).

4. Substitute the given solubility value for x and solve for Ksp.

Example: Calculating Ksp for Silver Phosphate

Given: The solubility of Ag3PO4 is 1.8 × 10-8 M at 25°C.

1. Write the dissociation equation:

2. $Ag_3PO_4 (s) \rightleftharpoons 3Ag^+ (aq) + PO_4^{3-} (aq)$

3. Write the Ksp expression:

4. $K_{sp} = [Ag^+]^3 [PO_4^{3-}]$

5. Let x = solubility of Ag3PO4:

   • $[Ag^+] = 3x$

   • $[PO_4^{3-}] = x$

6. Substitute x = 1.8 × 10-8 M:

7. $K_{sp} = (3x)^3 (x) = 27x^4$

   Calculate Ksp by plugging in the value of x.

Calculating Solubility from Ksp

If the Ksp value is known, the solubility of an ionic solid can be determined.

1. Write the equilibrium equation for the solid dissolving into its aqueous ions.

2. Write the Ksp expression based on the equilibrium equation.

3. Solve for the solubility variable (x) using the given Ksp value.

Example: Calculating Solubility for Strontium Fluoride

Given: Ksp for SrF2 is 7.9 × 10-10 at 25°C. Calculate its solubility in M.

1. Write the dissociation equation:

2. $SrF_2 (s) \rightleftharpoons Sr^{2+} (aq) + 2F^- (aq)$

3. Write the Ksp expression:

4. $K_{sp} = [Sr^{2+}][F^-]^2$

5. Let x = solubility of SrF2:

   • $[Sr^{2+}] = x$

   • $[F^-] = 2x$

6. Substitute into Ksp expression:

7. $K_{sp} = (x)(2x)^2 = 4x^3$

8. Solve for x using the given Ksp value.

Practice Problems

• Determine the equilibrium expression of the barium nitride solid.

• Manganese (V) hydroxide has a measured solubility of 3.4 × 10-6 M at 25°C. Calculate its Ksp value.

• The Ksp value for strontium fluoride, SrF2, is 7.9 × 10-10 at 25°C. Calculate its solubility in g/L.

Additional info: For all Ksp calculations, always write the balanced dissolution equation, set up the Ksp expression, and use stoichiometry to relate solubility to ion concentrations.