The Solubility Product, Ksp

Solubility Equilibria

When ionic compounds dissolve in water, the solid compound dissociates into a solution of its ions. Ionic compounds are described as soluble, sparingly/partially soluble or insoluble. In reality, no ionic compound is completely insoluble in water. When considering insoluble salts, so little of the ionic compound dissolves before the solution is saturated.

When an ionic compound dissolves in water and forms a saturated solution, a dynamic equilibrium is achieved between the solid ionic compound and its ions in solution. For example:

BaSO4(s) ⇌ Ba2+(aq) + SO42-(aq)

Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH(aq)

The Solubility Product, Ksp

The equilibrium constant for this type of system is called a solubility product, Ksp. As these are heterogeneous equilibriums, only the ions are included in the equilibrium expression. The equilibrium expression for barium sulfate and magnesium hydroxide:

Ksp = [Ba2+] [SO42-]

Ksp = [Mg2+] [OH]2

Equilibrium expressions are only quantitatively correct for dilute solutions. Therefore, equilibrium expressions are only considered and analysed for ‘sparingly soluble’ and ‘insoluble’ salts.

Ksp and Units

When dealing with solubility product problems, it is common to see the unit of g/100mL used for concentration. This is often because the very low solubilities of the salts that form Ksp problems are difficult to conceptualise when using the usual mol/L. It also makes it easier to compare the solubility of different salts in terms of the mass that will dissolve.

When determining the Ksp value of a salt it is important to remember that we always use the mol/L concentration of the salt when determining the solubility product. We may need to convert g/100mL into mol/L to or vice versa depending on the problem.

Converting mol/L into g/100mL

Starting with mol/L:

• convert the mol/L into g/L by multiplying by the molar mass of the salt
• divide by 10 to get g/100mL

For example, the solubility of BaSO4 is 1.07 × 10−5 mol/L

• converting mol/L of BaSO4 into g/L = (1.07 × 10−5) × 233.4 = 0.0025 g/L
• converting g/L of BaSO4 into g/100mL = 0.0025 g/L ÷ 10 = 0.00025 g/100mL

Converting g/100mL into mol/L

Starting with g/100mL:

• convert the g/100mL into g/L by multiplying by 10
• divide the g of salt by its molar mass to get mol/L

For example, the solubility of Zn(OH)2 is 0.001 g/100mL

• converting g/100mL of Zn(OH)2 into g/L = 0.001 × 10 = 0.01 g/L
• converting g/L of Zn(OH)2 into mol/L = 0.01 g/L ÷ 99.4 g/mol = 1.01 × 10−4 mol/L

Determining Ksp from Molar Solubility

When determining a Ksp value from the molar solubility of a salt, there are some key steps to follow:

• Write a balanced equation for the dissociation of the salt
• Convert the solubility into mol/L (if not already)
• Determine the concentration of the ions in solution
• Write the equilibrium expression and substitute in the appropriate concentrations

Example:

Calcium carbonate has a solubility of 0.00058 g/100mL. Determine the Ksp value of calcium carbonate:

CaCO3(s) ⇌ Ca2+(aq) + CO32-(aq)

Determining the solubility in mol/L:

0.00058 g/100mL = 0.0058 g/L (multiply by 10)

= 0.0058/100.09 mol/L (divide soluble mass by molar mass of calcium carbonate)

= 5.8 × 10-5 mol/L

Therefore, 5.8 × 10-5 mol/L of CaCO3 ionises to form 5.8 × 10-5 mol/L of Ca2+ and 5.8 × 10-5 mol/L of CO32-

Ksp = [Ca2+][CO32-]

Ksp = [5.8 × 10-5][5.8 × 10-5]

Ksp = 3.36 × 10-9

Determining Solubility from Ksp

A similar process is used to determine the solubility from the Ksp value, however, the steps may be in the reverse order. It also involves some minor algebra skills and an understanding of ions in solution to solve.

Example:

Lead (II) hydroxide has a Ksp value of 1.43 × 10-15. Determine the solubility of lead hydroxide in mol/L and g/100mL:

Pb(OH)2(s) ⇌ Pb2+(aq) + 2OH(aq)

When writing the Ksp expression, the concentration of the ions will be the same or in some whole number proportion based on the concentration of the ions in solution. In any case, it is common to denote the concentration of the salt as s. In the above example: Pb(OH)2(s) ⇌ Pb2+(aq) + 2OH(aq), the concentration of each species can be thought of as: s ⇌ s + 2s.

The Ksp expression would be:

Ksp = [Pb2+][OH]2

We will use our notation of s:

Ksp = [s][2s]2

Now we can substitute in the given Ksp value and expand the right hand side:

1.43 × 10-15 = 4s3

s3 = 3.575 × 10-16

s = 7.1 × 10-6 mol/L

Now we can multiply by the molar mass of Pb(OH)2 to get the concentration in g/L:

(7.1 × 10-6) × 241.2 = 1.71 × 10-3 g/L

And dividing by 10 will give us the concentration in g/100mL:

1.71 × 10-3 g/L ÷ 10 = 1.71 × 10-4 g/100mL

Therefore, the solubility of lead hydroxide is 7.1 × 10-6 mol/L or 1.71 × 10-4 g/100mL