Introduction and Example 1—Molar Solubility of an AB type Compound. Molar Solubility from K sp

Compounds with low solubility always establish an equilibrium when they are placed in water.

Compounds with low solubility always establish an equilibrium when they are placed in water. This is called solubility equilibrium.

E.g., when lead(II) sulphate is added to water:

Compounds with low solubility always establish an equilibrium when they are placed in water. This is called solubility equilibrium. E.g., when lead(II) sulphate is added to water:

Compounds with low solubility always establish an equilibrium when they are placed in water. This is called solubility equilibrium. E.g., when lead(II) sulphate is added to water:

Compounds with low solubility always establish an equilibrium when they are placed in water. This is called solubility equilibrium. E.g., when lead(II) sulphate is added to water:

Compounds with low solubility always establish an equilibrium when they are placed in water. This is called solubility equilibrium. E.g., when lead(II) sulphate is added to water:

The equilibrium constant expression, K eq, for this is:

Solid

The equilibrium constant expression, K eq, for this is: For Solubility Equilibrium

The equilibrium constant expression, K eq, for this is: For Solubility Equilibrium

The equilibrium constant expression, K eq, for this is: The Solubility Product Constant

The equilibrium constant expression, K eq, for this is: Product

The equilibrium constant expression, K eq, for this is: Product The Solubility Product Constant

Solubility depends on the temperature.

Molar solubility is the number of moles of a substance that will dissolve to form 1.0 L of a saturated solution,

Molar solubility is the number of moles of a substance that will dissolve to form 1.0 L of a saturated solution, expressed in moles/L, or molarity (M).

A saturated solution is the solution that forms when a compound is at solubility equilibrium.

Molar solubility and Concentrations of Ions in a Saturated Solution

The concentrations of ions in a saturated solution (one at solubility equilibrium) depend on:

The molar solubility of the compound

The concentrations of ions in a saturated solution (one at solubility equilibrium) depend on: The molar solubility of the compound The coefficient of each ion in the equilibrium equation

For example, consider the compound lead(II) sulphate.

For example, consider the compound lead(II) sulphate. It’s solubility equilibrium equation is:

Solubility Equilibrium Equation

For example, consider the compound lead(II) sulphate. It’s solubility equilibrium equation is:

× 1/1

For example, consider the compound lead(II) sulphate. It’s solubility equilibrium equation is: × 1/1

Molar solubility is how much of this dissolves, in moles/L

= the molar solubility of PbSO 4 × 1/1 Molar solubility is how much of this dissolves, in moles/L

× 1/1 Molar solubility is how much of this dissolves, in moles/L = the molar solubility of PbSO 4 = the molar solubility of PbSO 4

× 1/1 One of each ion so PbSO 4 is called an “AB” type compound.

Lead(II) bromide

Solubility Equilibrium Equation

× 1/1 Molar solubility is how much of this dissolves, in moles/L = the molar solubility of PbBr 2

× 1/1 × 2/1 Molar solubility is how much of this dissolves, in moles/L = the molar solubility of PbBr 2 = 2 × the molar solubility of PbBr 2

× 1/1 × 2/1 1 of one ion and 2 of the other ion so PbBr 2 is called an “AB 2 ” type compound.

Finding the Molar Solubility of an AB Type Compound from K sp

Find the molar solubility of PbSO 4 at 25°C. Its K sp at 25°C is 1.8 × 10 –8.

1 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M Solubility Equilibrium Equation

1 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. s The molar solubility of PbSO 4 = 1.3 × 10 –4 M Let molar solubility of PbSO 4 = s

1 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. s The molar solubility of PbSO 4 = 1.3 × 10 –4 M Let molar solubility of PbSO 4 = s × 1/1 1/1 ratio

1 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s The molar solubility of PbSO 4 = 1.3 × 10 –4 M 1/1 ratio s × 1/1

1 2 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M Ksp Expression

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M Use this to find the molar solubility

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M s

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M s

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M s s

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M So Ksp is equal to s time s s s

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M Take the square root of both sides

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M Substitute 1.8 × 10 –8 in for K sp

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M M is the unit for molar concentration

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M 2 significant figures

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M round to 2 significant figures

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M 2 significant figures

1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 at 25°C = 1.3 × 10 –4 M