3.3 Solubility Product and Precipitation Reactions | Chemical Equilibria | Chemistry 12

Introduction

Many salts that are considered "insoluble" actually dissolve to a very small extent in water. These are called sparingly soluble salts. The equilibrium established between the undissolved solid and its ions in solution is described by the Solubility Product, $K_{s p}$ .

The Solubility Product ( $K_{s p}$ )

For a sparingly soluble salt $A_{x} B_{y}$ dissolving in water:

$A_{x} B_{y} (s) ⇌ x A^{y +} (a q) + y B^{x -} (a q)$

The solubility product expression is:

$K_{s p} = [A^{y +}]^{x} [B^{x -}]^{y}$

Note: The concentration of the pure solid $A_{x} B_{y}$ is constant and is incorporated into $K_{s p}$ , so it does not appear in the expression.

Key Points

$K_{s p}$ is a constant at a given temperature.
A smaller $K_{s p}$ means the salt is less soluble.
$K_{s p}$ only applies to saturated solutions of sparingly soluble salts.

Examples of $K_{s p}$ Expressions

Salt	Equilibrium	$K_{s p}$ Expression
$A g C l$	$A g C l ⇌ A g^{+} + C l^{-}$	$K_{s p} = [A g^{+}] [C l^{-}]$
$C a F_{2}$	$C a F_{2} ⇌ C a^{2 +} + 2 F^{-}$	$K_{s p} = [C a^{2 +}] [F^{-}]^{2}$
$A g_{2} C r O_{4}$	$A g_{2} C r O_{4} ⇌ 2 A g^{+} + C r O_{4}^{2 -}$	$K_{s p} = [A g^{+}]^{2} [C r O_{4}^{2 -}]$
$C a_{3} (P O_{4})_{2}$	$C a_{3} (P O_{4})_{2} ⇌ 3 C a^{2 +} + 2 P O_{4}^{3 -}$	$K_{s p} = [C a^{2 +}]^{3} [P O_{4}^{3 -}]^{2}$

Molar Solubility and $K_{s p}$

Molar solubility ( $S$ ) is the number of moles of a salt that dissolves per litre of solution to form a saturated solution.

Worked Example 1: $A g C l$

$A g C l (s) ⇌ A g^{+} (a q) + C l^{-} (a q)$

If solubility = $S$ : $[A g^{+}] = S$ , $[C l^{-}] = S$

$K_{s p} = S \times S = S^{2} ⟹ S = K_{s p}$

Worked Example 2: $A g_{2} C r O_{4}$

$A g_{2} C r O_{4} (s) ⇌ 2 A g^{+} (a q) + C r O_{4}^{2 -} (a q)$

If solubility = $S$ : $[A g^{+}] = 2 S$ , $[C r O_{4}^{2 -}] = S$

$K_{s p} = (2 S)^{2} (S) = 4 S^{3} ⟹ S = (\frac{K _{s p}}{4})^{1/3}$

Worked Example 3: $M X_{2}$ type salt

Given $K_{s p} = 4 \times 1 0^{- 12}$ :

$K_{s p} = 4 S^{3} ⟹ S^{3} = \frac{4 \times 1 0 ^{- 12}}{4} = 1 0^{- 12} ⟹ S = 1 0^{- 4} m o l d m^{- 3}$

The Ionic Product ( $Q_{s p}$ ) and Precipitation

The Ionic Product $Q_{s p}$ is calculated the same way as $K_{s p}$ , but using the actual concentrations of ions in any solution (not necessarily saturated).

Precipitation Rules

Condition	Meaning	Result
$Q_{s p} < K_{s p}$	Unsaturated	No precipitate; more salt can dissolve
$Q_{s p} = K_{s p}$	Saturated	Equilibrium; no change
$Q_{s p} > K_{s p}$	Supersaturated	Precipitate forms

Worked Example: Will a precipitate form?

If $[B a^{2 +}] = 0.001 m o l d m^{- 3}$ and $[S O_{4}^{2 -}] = 0.002 m o l d m^{- 3}$ are mixed, and $K_{s p} (B a S O_{4}) = 1.1 \times 1 0^{- 10}$ :

$Q_{s p} = [B a^{2 +}] [S O_{4}^{2 -}] = (0.001) (0.002) = 2 \times 1 0^{- 6}$

Since $Q_{s p} = 2 \times 1 0^{- 6} ≫ K_{s p} = 1.1 \times 1 0^{- 10}$ , a precipitate of $B a S O_{4}$ will form.

The Common Ion Effect on Solubility

When a soluble salt sharing a common ion with a sparingly soluble salt is added to the solution, the solubility of the sparingly soluble salt decreases. This is the Common Ion Effect, explained by Le Chatelier's Principle.

Example: $A g C l$ in $N a C l$ solution

$A g C l (s) ⇌ A g^{+} (a q) + C l^{-} (a q)$

Adding $N a C l$ increases $[C l^{-}]$ . By Le Chatelier's Principle, the equilibrium shifts left, reducing the dissolution of $A g C l$ . Therefore, the solubility of $A g C l$ decreases.

Quantitative Treatment

If $A g C l$ is dissolved in $0.1 m o l d m^{- 3}$ $N a C l$ solution ( $K_{s p} (A g C l) = 1.8 \times 1 0^{- 10}$ ):

$K_{s p} = [A g^{+}] [C l^{-}] = S (0.1 + S) \approx S \times 0.1$

$S = \frac{K _{s p}}{0.1} = \frac{1.8 \times 1 0 ^{- 10}}{0.1} = 1.8 \times 1 0^{- 9} m o l d m^{- 3}$

Compare this to the solubility in pure water: $S = 1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5} m o l d m^{- 3}$ . The common ion reduces solubility by a factor of ~7000.

Definition of $K_{s p}$

Applications of Solubility Product

1. Barium Meal in Medical Imaging

$B a S O_{4}$ has an extremely low $K_{s p}$ ( $\approx 1.1 \times 1 0^{- 10}$ ), making it virtually insoluble in body fluids. Although barium ions ( $B a^{2 +}$ ) are toxic, the concentration of dissolved $B a^{2 +}$ from $B a S O_{4}$ is negligibly small and harmless. Its high atomic mass makes it opaque to X-rays, providing excellent contrast for imaging the gastrointestinal tract.

2. Qualitative Analysis (Selective Precipitation)

In qualitative analysis, specific reagents are added to selectively precipitate certain ions while leaving others in solution. For example:

Adding $H_{2} S$ in acidic conditions precipitates Group II cations (e.g., $P b S$ , $C u S$ ) because their $K_{s p}$ values are extremely small.
Group III cations (e.g., $Z n S$ ) have larger $K_{s p}$ values and only precipitate in basic conditions.

3. Water Treatment

Phosphate ions ( $P O_{4}^{3 -}$ ) in wastewater can be removed by adding $C a^{2 +}$ ions, precipitating insoluble $C a_{3} (P O_{4})_{2}$ :

$3 C a^{2 +} (a q) + 2 P O_{4}^{3 -} (a q) \to C a_{3} (P O_{4})_{2} (s)$

4. Salt Purification (Common Ion Effect)

Impure $N a C l$ dissolved in water can be purified by passing $H C l$ gas through the solution. The common ion $C l^{-}$ from $H C l$ causes $N a C l$ to precipitate out, leaving more soluble impurities in solution.

Summary

Concept	Key Relationship
$K_{s p}$ definition	$K_{s p} = [A^{y +}]^{x} [B^{x -}]^{y}$
Molar solubility (1:1 salt)	$S = K_{s p}$
Molar solubility ( $M X_{2}$ type)	$K_{s p} = 4 S^{3}$
Precipitation condition	$Q_{s p} > K_{s p}$
Common ion effect	Decreases solubility

Introduction

The Solubility Product ( $K_{s p}$ )

For a sparingly soluble salt $A_{x} B_{y}$ dissolving in water:

$A_{x} B_{y} (s) ⇌ x A^{y +} (a q) + y B^{x -} (a q)$

The solubility product expression is:

$K_{s p} = [A^{y +}]^{x} [B^{x -}]^{y}$

Note: The concentration of the pure solid $A_{x} B_{y}$ is constant and is incorporated into $K_{s p}$ , so it does not appear in the expression.

Key Points

$K_{s p}$ is a constant at a given temperature.
A smaller $K_{s p}$ means the salt is less soluble.
$K_{s p}$ only applies to saturated solutions of sparingly soluble salts.

Examples of $K_{s p}$ Expressions

Salt	Equilibrium	$K_{s p}$ Expression
$A g C l$	$A g C l ⇌ A g^{+} + C l^{-}$	$K_{s p} = [A g^{+}] [C l^{-}]$
$C a F_{2}$	$C a F_{2} ⇌ C a^{2 +} + 2 F^{-}$	$K_{s p} = [C a^{2 +}] [F^{-}]^{2}$
$A g_{2} C r O_{4}$	$A g_{2} C r O_{4} ⇌ 2 A g^{+} + C r O_{4}^{2 -}$	$K_{s p} = [A g^{+}]^{2} [C r O_{4}^{2 -}]$
$C a_{3} (P O_{4})_{2}$	$C a_{3} (P O_{4})_{2} ⇌ 3 C a^{2 +} + 2 P O_{4}^{3 -}$	$K_{s p} = [C a^{2 +}]^{3} [P O_{4}^{3 -}]^{2}$

Molar Solubility and $K_{s p}$

Molar solubility ( $S$ ) is the number of moles of a salt that dissolves per litre of solution to form a saturated solution.

Worked Example 1: $A g C l$

$A g C l (s) ⇌ A g^{+} (a q) + C l^{-} (a q)$

If solubility = $S$ : $[A g^{+}] = S$ , $[C l^{-}] = S$

$K_{s p} = S \times S = S^{2} ⟹ S = K_{s p}$

Worked Example 2: $A g_{2} C r O_{4}$

$A g_{2} C r O_{4} (s) ⇌ 2 A g^{+} (a q) + C r O_{4}^{2 -} (a q)$

If solubility = $S$ : $[A g^{+}] = 2 S$ , $[C r O_{4}^{2 -}] = S$

$K_{s p} = (2 S)^{2} (S) = 4 S^{3} ⟹ S = (\frac{K _{s p}}{4})^{1/3}$

Worked Example 3: $M X_{2}$ type salt

Given $K_{s p} = 4 \times 1 0^{- 12}$ :

$K_{s p} = 4 S^{3} ⟹ S^{3} = \frac{4 \times 1 0 ^{- 12}}{4} = 1 0^{- 12} ⟹ S = 1 0^{- 4} m o l d m^{- 3}$

The Ionic Product ( $Q_{s p}$ ) and Precipitation

The Ionic Product $Q_{s p}$ is calculated the same way as $K_{s p}$ , but using the actual concentrations of ions in any solution (not necessarily saturated).

Precipitation Rules

Condition	Meaning	Result
$Q_{s p} < K_{s p}$	Unsaturated	No precipitate; more salt can dissolve
$Q_{s p} = K_{s p}$	Saturated	Equilibrium; no change
$Q_{s p} > K_{s p}$	Supersaturated	Precipitate forms

Worked Example: Will a precipitate form?

If $[B a^{2 +}] = 0.001 m o l d m^{- 3}$ and $[S O_{4}^{2 -}] = 0.002 m o l d m^{- 3}$ are mixed, and $K_{s p} (B a S O_{4}) = 1.1 \times 1 0^{- 10}$ :

$Q_{s p} = [B a^{2 +}] [S O_{4}^{2 -}] = (0.001) (0.002) = 2 \times 1 0^{- 6}$

Since $Q_{s p} = 2 \times 1 0^{- 6} ≫ K_{s p} = 1.1 \times 1 0^{- 10}$ , a precipitate of $B a S O_{4}$ will form.

The Common Ion Effect on Solubility

Example: $A g C l$ in $N a C l$ solution

$A g C l (s) ⇌ A g^{+} (a q) + C l^{-} (a q)$

Adding $N a C l$ increases $[C l^{-}]$ . By Le Chatelier's Principle, the equilibrium shifts left, reducing the dissolution of $A g C l$ . Therefore, the solubility of $A g C l$ decreases.

Quantitative Treatment

If $A g C l$ is dissolved in $0.1 m o l d m^{- 3}$ $N a C l$ solution ( $K_{s p} (A g C l) = 1.8 \times 1 0^{- 10}$ ):

$K_{s p} = [A g^{+}] [C l^{-}] = S (0.1 + S) \approx S \times 0.1$

$S = \frac{K _{s p}}{0.1} = \frac{1.8 \times 1 0 ^{- 10}}{0.1} = 1.8 \times 1 0^{- 9} m o l d m^{- 3}$

Compare this to the solubility in pure water: $S = 1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5} m o l d m^{- 3}$ . The common ion reduces solubility by a factor of ~7000.

Adding $H_{2} S$ in acidic conditions precipitates Group II cations (e.g., $P b S$ , $C u S$ ) because their $K_{s p}$ values are extremely small.
Group III cations (e.g., $Z n S$ ) have larger $K_{s p}$ values and only precipitate in basic conditions.

Concept	Key Relationship
$K_{s p}$ definition	$K_{s p} = [A^{y +}]^{x} [B^{x -}]^{y}$
Molar solubility (1:1 salt)	$S = K_{s p}$
Molar solubility ( $M X_{2}$ type)	$K_{s p} = 4 S^{3}$
Precipitation condition	$Q_{s p} > K_{s p}$
Common ion effect	Decreases solubility