Ksp: Common Ion Effect: Videos & Practice Problems

The common ion effect reduces the solubility of an ionic solid in a solution that already contains an ion or ions similar to those in the solid. This is due to Le Chatelier's principle, which states that a system in equilibrium will adjust to minimize the impact of any changes. When an ionic solid is introduced into pure water, it dissolves up to its solubility limit as determined by its solubility product constant (K_SP). However, if the solution already contains common ions, the solid's solubility decreases because the equilibrium shifts to favor the reverse reaction, thus less of the solid dissolves. This effect also applies to acids and bases, where the presence of hydroxide ions or hydrogen ions (H⁺ or H₃O⁺) in a solution decreases the solubility of bases or acids, respectively. Understanding the common ion effect is essential for predicting solubility changes in various chemical contexts.

Common Ion Effect

The common ion effect is a significant concept in solubility and equilibrium, particularly in the context of ionic solids dissolving in solutions. This phenomenon occurs when an ionic solid is introduced into a solution that already contains one or more ions that are common to the solid. The presence of these common ions leads to a decrease in the solubility of the solid, which can be explained through Le Chatelier's principle. This principle states that if a system at equilibrium experiences a disturbance, the equilibrium will shift in a direction that counteracts the disturbance.

For instance, consider barium sulfide (BaS) dissolving in pure water. In this scenario, BaS dissociates into barium ions (Ba²⁺) and sulfide ions (S^2-) without any initial presence of these ions in the solution. The solubility of BaS in pure water is determined by its solubility product constant (K_sp), which indicates the maximum concentration of ions that can exist in equilibrium with the solid.

Now, if we introduce barium sulfate (BaSO₄) into a solution that already contains barium and sulfide ions, the situation changes. The existing ions in the solution create a common ion effect, which limits the further dissolution of BaSO₄. Since the solution already has Ba²⁺ and S^2- ions, the equilibrium shifts to favor the formation of the solid, thereby reducing the amount of BaSO₄ that can dissolve. This shift occurs because the system attempts to reduce the concentration of the common ions, thus maintaining equilibrium.

In summary, the common ion effect illustrates how the solubility of an ionic solid is influenced by the presence of common ions in the solution. When these ions are present, the dissolution of the solid is restricted, leading to a lower solubility than would be observed in pure water. Understanding this effect is crucial for predicting the behavior of ionic compounds in various chemical environments, particularly in relation to their K_sp values.

Common Ion Effect Example

To determine the molar solubility of copper(II) carbonate in a 0.15 M magnesium carbonate solution, we start by recognizing that copper(II) carbonate dissociates into copper(II) ions and carbonate ions in solution. The dissociation can be represented as:

\[\text{CuCO}_3 (s) \rightleftharpoons \text{Cu}^{2+} (aq) + \text{CO}_3^{2-} (aq)\]

Given the solubility product constant (K_sp) for copper(II) carbonate is 2.4 × 10^-10, we can set up an ICE (Initial, Change, Equilibrium) chart to analyze the system. In this case, we ignore the solid reactant in the ICE chart, focusing only on the ions in solution.

Initially, the concentration of carbonate ions from magnesium carbonate is 0.15 M, while the concentration of copper(II) ions is 0 M. The ICE chart can be filled out as follows:

Species	Initial (M)	Change (M)	Equilibrium (M)
Cu2+	0	+x	x
CO32-	0.15	+x	0.15 + x

At equilibrium, the expression for K_sp is given by:

\[K_{sp} = [\text{Cu}^{2+}][\text{CO}_3^{2-}] = x(0.15 + x)\]

Since x is expected to be very small compared to 0.15, we can simplify the expression to:

\[K_{sp} \approx x(0.15)\]

Substituting the known value of K_sp:

\[2.4 \times 10^{-10} = x(0.15)\]

To find x, we divide both sides by 0.15:

\[x = \frac{2.4 \times 10^{-10}}{0.15} \approx 1.6 \times 10^{-9} \text{ M}\]

This value of x represents the molar solubility of copper(II) carbonate in the magnesium carbonate solution. Therefore, the molar solubility of copper(II) carbonate is approximately 1.6 × 10^-9 M.

Common Ion Effect in Acids and Bases

The common ion effect is a significant concept in chemistry that describes how the solubility of a substance is influenced by the presence of a common ion in solution. This effect is not limited to ionic solids; it also applies to acids and bases. When a base is dissolved in a solution that already contains hydroxide ions (OH^-), its solubility decreases. This is because the equilibrium shifts to favor the formation of the undissociated base, reducing the amount that can dissolve.

Similarly, for acids, the presence of hydronium ions (H₃O⁺), which can also be represented simply as H⁺, leads to a decrease in solubility. The equilibrium of the acid dissociation reaction shifts to the left, favoring the formation of the undissociated acid. This illustrates that the common ion effect plays a crucial role in the behavior of both acids and bases in solution, highlighting the interconnectedness of chemical equilibria.

Solubility at Buffered pH Example

To determine the solubility of chromium(III) hydroxide in grams per milliliter, we start with the solubility product constant (K_sp) value of 6.7 × 10⁻³¹ at a buffered pH of 8.4. The dissociation of chromium(III) hydroxide in water can be represented as:

Cr(OH)₃ (s) ⇌ Cr³⁺ (aq) + 3 OH⁻ (aq)

Since K_sp is involved, we set up an ICE (Initial, Change, Equilibrium) chart. The initial concentration of hydroxide ions (OH⁻) is calculated from the pH. First, we convert pH to pOH:

pOH = 14 - pH = 14 - 8.4 = 5.6

Next, we find the concentration of hydroxide ions:

[OH⁻] = 10^−pOH = 10^−5.6 ≈ 2.51 × 10⁻⁶ M

In the ICE chart, the initial concentration of Cr³⁺ is 0, while the initial concentration of OH⁻ is 2.51 × 10⁻⁶ M. As the reaction proceeds, we denote the change in concentration of Cr³⁺ as +x and for OH⁻ as +3x. The equilibrium concentrations are then:

[Cr³⁺] = x

[OH⁻] = 2.51 × 10⁻⁶ + 3x

Using the K_sp expression:

K_sp = [Cr³⁺][OH⁻]³

Substituting the equilibrium concentrations into the K_sp expression gives:

6.7 × 10⁻³¹ = x(2.51 × 10⁻⁶ + 3x)³

Assuming that 3x is negligible compared to 2.51 × 10⁻⁶, we simplify to:

6.7 × 10⁻³¹ = x(2.51 × 10⁻⁶)³

Calculating (2.51 × 10⁻⁶)³ yields approximately 1.58 × 10⁻¹⁷. Thus, we can solve for x:

x = (6.7 × 10⁻³¹) / (1.58 × 10⁻¹⁷) ≈ 4.227 × 10⁻¹⁴ M

This value represents the molar solubility of chromium(III) hydroxide. To convert this to grams per milliliter, we use the molar mass of chromium(III) hydroxide, which is approximately 103.02 g/mol:

Solubility (g/mL) = (4.227 × 10⁻¹⁴ mol/L) × (103.02 g/mol) × (1 L / 1000 mL)

Calculating this gives:

Solubility ≈ 4.35 × 10⁻¹⁵ g/mL

This final value represents the solubility of chromium(III) hydroxide in a buffered solution at the specified conditions.

Which of the following compounds will become less soluble in a chloride containing solution?
a) Na₂S
b) AgCl
c) BaI₂
d) CaF₂

• A. Na₂S
• B. AgCl
• C. BaI₂
• D. CaF₂

Answer: (B)

A solution of Ba(OH)₂ has a K_sp of 5.0 x 10⁻³.
i) Determine the pH of this solution.
ii) Determine the pH if Ba(OH)₂ was added to a solution containing 3.2 M of BaF₂ and 0.94 M of Al(OH)₃.

• A. i. 2.30
  ii. 14.03
• B. i. 0.67
  ii. 13.55
• C. i. 13.33
  ii. 14.45
• D. i. 2.30
  ii. 13.65

Answer: (C)

Here’s what students ask on this topic:

What is the common ion effect?

The common ion effect refers to the phenomenon where the solubility of a salt is reduced when another compound that shares a common ion is added to the same solution. This effect is a consequence of Le Chatelier's principle, which states that a system at equilibrium will react to counteract any changes imposed upon it.

For example, consider a solution of silver chloride (AgCl), which is only slightly soluble in water. If you add sodium chloride (NaCl) to the solution, the concentration of chloride ions (Cl^-) will increase because NaCl is highly soluble and dissociates completely. Since AgCl and NaCl share the common ion Cl^-, the increased concentration of Cl^- from the NaCl will shift the equilibrium of the slightly soluble AgCl to the left, according to the equilibrium equation:

AgCl(s) → Ag⁺(aq) + Cl^-(aq)

As a result, more AgCl will precipitate out of the solution to re-establish equilibrium by reducing the concentration of Cl^- ions. This is the common ion effect in action, where the addition of a common ion decreases the solubility of an ionic compound in a solution.

The addition of which compound will cause a shift in equilibrium because of the common ion effect?

The addition of a compound that contains a common ion will cause a shift in equilibrium due to the common ion effect. This effect is based on Le Chatelier's Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

For example, consider the equilibrium established in a saturated solution of calcium sulfate (CaSO₄):

CaSO₄ (s) → Ca²⁺ (aq) + SO₄^2- (aq)

If you add a compound like sodium sulfate (Na₂SO₄), which also contains the sulfate ion (SO₄^2-), you introduce more sulfate ions into the solution. According to the common ion effect, the equilibrium will shift to the left to reduce the concentration of the common sulfate ions by precipitating more calcium sulfate. This results in a decrease in the solubility of calcium sulfate in the solution. The common ion effect can be applied to any equilibrium involving ions in solution, where the addition of a common ion will always shift the equilibrium away from the side with the common ion.

Why does the common ion effect decrease solubility?

The common ion effect refers to the decrease in solubility of an ionic compound when a solution already contains one of the ions present in the compound. This occurs due to Le Chatelier's principle, which states that if a dynamic equilibrium is disturbed, the system will adjust to minimize that disturbance.

When you introduce a common ion into the solution, you are effectively increasing the concentration of that ion. For a saturated solution in equilibrium, this addition shifts the equilibrium to favor the formation of the solid precipitate in order to reduce the concentration of the common ion in the solution, thus decreasing the solubility of the compound.

This is because the solubility product (K_sp) of the ionic compound—a constant value at a given temperature—must be maintained. The presence of the additional common ion means that less of the ionic compound can dissolve to reach the required product of ion concentrations dictated by the K_sp. Consequently, the solubility of the compound decreases in the presence of a common ion.