Common Ion Effect — Apply common ion effect to solubility equilibria - Chemistry 2

SECTION 1

Historical Context & Motivation

Understanding why some salts barely dissolve while others flood into solution has fascinated chemists for centuries. Early experimenters noticed something puzzling: a salt that dissolved readily in pure water became far less soluble when a related salt was already present. This observation hinted at an invisible tug-of-war between ions in solution, and it eventually led to one of the most practical rules in aqueous chemistry — the common ion effect.

1864 • Law of Mass Action

Norwegian chemists Cato Guldberg and Peter Waage proposed that the rate of a reaction depends on the concentrations of the reactants. Their Law of Mass Action provided the mathematical language needed to describe equilibrium, laying the groundwork for every equilibrium expression that followed.

1884 • Le Chatelier's Principle

Henry Louis Le Chatelier articulated his famous principle: when a system at equilibrium is disturbed, it shifts to partially counteract the disturbance. This idea explains why adding an ion already present in solution pushes the equilibrium back toward the solid.

1889 • Walther Nernst & Solubility Product

Walther Nernst formalized the solubility product constant (Ksp), giving chemists a single number to quantify how much of a sparingly soluble salt dissolves. With Ksp in hand, calculating the common ion effect became straightforward algebra.

1923 • Debye–Hückel Theory

Peter Debye and Erich Hückel showed that ion–ion interactions in solution cause real behavior to deviate from ideal predictions. Their work introduced activity coefficients, refining common-ion calculations for more concentrated solutions.

The central question this lesson addresses is: How does the presence of an ion that already exists in solution change the solubility of a sparingly soluble salt? By combining Ksp expressions with Le Chatelier's principle, you'll be able to predict — and calculate — exactly how much less solid dissolves when a common ion is introduced.

SECTION 2

Core Principles & Definitions

Before diving into calculations, you need a firm grip on a few foundational ideas. Each one builds on the last, so take them in order.

Solubility Equilibrium

When a sparingly soluble ionic solid sits in water, it establishes a dynamic equilibrium between the solid phase and its dissolved ions. The rate of dissolving equals the rate of precipitation, so the ion concentrations remain constant.

Solubility Product (Ksp)

The Ksp is a special equilibrium constant that equals the product of the ion concentrations, each raised to the power of its coefficient. A smaller Ksp means lower solubility.

Common Ion

A common ion is any ion that appears in two different solute sources in the same solution. For example, Cl⁻ is common if you dissolve AgCl in a solution that already contains NaCl.

Le Chatelier's Shift

Adding a common ion increases the product of ion concentrations (the ion product, Q) above Ksp. The system responds by shifting left — toward the solid — until Q = Ksp again. The result: less solid dissolves.

Molar Solubility

The molar solubility (often written as s) is the number of moles of a salt that dissolve per liter of solution. The common ion effect always decreases the molar solubility compared to dissolving in pure water.

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KEY TAKEAWAY

Think of a crowded concert venue with a maximum capacity (Ksp). If half the seats are already taken by people who arrived early (the common ion from another source), there are fewer seats left for the band's fans (the dissolving salt). The venue's capacity hasn't changed, but the number of new fans that can enter is reduced. Similarly, Ksp stays the same, but the molar solubility of the salt drops because some of the ion 'seats' are already occupied.

SECTION 3

Visualizing the Common Ion Effect

The diagram below compares what happens when silver chloride (AgCl) dissolves in pure water versus in a solution that already contains sodium chloride (NaCl). Notice how the common Cl⁻ ion shifts the equilibrium to the left, reducing the amount of AgCl that dissolves.

Beaker A shows AgCl dissolving in pure water, producing equal amounts of Ag⁺ (cyan) and Cl⁻ (pink). Beaker B already contains abundant Cl⁻ from NaCl (plus Na⁺ in gold), so the equilibrium shifts left, and far fewer Ag⁺ ions enter solution. Notice the molar solubility drops from 1.3 × 10⁻⁵ M to 1.8 × 10⁻⁹ M — nearly 10,000 times smaller.

In Beaker A, the only source of Cl⁻ is the dissolving AgCl itself, so the concentrations of Ag⁺ and Cl⁻ are equal. In Beaker B, NaCl supplies a large concentration of Cl⁻ before any AgCl dissolves. Because Ksp = [Ag⁺][Cl⁻] must remain constant at a given temperature, the extra Cl⁻ forces [Ag⁺] to shrink dramatically. This is the common ion effect in action: the shared Cl⁻ ion suppresses the dissolution of AgCl.

SECTION 4

Mathematical Framework

The math behind the common ion effect flows directly from the Ksp expression. You set up the same equilibrium you would for pure water, but one of the ion concentrations now has an additional contribution from the second solute.

GENERAL DISSOLUTION

MₐXᵦ (s) ⇌ a M^(m+) (aq) + b X^(n−) (aq)

M = cation, X = anion, a and b = stoichiometric coefficients, m+ and n− = ion charges.

SOLUBILITY PRODUCT EXPRESSION

Ksp = [M^(m+)]ᵃ × [X^(n−)]ᵇ

Ksp is a constant at a given temperature. Concentrations are in molarity (mol/L). The pure solid does not appear in the expression.

WITH A COMMON ION

Ksp = (s)ᵃ × (C₀ + s)ᵇ

s = molar solubility of the sparingly soluble salt. C₀ = initial concentration of the common ion from the second solute. When C₀ >> s, the approximation (C₀ + s) ≈ C₀ simplifies the algebra considerably.

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The Approximation Shortcut

In most common-ion problems the added ion concentration (C₀) is thousands of times larger than the molar solubility (s). That means C₀ + s ≈ C₀. This simplification lets you solve for s in one step instead of expanding a polynomial. Always check at the end: if s is less than 5% of C₀, the approximation is valid.

EXAMPLE: AgCl IN NaCl SOLUTION

1.8 × 10⁻¹⁰ = (s)(0.10 + s) ≈ (s)(0.10) → s = 1.8 × 10⁻⁹ M

Ksp of AgCl = 1.8 × 10⁻¹⁰. Here 0.10 M is the Cl⁻ from NaCl. Since s = 1.8 × 10⁻⁹ is negligible compared to 0.10, the approximation holds.

SECTION 5

Setting Up ICE Tables with a Common Ion

An ICE table (Initial–Change–Equilibrium) is the most organized way to set up common-ion problems. The only difference from a standard Ksp ICE table is that the initial concentration of the common ion is not zero — it equals C₀, the concentration supplied by the other solute.

	PbCl₂ (s)	Pb²⁺ (aq)	Cl⁻ (aq)
I (Initial)	— (solid)	0	0.20 M (from NaCl)
C (Change)	—	+ s	+ 2s
E (Equilibrium)	—	s	0.20 + 2s

ICE table for PbCl₂ dissolving in 0.20 M NaCl

For PbCl₂, the dissolution is: PbCl₂ (s) ⇌ Pb²⁺ (aq) + 2 Cl⁻ (aq). The Ksp expression becomes Ksp = [Pb²⁺][Cl⁻]². Substituting the equilibrium row: Ksp = (s)(0.20 + 2s)². If 2s is much smaller than 0.20, we approximate (0.20 + 2s) ≈ 0.20, giving Ksp ≈ s × (0.20)². Always write the ICE table first — it prevents sign errors and ensures correct stoichiometry.

Flowchart showing the three stages: (1) equilibrium in pure water where Q = Ksp, (2) the moment a common ion is added and Q exceeds Ksp, and (3) the new equilibrium established with a much smaller molar solubility.

SECTION 6

Worked Example

Let's calculate the molar solubility of lead(II) chloride (PbCl₂) in a 0.20 M NaCl solution. The Ksp of PbCl₂ at 25 °C is 1.7 × 10⁻⁵.

Molar Solubility of PbCl₂ in 0.20 M NaCl

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Step 1 — Write the Dissolution Equation

PbCl₂ (s) ⇌ Pb²⁺ (aq) + 2 Cl⁻ (aq). Each formula unit that dissolves produces one Pb²⁺ ion and two Cl⁻ ions. The Ksp expression is Ksp = [Pb²⁺][Cl⁻]².

Ksp = [Pb²⁺][Cl⁻]² = 1.7 × 10⁻⁵
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Step 2 — Identify the Common Ion

NaCl dissociates completely: NaCl → Na⁺ + Cl⁻. The Cl⁻ ion is common to both NaCl and PbCl₂. Before any PbCl₂ dissolves, [Cl⁻] is already 0.20 M from NaCl.

Common ion: Cl⁻, initial [Cl⁻] = 0.20 M
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Step 3 — Build the ICE Table

Let s = molar solubility of PbCl₂. Initial [Pb²⁺] = 0, initial [Cl⁻] = 0.20 M. Change: [Pb²⁺] increases by s, [Cl⁻] increases by 2s. Equilibrium: [Pb²⁺] = s, [Cl⁻] = 0.20 + 2s.

[Pb²⁺] = s, [Cl⁻] = 0.20 + 2s
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Step 4 — Apply the Approximation

Assume 2s << 0.20, so 0.20 + 2s ≈ 0.20. Substitute into the Ksp expression: 1.7 × 10⁻⁵ = (s)(0.20)². This gives 1.7 × 10⁻⁵ = s × 0.040.

s = (1.7 × 10⁻⁵) ÷ 0.040 = 4.3 × 10⁻⁴ M
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Step 5 — Validate the Approximation

Check: 2s = 2 × 4.3 × 10⁻⁴ = 8.6 × 10⁻⁴ M. Compare to 0.20 M: (8.6 × 10⁻⁴ ÷ 0.20) × 100% = 0.43%. This is well below 5%, so the approximation is valid.

s = 4.3 × 10⁻⁴ M (valid approximation — 0.43%)
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Step 6 — Compare with Pure Water

In pure water (no common ion), Ksp = (s)(2s)² = 4s³, so s = ³√(Ksp/4) = ³√(4.25 × 10⁻⁶) ≈ 1.62 × 10⁻² M. The common ion reduced the solubility from about 0.016 M to 0.00043 M — roughly 38 times less soluble!

Solubility decreased from 0.016 M to 4.3 × 10⁻⁴ M — a ~38× reduction

SECTION 7

Common Ion Effect vs. No Common Ion

It is easy to confuse the common ion effect with other factors that influence solubility. The table below highlights key contrasts between dissolving a salt in pure water versus in a solution that contains a common ion, and also distinguishes this concept from the effect of an unrelated salt.

Feature	Pure Water	Common Ion Present	Non-Common Ion (Salt Effect)
Initial [common ion]	0 M	C₀ > 0	0 M (different ion)
Effect on solubility	Normal (maximum)	Decreases solubility	Slightly increases solubility
Equilibrium shift	No shift (starts from scratch)	Shifts left → more solid	Slightly right (ionic atmosphere)
Ksp value	Constant	Constant	Constant (activity changes)
Typical magnitude	—	Can reduce s by 10–10,000×	Usually < 2× increase

Comparing solubility conditions

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KEY TAKEAWAY

The common ion effect is one of the most powerful tools for controlling precipitation in real life. Water-treatment plants exploit it to remove toxic heavy-metal ions, and your body uses it to regulate calcium phosphate levels in blood. Whenever you see a shared ion between two solutes, expect solubility to drop — often dramatically.

SECTION 8

Connection to Advanced Theory

At the high-school level, we assume that Ksp describes real solution behavior perfectly. In reality, ions interact with each other electrostatically, and those interactions become significant at higher concentrations. Advanced courses introduce activity (effective concentration) and activity coefficients (γ) to account for these non-ideal effects. The table below gives you a preview of how the ideas connect.

Concept	High School Level	College / AP Level
Equilibrium constant	Ksp uses molar concentrations	Ksp uses activities (a = γ × [ion])
Ionic strength	Not considered	I = ½ Σ cᵢzᵢ² affects γ
Common ion prediction	Solubility always decreases	Decreases, but actual s may be slightly higher than calculated due to γ < 1
Complex ion formation	Usually ignored	Can increase solubility (competing equilibrium)

High school vs. advanced treatment of solubility equilibria

Don't worry about mastering activities right now. The key takeaway is that the common ion effect always suppresses solubility, and the concentration-based Ksp approach gives you an excellent approximation for dilute solutions. When you encounter very concentrated solutions in future courses, you'll refine these calculations with activity coefficients.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Silver chromate (Ag₂CrO₄) is a sparingly soluble salt. If you dissolve it in a solution that already contains 0.050 M AgNO₃, will more or less Ag₂CrO₄ dissolve compared to pure water? Explain your reasoning using Le Chatelier's principle.

Show answer

Less Ag₂CrO₄ will dissolve. The dissolution equilibrium is Ag₂CrO₄ (s) ⇌ 2 Ag⁺ (aq) + CrO₄²⁻ (aq). The AgNO₃ supplies Ag⁺, a common ion. According to Le Chatelier's principle, adding Ag⁺ increases the ion product Q above Ksp, so the equilibrium shifts to the left (toward the solid). This means fewer moles of Ag₂CrO₄ dissolve, and the molar solubility is reduced.

Question 2: BASIC CALCULATION

Calculate the molar solubility of CaF₂ in pure water and in a 0.10 M NaF solution. Ksp of CaF₂ = 3.9 × 10⁻¹¹.

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In pure water: CaF₂ ⇌ Ca²⁺ + 2 F⁻. Ksp = (s)(2s)² = 4s³. So s = ³√(3.9 × 10⁻¹¹ / 4) = ³√(9.75 × 10⁻¹²) ≈ 2.1 × 10⁻⁴ M. In 0.10 M NaF: initial [F⁻] = 0.10 M. Ksp = (s)(0.10 + 2s)² ≈ (s)(0.10)² = s × 0.010. So s = 3.9 × 10⁻¹¹ / 0.010 = 3.9 × 10⁻⁹ M. The common ion reduced solubility by about 50,000 times.

Question 3: INTERMEDIATE

BaSO₄ has Ksp = 1.1 × 10⁻¹⁰. What is the molar solubility of BaSO₄ in a 0.025 M Na₂SO₄ solution? Verify that your approximation is valid.

Show answer

BaSO₄ ⇌ Ba²⁺ + SO₄²⁻. Ksp = [Ba²⁺][SO₄²⁻]. Let s = molar solubility. Initial [SO₄²⁻] = 0.025 M from Na₂SO₄. Equilibrium: [Ba²⁺] = s, [SO₄²⁻] = 0.025 + s. Assuming s << 0.025: 1.1 × 10⁻¹⁰ = s × 0.025, so s = 4.4 × 10⁻⁹ M. Check: s / 0.025 = 1.76 × 10⁻⁷ = 0.0000176%, which is far less than 5%. The approximation is excellent. Compare to pure water where s = √(1.1 × 10⁻¹⁰) = 1.05 × 10⁻⁵ M — a roughly 2,400-fold decrease.

Question 4: APPLIED

A wastewater treatment plant needs to remove Pb²⁺ from industrial effluent by precipitating PbCl₂. The current [Pb²⁺] is 0.050 M. If the plant adds enough NaCl to bring [Cl⁻] to 0.50 M, will PbCl₂ precipitate? (Ksp PbCl₂ = 1.7 × 10⁻⁵.) If so, what will the final [Pb²⁺] be once equilibrium is reached?

Show answer

First, calculate the ion product Q: Q = [Pb²⁺][Cl⁻]² = (0.050)(0.50)² = (0.050)(0.25) = 0.0125 = 1.25 × 10⁻². Since Q = 1.25 × 10⁻² >> Ksp = 1.7 × 10⁻⁵, yes, PbCl₂ will precipitate. At the new equilibrium with [Cl⁻] ≈ 0.50 M (excess Cl⁻ barely consumed), Ksp = [Pb²⁺](0.50)² gives [Pb²⁺] = 1.7 × 10⁻⁵ / 0.25 = 6.8 × 10⁻⁵ M. The plant reduced [Pb²⁺] from 0.050 M to about 6.8 × 10⁻⁵ M — removing over 99.8% of the lead.

Question 5: CRITICAL THINKING

Suppose you dissolve SrSO₄ (Ksp = 3.2 × 10⁻⁷) in a solution containing 0.10 M Na₂SO₄ and simultaneously add SrCl₂ to make [Sr²⁺] = 0.010 M. Both ions — Sr²⁺ and SO₄²⁻ — are now common ions. Calculate Q and determine whether SrSO₄ will precipitate further or dissolve. Then calculate the molar solubility of additional SrSO₄ in this mixed solution and explain the practical limitation of adding both ions.

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SrSO₄ ⇌ Sr²⁺ + SO₄²⁻. Q = [Sr²⁺][SO₄²⁻] = (0.010)(0.10) = 1.0 × 10⁻³. Since Q = 1.0 × 10⁻³ >> Ksp = 3.2 × 10⁻⁷, the solution is highly supersaturated and SrSO₄ will precipitate until Q = Ksp. Equilibrium concentrations: [Sr²⁺] = 0.010 − x and [SO₄²⁻] = 0.10 − x, where x is the amount precipitated. Since 0.010 M is the limiting reagent, most precipitation consumes Sr²⁺. At equilibrium, essentially all excess Sr²⁺ precipitates, and we solve Ksp = [Sr²⁺](0.10) ≈ [Sr²⁺] × 0.10, giving [Sr²⁺] ≈ 3.2 × 10⁻⁶ M. The practical limitation is that when both ions are common, any additional SrSO₄ has essentially zero molar solubility — both equilibrium ion concentrations are constrained by outside sources, leaving no room for additional dissolution.

SUMMARY

Lesson Summary

Summary

The common ion effect states that the solubility of a sparingly soluble salt decreases when a solution already contains one of its ions from another source. This occurs because Le Chatelier's principle drives the equilibrium toward the solid (left) when the ion product Q exceeds Ksp. The Ksp value itself never changes at a given temperature — only the molar solubility changes.

To solve common-ion problems, write the dissolution equation, set up an ICE table with the common ion's initial concentration (C₀) already filled in, substitute into the Ksp expression, and apply the approximation C₀ + s ≈ C₀ when s is small. Always validate by checking that s is less than 5% of C₀. Real-world applications include water treatment, medical chemistry, and qualitative analysis in the lab.