Chemistry: Atoms First 2e · Equilibria of Other Reaction Classes

Coupled Equilibria

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

A single equilibrium can be misleading: silver chloride looks "insoluble" (K_sp = 1.8 × 10^-10) and calcium carbonate barely soluble, yet both dissolve readily under the right conditions. The explanation is that equilibria rarely act alone. are two or more equilibria that share a species, so shifting one shifts the other. When reactions are added, their equilibrium constants multiply, producing a net constant that can be enormously larger than any single one. This is how acid dissolves limestone (K_sp × 1/K_a2 turns a tiny constant into a favorable one), how ammonia dissolves silver chloride (K_sp × K_f couples solubility to ), and how hydroxides like Al(OH)₃ dissolve in both acid and strong base. The same logic underlies buffers — the acid and its conjugate base are two coupled equilibria sharing H⁺ — and ties together the K_sp and Lewis acid–base chemistry of the previous topics.

Why this matters

Coupled equilibria explain real chemistry that single-constant thinking gets wrong. Rainwater (slightly acidic from dissolved CO₂) dissolves limestone, carving caves and creating sinkholes; acid rain accelerates the same coupled reaction, weathering carbonate rock and monuments. As oceans absorb CO₂, the extra H⁺ shifts the carbonate equilibria, making it harder for shell-building organisms to deposit CaCO₃. Antacids like calcium carbonate and magnesium hydroxide must dissolve in the strongly acidic stomach, and chronic acidosis weakens bone because bone mineral dissolves when local pH drops — both coupled-equilibria phenomena. In the lab, chemists dissolve "insoluble" precipitates on purpose: AgCl in ammonia is a classic qualitative-analysis step, and scale removal from boilers and pipes uses acid on carbonate and hydroxide deposits.

The college version

Core Concepts

What "coupled" means

Two equilibria are coupled when a species produced by one is consumed by the other. CaCO₃(s) ⇌ Ca²⁺ + CO₃^2- produces carbonate; if acid is present, CO₃^2- + H⁺ ⇌ HCO₃^- consumes it. Removing the product of the first equilibrium pulls it right by Le Châtelier's principle, so more solid dissolves than K_sp alone suggests. The acid works through the shared carbonate ion, not by attacking the solid in one step.

The net reaction and the net constant

Add the two reactions and cancel the shared species:

$CaCO 3 (s) ⇌ Ca 2+ + CO 3 2- K 1 = K sp$

$CO 3 2- + H + ⇌ HCO 3 - K 2 = 1 K a2 = 1 4.7 \times 10 -11 = 2.1 \times 1010$

Sum: CaCO₃(s) + H⁺ ⇌ Ca²⁺ + HCO₃^- with:

$K net = K 1 \times K 2 = K sp \times 1 K a2 = (3.3 \times 10 -9)(2.1 \times 1010) = 70$

Why multiply? The mass-action expression for the summed reaction is the product of the two individual expressions. Two rules follow: adding reactions multiplies their constants, and reversing a reaction inverts its constant (K_rev = 1/K). With K_net = 70, dissolution is favorable even though K_sp alone is tiny — dividing by the small K_a2 amplified the constant by ten orders of magnitude.

Dissolving precipitates with acid

Any salt whose anion is the conjugate base of a weak acid can be dissolved by strong acid, which removes the anion from solution: carbonates, hydroxides, sulfides, and phosphates all behave this way. The more basic the anion, the more effective acid is at dissolving the salt — which is why CaCO₃, Mg(OH)₂, and ZnS all dissolve in strong acid despite tiny K_sp values.

Dissolving precipitates by complexation

A precipitating ion can also be removed by turning it into a complex. Silver chloride dissolves in ammonia:

$AgCl (s) + 2 NH 3 ⇌ [Ag(NH 3) 2] + + Cl - K net = K sp \times K f = 3.1 \times 10 -3$

The free Ag⁺ produced by dissolution is captured by ammonia (a Lewis base), starving the equilibrium of product and pulling it right — the previous topic's Lewis chemistry applied to solubility.

Amphoteric hydroxides

Hydroxides like Al(OH)₃ and Zn(OH)₂ dissolve in both acid and strong base, because two different coupled paths remove their ions:

In acid: Al(OH)₃(s) + 3H⁺ ⇌ Al³⁺ + 3H₂O (OH⁻ is consumed by H⁺).
In strong base: Al(OH)₃(s) + OH^- ⇌ [Al(OH)₄]^- (Al³⁺ is captured as a hydroxo complex).

A substance that reacts as both acid and base this way is amphoteric. Between these extremes, at intermediate pH, Al(OH)₃ precipitates — the "amphoteric minimum" used to separate aluminum in qualitative analysis.

Buffers as coupled equilibria

Even the buffer of Chapter 14 is a coupled system: the weak acid and its conjugate base are two equilibria sharing H⁺, and the Henderson–Hasselbalch equation is the algebraic result of coupling HA ⇌ H⁺ + A^- with water autoionization. Viewing buffers, solubility, and complexation through the same "shared species" lens is the payoff of this chapter.

How It Works / Step-by-Step Process

Combining two equilibria

Write both equilibria, making sure the shared species appears as a product of one and a reactant of the other.
Add the reactions; cancel the shared species to get the net reaction.
Multiply the constants (invert any reaction you reversed) to get K_net.
Use K_net like any equilibrium constant — with an ICE table if concentrations are needed.

Predicting whether a precipitate dissolves

Identify the ion the dissolving agent removes (anion by acid, cation by ligand).
Write the removal equilibrium (protonation: K = 1/K_a; complexation: K = K_f).
Combine: K_net = K_sp × (removal constant).
If K_net is not tiny (say 10^-3 or larger), dissolution is significant.

Common Confusions

Do Not Confuse	With	The Difference
Coupled equilibria	Independent simultaneous equilibria	Coupled equilibria share a species, so shifting one shifts the other; independent ones do not interact
K_net = K₁ × K₂	K_net = K₁ + K₂	Constants multiply when reactions add (mass-action expressions multiply); they never add
Forward constant	Reverse constant	Reversing a reaction inverts its constant: K_rev = 1/K
Common-ion effect	Coupled equilibria	Common ion: one equilibrium, product added directly (AgCl in NaCl). Coupled: two equilibria, shared species removed (AgCl in NH₃)
Amphoteric	Amphiprotic	Amphoteric = reacts as acid and base generally; amphiprotic = specifically donates and accepts a proton (HCO₃^-)
"Insoluble" salt	A salt that never dissolves	Every salt has finite solubility; coupling can multiply effective solubility thousands of times (AgCl in NH₃) or make it essentially complete (Mg(OH)₂ in acid)
K_sp alone	Whether the salt dissolves in a real system	Real systems contain acids, ligands, and other ions; always check for coupling partners

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine a bucket with a small hole: water leaks out slowly (the salt dissolving). If you put a sponge under the hole that soaks up every drop (the acid or ammonia grabbing the ions), the bucket drains much faster — even though the hole is the same size. Coupled equilibria work like that: one reaction keeps removing what the other makes, so things dissolve that looked impossible.

Worked example

Example 1: Calcium carbonate in acid — the limestone problem

Compute the net constant for dissolving calcite in acid. Constants: K_sp(CaCO₃) = 3.3 × 10^-9 (calcite; values vary slightly with polymorph), and the second acid dissociation of carbonic acid, K_a2(H₂CO₃) = 4.7 × 10^-11.

Reaction 1 — dissolution:

$CaCO 3 (s) ⇌ Ca 2+ (aq) + CO 3 2- (aq) K 1 = 3.3 \times 10 -9$

Reaction 2 — protonation of carbonate (the reverse of the K_a2 step):

$CO 3 2- (aq) + H + (aq) ⇌ HCO 3 - (aq) K 2 = 1 K a2 = 1 4.7 \times 10 -11 = 2.1 \times 1010$

Add them (carbonate cancels) and multiply the constants:

$K net = K sp \times 1 K a2 = (3.3 \times 10 -9)(2.1 \times 1010) = 69 \approx 7 \times 101$

With K_net ≈ 70, the reaction is favorable: in acid, calcium carbonate dissolves steadily — dissolving antacid tablets in the stomach, carving limestone caves, and eroding carbonate rock. In neutral water (little H⁺), only the tiny K_sp path operates and dissolution is negligible.

Example 2: Silver chloride in ammonia — how much dissolves?

Constants: K_sp(AgCl) = 1.8 × 10^-10, K_f([Ag(NH₃)₂]⁺) = 1.7 × 10⁷.

Combine the two equilibria:

$AgCl (s) ⇌ Ag + + Cl - K sp = 1.8 \times 10 -10$

$Ag + + 2 NH 3 ⇌ [Ag(NH 3) 2] + K f = 1.7 \times 107$

Net:

$AgCl (s) + 2 NH 3 (aq) ⇌ [Ag(NH 3) 2] + (aq) + Cl - (aq)$

$K net = K sp \times K f = (1.8 \times 10 -10)(1.7 \times 107) = 3.1 \times 10 -3$

Now find how much AgCl dissolves in 1.0 M ammonia. Let x = molar solubility; then [[Ag(NH₃)₂]⁺] = x, [Cl^-] = x, and [NH₃] = 1.0 - 2x:

$K net = x \cdot x (1.0 - 2x) 2 = 3.1 \times 10 -3$

Take the square root of both sides:

$x 1.0 - 2x = \sqrt 3.1 \times 10 -3 = 0.056$

$x = 0.056(1.0 - 2x) \Rightarrow x + 0.112x = 0.056 \Rightarrow x = 0.050 M$

About 0.05 mol of AgCl dissolves per liter of 1.0 M ammonia — roughly 4,000 times the pure-water solubility of 1.3 × 10^-5 M. This is the classic qualitative-analysis step: precipitate AgCl, then confirm its identity by dissolving it in ammonia.

Example 3: Why milk of magnesia dissolves in the stomach

Magnesium hydroxide has K_sp ≈ 8.9 × 10^-12 (values near 10^-12 appear in standard tables). In the strongly acidic stomach, the coupled path is:

$Mg(OH) 2 (s) ⇌ Mg 2+ + 2 OH - K sp$

$2 OH - + 2 H + ⇌ 2 H 2 O K = (1 K w) 2$

Net: Mg(OH)₂(s) + 2H⁺ ⇌ Mg²⁺ + 2H₂O, with:

$K net = K sp \times 1 K w 2 = 8.9 \times 10 -12 (1.0 \times 10 -14) 2 = 8.9 \times 1016$

A net constant on the order of 10¹⁷ means the reaction runs essentially to completion: the solid dissolves, releasing Mg²⁺ and consuming H⁺ — exactly how the antacid neutralizes stomach acid. The same logic applies to amphoteric Al(OH)₃, which dissolves in acid by this path and in strong base via [Al(OH)₄]^- formation; its "insolubility" is only real in the middle pH range.

Key takeaways

Coupled equilibria share a species; removing a product of one drives the other via Le Châtelier.
Adding reactions multiplies their constants: K_net = K₁ × K₂. Reversing a reaction inverts its constant.
Acid dissolves salts whose anion is a weak-acid conjugate base: carbonates, hydroxides, sulfides, phosphates.
For carbonate salts: K_net = K_sp/K_a2 — dividing by a tiny number turns "insoluble" into "dissolves readily" (CaCO₃: K_net ≈ 70).
Complexation dissolves precipitates: AgCl + 2NH₃ ⇌ [Ag(NH₃)₂]⁺ + Cl^-, K_net = K_sp × K_f = 3.1 × 10^-3.
Amphoteric hydroxides (Al(OH)₃, Zn(OH)₂) dissolve in both acid and strong base via different coupled paths.
Antacids work because acid-coupled dissolution is favorable; chronic acidosis dissolves bone mineral for the same reason.
Buffers are coupled equilibria sharing H⁺ — one framework ties the whole chapter together.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

State the rule for combining equilibrium constants when two reactions are added. What about when a reaction is reversed?
Show answer
Adding reactions multiplies their constants: K_net = K₁ × K₂. Reversing a reaction inverts its constant: K_rev = 1/K.
Write the two equilibria and the net constant for dissolving AgCl in ammonia.
Show answer
AgCl(s) ⇌ Ag⁺ + Cl^- (K_sp) and Ag⁺ + 2NH₃ ⇌ [Ag(NH₃)₂]⁺ (K_f); net: AgCl(s) + 2NH₃ ⇌ [Ag(NH₃)₂]⁺ + Cl^-, K_net = K_sp × K_f = 3.1 × 10^-3.
Why does acid dissolve CaCO₃ even though its K_sp is only 3.3 × 10^-9?
Show answer
Acid consumes the carbonate ion produced by dissolution (CO₃^2- + H⁺ ⇌ HCO₃^-), removing the product so the equilibrium shifts right. The combined constant K_sp/K_a2 ≈ 70 is favorable even though K_sp alone is tiny.
Compute K_net for CaCO₃(s) + H⁺ ⇌ Ca²⁺ + HCO₃^- using K_sp = 3.3 × 10^-9 and K_a2 = 4.7 × 10^-11.
Show answer
K_net = 3.3 × 10^-9 / (4.7 × 10^-11) = 7.0 × 10¹ ≈ 70.
Explain why Al(OH)₃ dissolves in both acid and strong base.
Show answer
In acid, H⁺ consumes the OH⁻ product (dissolution via Al³⁺). In strong base, OH⁻ complexes the Al³⁺ product as [Al(OH)₄]^-. Both paths remove a product of the dissolution equilibrium, pulling it right by different coupled reactions.
Why is it wrong to call a salt with a very small K_sp "completely insoluble" in a real sample?
Show answer
Because real samples contain coupling partners — acids, ligands, or other ions — that remove the dissolved ions and can multiply effective solubility by orders of magnitude (e.g., AgCl dissolves ~4,000× better in 1 M NH₃).

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

coupled equilibria: Two or more equilibria connected by a shared species
net (overall) reaction: The sum of coupled reactions after canceling shared species
net equilibrium constant: K_net = K₁ × K₂ for summed reactions
amphoteric: Able to react as both an acid and a base
complexation: A Lewis base (ligand) binding a metal ion to form a complex
hydrolysis: Reaction of an ion with water, e.g., CO₃^2- with H⁺ from water
Kf: Formation constant for complex-ion formation

Sources & references

openstax.org — Chemistry Atoms First 2e

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