Chemistry 2e · Equilibria of Other Reaction Classes

Coupled Equilibria

8 min read

Constants are standard textbook values at 25 °C (AgCl K_sp = 1.8 × 10^-10; [ Ag(NH₃)₂]⁺ K_f = 1.7 × 10⁷; K_w = 1.0 × 10^-14; K_a2(H₂CO₃) = 4.7 × 10^-11; Mg(OH)₂ K_sp ≈ 5.6 × 10^-12; CaCO₃ K_sp ≈ 4.5 × 10^-9). Tabulated values vary slightly among references.

On this page 9 sections

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

In 30 seconds

Rarely does a single equilibrium operate alone. A sparingly soluble solid sits in water containing acids, complexing agents, or other ions, and each of those species sets up its own equilibrium with the ions from the solid. When two or more equilibria share a species, they are coupled: changing one shifts the others. The key tools for handling are simple rules: add the individual reactions to get the , and multiply their equilibrium constants to get the net constant.

These rules explain remarkable chemistry. Silver chloride, nearly insoluble in pure water (K_sp = 1.8 × 10^-10), dissolves readily in ammonia because removes Ag⁺ from solution. Magnesium hydroxide, the antacid in milk of magnesia, dissolves in stomach acid because H⁺ removes OH⁻. hydroxides such as Al(OH)₃ dissolve in both acid and excess base. In every case, one equilibrium "pulls" another along.

Why this matters

Coupled equilibria control natural waters, biology, and industry. Limestone caves and sinkholes form because CO₂-laden rainwater acidifies groundwater, which dissolves CaCO₃ — a coupled dissolution/acid reaction. Ocean acidification weakens coral and shell formation through the same chemistry. In the body, the carbonate buffer, calcium phosphate in bone, and kidney-stone formation all involve coupled equilibria among solubility, acid–base, and complexation reactions. Industrially, gold is leached from ore with cyanide because Au⁺ forms an extremely stable complex ion, and water-treatment plants rely on coupled equilibria to precipitate and remove metals. Understanding how to combine reactions and constants lets you predict when a "insoluble" substance will, in fact, dissolve.

The college version

Core Concepts

Adding reactions multiplies their constants

Suppose reaction 1 has constant K₁ and reaction 2 has constant K₂. If the net reaction is reaction 1 plus reaction 2, then

$K net = K 1 \times K 2$

This follows directly from the mass-action expressions: when you multiply two equilibrium expressions, species that appear as products in one and reactants in the other cancel. If a reaction must be reversed before adding, use its reciprocal constant, 1/K. If it must be doubled, square the constant.

Complexation dissolves precipitates

Silver chloride in ammonia involves the dissolution equilibrium

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

coupled to the complexation equilibrium

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

Adding the two equations cancels Ag⁺ and gives the net reaction

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

with

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

The net constant is small, but it is not zero: ammonia dissolves a little AgCl, and in concentrated ammonia enough dissolves to matter — the basis of the classic lab test that distinguishes AgCl from AgI.

Acids dissolve insoluble hydroxides and carbonates

Acid removes OH⁻ or CO₃²⁻ as fast as the solid releases them, dragging the dissolution equilibrium forward. For magnesium hydroxide,

$Mg(OH) 2 (s) ⇌ Mg 2+ (aq) + 2 OH - (aq) K sp$

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

so the net reaction Mg(OH)₂(s) + 2H⁺(aq) ⇌ Mg²⁺(aq) + 2H₂O(l) has

$K net = K sp K w 2$

Because K_w = 1.0 × 10^-14 is tiny, dividing by K_w² produces an enormous net constant: essentially any hydroxide dissolves in strong acid. Sulfides dissolve in acid by the same logic, except that the S²⁻ released is consumed by two protonation steps, so K_net = K_sp/(K_a1K_a2) of H₂S.

Amphoteric hydroxides dissolve in both acid and base

Al(OH)₃ and Zn(OH)₂ react with H⁺ (base dissolves them) and with excess OH⁻ (forming aluminate or zincate ions):

$Al(OH) 3 (s) + OH - (aq) ⇌ [Al (OH) 4] - (aq)$

$Zn(OH) 2 (s) + 2 OH - (aq) ⇌ [Zn (OH) 4] 2- (aq)$

The result is a U-shaped solubility curve: solubility is high in acid, drops to a minimum near neutral pH, and rises again in concentrated base. This behavior is used to separate Al³⁺ from other metal ions in qualitative analysis.

Coupled equilibria in the environment

Dissolved CO₂ sets up a chain of coupled equilibria:

$CO 2 (g) ⇌ CO 2 (aq) ⇌ H 2 CO 3 (aq) ⇌ H + + HCO 3 - ⇌ 2 H + + CO 3 2-$

Adding H⁺ (from more CO₂, as in ocean acidification) drives the last step left, lowering CO₃²⁻ concentration and forcing CaCO₃ to dissolve — the same net reaction that erodes limestone and threatens coral reefs.

How It Works / Step-by-Step Process

Write every equilibrium that involves the species of interest (dissolution, complexation, acid–base).
Decide which direction each reaction must run so that the unwanted ion (e.g., Ag⁺, OH⁻) cancels.
Reverse or multiply reactions as needed, adjusting constants (1/K for reversal, powers for multiples).
Add the reactions and multiply the adjusted constants to get K_net.
Interpret K_net: if it is comfortably larger than about 10^-3, the coupled process is significant at typical concentrations.

Common Confusions

Common Confusion	Correct Understanding
"A tiny K_sp means the solid can never dissolve."	Coupling with complexation or acid can produce large K_net; AgCl and Mg(OH)₂ are everyday counterexamples.
"To combine equilibria, add the constants."	Constants multiply when reactions are added; only the reactions add.
"When I reverse a reaction, the constant stays the same."	Reversing inverts the constant: K → 1/K.
"K_net tells me the concentration of the canceled intermediate."	The intermediate's concentration is controlled by the individual equilibria, not the net constant.
"Amphoteric means a substance dissolves only in acid."	Amphoteric species dissolve in acid and in excess base; that is why Al(OH)₃ is soluble at both ends of the pH scale.
"The product rule only works for equilibria in gases."	It works for any reactions that can be added, including solubility, complexation, and acid–base steps.

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine two people holding opposite ends of the same rope, both pulling. One person (the solid dissolving) pulls weakly; the other (acid or ammonia grabbing the ions) pulls hard. The rope moves toward the strong puller, so the solid ends up dissolved. Coupled equilibria are just two pulls on the same rope, and you can predict who wins by multiplying the two pulling strengths.

Worked examples

For AgCl(s) + 2NH₃(aq) ⇌ [Ag(NH₃)₂]⁺ + Cl^-, we already found K_net = K_sp × K_f = (1.8 × 10^-10)(1.7 × 10⁷) = 3.1 × 10^-3.

In 1.0 M NH₃, let x be the molar solubility of AgCl. Then [Ag(NH₃)₂⁺] = [Cl^-] = x, and the equilibrium expression is

$K net = [Ag (NH 3) 2 +][Cl -] [NH 3] 2 = x 2 (1.0 - 2x) 2$

Take the square root of both sides and solve:

$\sqrt K net = x 1.0 - 2x \Rightarrow 0.0557 = x 1.0 - 2x$

$x = 0.0557(1.0 - 2x) \Rightarrow x = 0.0557 - 0.1114x$

$x(1.1114) = 0.0557 \Rightarrow x = 0.050 mol/L$

That is roughly 4000 times more silver dissolved than the 1.3 × 10^-5 mol/L that pure water dissolves. The chemistry of photographic fixing baths and silver recovery exploits exactly this coupling.

Mg(OH)₂ has K_sp = 5.6 × 10^-12 (values near 10^-12 appear across reference tables). In acid, the net reaction is

$Mg(OH) 2 (s) + 2 H + (aq) ⇌ Mg 2+ (aq) + 2 H 2 O (l)$

with

$K net = K sp K w 2 = 5.6 \times 10 -12 (1.0 \times 10 -14) 2 = 5.6 \times 10 -12 1.0 \times 10 -28 = 5.6 \times 1016$

An equilibrium constant of 10¹⁶ means the solid is consumed essentially completely as long as acid is available. The hydroxide is "insoluble" in pure water yet fully reactive in stomach acid — which is precisely why it can neutralize excess acid without being absorbed.

Calcite, CaCO₃ (K_sp = 4.5 × 10^-9), dissolves in acidic water. The carbonate released is protonated: CO₃^2- + H⁺ ⇌ HCO₃^-, whose constant is the reciprocal of the second acid dissociation constant of carbonic acid, 1/K_a2 = 1/(4.7 × 10^-11).

Combining dissolution with protonation:

$K net = K sp K a 2 = 4.5 \times 10 -9 4.7 \times 10 -11 \approx 96$

A net constant near 10² means acid water dissolves calcite readily — enough to carve caves over time and to erode marble statues and building stone exposed to acid rain.

Key takeaways

Net reaction = sum of individual reactions; net constant = product of individual constants.
Reverse a reaction → use 1/K. Double it → square K. Triple → cube, and so on.
Solubility + complexation: K_net = K_sp × K_f.
Solubility + acid neutralization of OH⁻: K_net = K_sp/K_w² (hydroxides).
Solubility + acid protonation of S²⁻: K_net = K_sp/(K_a1K_a2) (sulfides).
Amphoteric hydroxides (Al³⁺, Zn²⁺) dissolve in acid and excess base; solubility is minimum at intermediate pH.
A small K_sp does not mean "never dissolves" — coupling can make it dissolve completely.
Intermediates that appear and cancel (like Ag⁺ above) are never given concentrations by the net constant; each step still obeys its own equilibrium.

Check yourself

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

If reaction A has K = 10^-5 and reaction B has K = 10⁸, and the net reaction is A + B, what is K_net?
Show answer
K_net = 10^-5 × 10⁸ = 10³ — the coupled process is strongly favorable.
Why does adding H⁺ dissolve Mg(OH)₂ even though K_sp is tiny?
Show answer
H⁺ consumes OH⁻ as it is released (K = 1/K_w² = 10²⁸), so K_net = K_sp/K_w² ≈ 10¹⁶, overwhelming the small K_sp.
What is the net constant for dissolving AgI (K_sp = 8.5 × 10^-17) in ammonia, given K_f = 1.7 × 10⁷ for [Ag(NH₃)₂]⁺?
Show answer
K_net = (8.5 × 10^-17)(1.7 × 10⁷) = 1.4 × 10^-9 — far too small to dissolve AgI appreciably, which is why ammonia distinguishes AgCl from AgI.
Why does Al(OH)₃ dissolve in concentrated NaOH?
Show answer
Excess OH⁻ forms the soluble aluminate complex [Al(OH)₄]^-: dissolution coupled to complexation (K_net = K_sp × K_f).
What happens to CaCO₃ solubility as ocean water absorbs more CO₂?
Show answer
More CO₂ lowers pH and drives CO₃^2- + H⁺ → HCO₃^-, reducing carbonate concentration and increasing CaCO₃ dissolution — ocean acidification.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

coupled equilibria: Two or more equilibria sharing a species, so that shifting one shifts the others
net reaction: Sum of the individual reactions after canceling shared species
Kₙet: Product (or quotient) of the individual constants
complexation: Lewis acid–base reaction forming a complex ion
amphoteric: Able to react as acid and base; dissolves in both acid and base
formation constant (Kf): Equilibrium constant for complex-ion formation

Sources & references

openstax.org — Chemistry 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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In 30 seconds

Why this matters

The college version

Core Concepts

Adding reactions multiplies their constants

Complexation dissolves precipitates

Acids dissolve insoluble hydroxides and carbonates

Amphoteric hydroxides dissolve in both acid and base

Coupled equilibria in the environment

How It Works / Step-by-Step Process

Common Confusions

The same idea, in plain words

Worked examples

Key takeaways

Check yourself

Keep learning

Key vocabulary

Related lessons

Sources & references