Coupled Equilibria, Equilibria of Other Reaction Classes | Chemistry 2e

Learning objectives

By the end of this section, you will be able to:

Describe examples of systems involving two (or more) coupled chemical equilibria
Calculate reactant and product concentrations for coupled equilibrium systems

As discussed in preceding chapters on equilibrium, coupled equilibria involve two or more separate chemical reactions that share one or more reactants or products. This section of this chapter will address solubility equilibria coupled with acid-base and complex-formation reactions.

An environmentally relevant example illustrating the coupling of solubility and acid-base equilibria is the impact of ocean acidification on the health of the ocean’s coral reefs. These reefs are built upon skeletons of sparingly soluble calcium carbonate excreted by colonies of corals (small marine invertebrates). The relevant dissolution equilibrium is

{CaCO}_{3} (s) ⇌ {Ca}^{2 +} (a q) + {CO}_{3}^{- 2} (a q) K_{sp} = 8.7 \times 10^{- 9}

Rising concentrations of atmospheric carbon dioxide contribute to an increased acidity of ocean waters due to the dissolution, hydrolysis, and acid ionization of carbon dioxide:

{CO}_{2} (g) ⇌ {CO}_{2} (a q)

{CO}_{2} (a q) + H_{2} O (l) ⇌ H_{2} {CO}_{3} (a q)

\begin{matrix} H_{2} {CO}_{3} (a q) + H_{2} O (l) ⇌ {HCO}_{3}^{-} (a q) + H_{3} O^{+} (a q) & K_{a 1} = 4.3 \times 10^{- 7} \end{matrix}

\begin{matrix} {HCO}_{3}^{-} (a q) + H_{2} O (l) ⇌ {CO}_{3}^{2 -} (a q) + H_{3} O^{+} (a q) & K_{a 2} = 4.7 \times 10^{- 11} \end{matrix}

Inspection of these equilibria shows the carbonate ion is involved in the calcium carbonate dissolution and the acid hydrolysis of bicarbonate ion. Combining the dissolution equation with the reverse of the acid hydrolysis equation yields

\begin{matrix} {CaCO}_{3} (s) + H_{3} O^{+} (a q) ⇌ {Ca}^{2 +} (a q) + {HCO}_{3}^{-} (a q) + H_{2} O (l) & K = K_{sp} / K_{a 2} = 180 \end{matrix}

The equilibrium constant for this net reaction is much greater than the K_sp for calcium carbonate, indicating its solubility is markedly increased in acidic solutions. As rising carbon dioxide levels in the atmosphere increase the acidity of ocean waters, the calcium carbonate skeletons of coral reefs become more prone to dissolution and subsequently less healthy (CNX_Chem_15_03_CoralReef).

View original figure (external; image not reproduced in this text edition)

Original figure description: This figure contains two photographs of coral reefs. In a, a colorful reef that includes hues of purple and pink corals is shown in blue green water with fish swimming in the background. In b, grey-green mossy looking coral is shown in a blue aquatic environment. This photo does not have the colorful appearance or fish that were shown in figure a.

Healthy coral reefs (a) support a dense and diverse array of sea life across the ocean food chain. But when coral are unable to adequately build and maintain their calcium carbonate skeletons because of excess ocean acidification, the unhealthy reef (b) is only capable of hosting a small fraction of the species as before, and the local food chain starts to collapse. (credit a: modification of work by NOAA Photo Library; credit b: modification of work by “prilfish”/Flickr)

The dramatic increase in solubility with increasing acidity described above for calcium carbonate is typical of salts containing basic anions (e.g., carbonate, fluoride, hydroxide, sulfide). Another familiar example is the formation of dental cavities in tooth enamel. The major mineral component of enamel is calcium hydroxyapatite (CNX_Chem_15_03_Apatite), a sparingly soluble ionic compound whose dissolution equilibrium is

{Ca}_{5} ({P O_{4})}_{3} OH (s) ⇌ 5 {Ca}^{2+} (a q) + 3 {PO}_{4}^{3−} (a q) + {OH}^{−} (a q)

View original figure (external; image not reproduced in this text edition)

Original figure description: This figure includes an image of two large light blue apatite crystals in a mineral conglomerate that includes white, grey, and tan crystals. The blue apatite crystals have a dull, dusty, or powdered appearance.

Crystal of the mineral hydroxyapatite, Ca₅(PO₄)₃OH, is shown here. The pure compound is white, but like many other minerals, this sample is colored because of the presence of impurities.

This compound dissolved to yield two different basic ions: triprotic phosphate ions

{PO}_{4}^{3−} (aq) + H_{3} O^{+} (aq) ⟶ {HPO}_{4}^{2−} (aq) + H_{2} O (l)

{HPO}_{4}^{2−} (a q) + H_{3} O^{+} (aq) ⟶ H_{2} {PO}_{4}^{−} (aq) + H_{2} O (l)

H_{2} {PO}_{4}^{−} (aq) + H_{3} O^{+} (aq) ⟶ H_{3} {PO}_{4} (aq) + H_{2} O (l)

and monoprotic hydroxide ions:

{OH}^{−} (a q) + H_{3} O^{+} ⟶ 2 H_{2} O

Of the two basic productions, the hydroxide is, of course, by far the stronger base (it’s the strongest base that can exist in aqueous solution), and so it is the dominant factor providing the compound an acid-dependent solubility. Dental cavities form when the acid waste of bacteria growing on the surface of teeth hastens the dissolution of tooth enamel by reacting completely with the strong base hydroxide, shifting the hydroxyapatite solubility equilibrium to the right. Some toothpastes and mouth rinses contain added NaF or SnF₂ that make enamel more acid resistant by replacing the strong base hydroxide with the weak base fluoride:

NaF + {Ca}_{5} ({P O_{4})}_{3} OH ⇌ {Ca}_{5} ({P O_{4})}_{3} F + {Na}^{+} + {OH}^{−}

The weak base fluoride ion reacts only partially with the bacterial acid waste, resulting in a less extensive shift in the solubility equilibrium and an increased resistance to acid dissolution. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information.

Role of Fluoride in Preventing Tooth Decay

As we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, Ca₅(PO₄)₃F. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride (CNX_Chem_15_03_Toothpaste).

View original figure (external; image not reproduced in this text edition)

Original figure description: Replace with updated art; figure in “99” art folder is current and correct

Fluoride, found in many toothpastes, helps prevent tooth decay (credit: Kerry Ceszyk).

Unfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the US Environmental Protection Agency sets a maximum level of 4 ppm (4 mg/L) of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater.

The solubility of ionic compounds may also be increased when dissolution is coupled to the formation of a complex ion. For example, aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion $Al {(OH)}_{4}^{−} .$

View original figure (external; image not reproduced in this text edition)

Original figure description: An H atom is bonded to an O atom. The O atom has 2 dots above it and 2 dots below it. The O atom is bonded to an A l atom, which has three additional O atoms bonded to it as well. Each of these additional O atoms has 4 dots arranged around it, and is bonded to an H atom. This entire molecule is contained in brackets, to the right of which is a superscripted negative sign.

The equations for the dissolution of aluminum hydroxide, the formation of the complex ion, and the combined (net) equation are shown below. As indicated by the relatively large value of K for the net reaction, coupling complex formation with dissolution drastically increases the solubility of Al(OH)₃.

\begin{matrix} Al {(OH)}_{3} (s) ⇌ {Al}^{3+} (a q) + 3 {OH}^{−} (a q) & K_{sp} = 2 \times 10^{- 32} \\ {Al}^{3+} (a q) + 4 {OH}^{−} (a q) ⇌ Al {(OH)}_{4}^{−} (a q) & K_{f} = 1.1 \times 10^{33} \\ Net: Al {(OH)}_{3} (s) + {OH}^{−} (a q) ⇌ Al {(OH)}_{4}^{−} (a q) & K = K_{sp} K_{f} = 22 \end{matrix}

Increased Solubility in Acidic Solutions

Compute and compare the molar solublities for aluminum hydroxide, Al(OH)₃, dissolved in (a) pure water and (b) a buffer containing 0.100 M acetic acid and 0.100 M sodium acetate.

Solution

(a) The molar solubility of aluminum hydroxide in water is computed considering the dissolution equilibrium only as demonstrated in several previous examples:

\begin{matrix} {Al(OH)}_{3} (s) ⇌_{}^{} {Al}_{}^{3 +} (a q) + {3OH}^{-} (a q) K_{sp} = 2 \times 10^{- 32} \\ molar solubility in water = [{Al}^{3 +}] = {(2 \times 10^{- 32} / 27)}^{1 / 4} = 5 \times 10^{- 9} M \end{matrix}

(b) The concentration of hydroxide ion of the buffered solution is conveniently calculated by the Henderson-Hasselbalch equation:

\begin{matrix} pH = {pK}_{a} + log [{CH}_{3} {COO}^{-}] / [{CH}_{3} COOH] \\ pH = 4.74 + log (0.100 / 0.100) = 4.74 \end{matrix}

At this pH, the concentration of hydroxide ion is

\begin{matrix} pOH = 14.00 - 4.74 = 9.26 \\ [{OH}^{-}] = 10^{- 9.26} = 5.5 \times 10^{- 10} \end{matrix}

The solubility of Al(OH)₃ in this buffer is then calculated from its solubility product expressions:

\begin{matrix} K_{sp} = [{Al}^{3 +}] {[{OH}^{-}]}^{3} \\ molar solubility in buffer = [{Al}^{3 +}] = K_{sp} / {[{OH}^{-}]}^{3} = (2 \times 10^{- 32}) / {(5.5 \times 10^{- 10})}^{3} = 1.2 \times 10^{- 4} M \end{matrix}

Compared to pure water, the solubility of aluminum hydroxide in this mildly acidic buffer is approximately ten million times greater (though still relatively low).

Check Your Learning

What is the solubility of aluminum hydroxide in a buffer comprised of 0.100 M formic acid and 0.100 M sodium formate?

Multiple Equilibria

Unexposed silver halides are removed from photographic film when they react with sodium thiosulfate (Na₂S₂O₃, called hypo) to form the complex ion

Ag {(S_{2} O_{3})}_{2}^{3−}

(K_f = 4.7

\times

10¹³).

View original figure (external; image not reproduced in this text edition)

Original figure description: A chemical reaction is shown using structural formulas. On the left, A g superscript plus is followed by a plus sign, the number 2, and a structure in brackets. The structure is composed of a central S atom which has O atoms single bonded above, right, and below. A second S atom is single bonded to the left. Each of these bonded atoms has 6 dots around it. Outside the brackets is a superscript 2 negative. Following a bidirectional arrow is a structure in brackets with a central A g atom. To the left and right, S atoms are single bonded to the A g atom. Each of these S atoms has four dots around it, and an S atom connected with a single bond moving out from the central A g atom, forming the ends of the structure. Each of these atoms has three O atoms attached with single bonds above, below, and at the end of the structure. Each O atom has six dots around it. Outside the brackets is a superscript 3 negative.

What mass of Na₂S₂O₃ is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of $Ag {(S_{2} O_{3})}_{2}^{3−} ?$

Solution

Two equilibria are involved when silver bromide dissolves in an aqueous thiosulfate solution containing the

S_{2} O_{3}^{2−}

ion:

dissolution: $AgBr (s) ⇌ {Ag}^{+} (a q) + {Br}^{−} (a q) K_{sp} = 5.0 \times 10^{−13}$

complexation: ${Ag}^{+} (a q) + {2S}_{2} O_{3}^{2−} (a q) ⇌ Ag {(S_{2} O_{3})}_{2}^{3−} (a q) K_{f} = 4.7 \times 10^{13}$

Combining these two equilibrium equations yields

$\begin{matrix} AgBr (s) + 2 S_{2} {O_{3}}^{2–} (aq) & ⇌ & Ag (S_{2} O_{3})_{2}^{3–} (aq) + {Br}^{–} (aq) \\ K = \frac{[Ag (S_{2} O_{3})_{2}^{3–}] [{Br}^{-}]}{[S_{2} O_{3}^{2–}]^{2}} = K_{sp} K_{f} = 24 \end{matrix}$

The concentration of bromide resulting from dissolution of 1.00 g of AgBr in 1.00 L of solution is

[{Br}^{-}] = \frac{1.00 g AgBr \times \frac{1 mol AgBr}{187.77 g/mol} \times \frac{1 {mol Br}^{-}}{1 mol AgBr}}{1.00 L} = 0.00532 M

The stoichiometry of the dissolution equilibrium indicates the same concentration of aqueous silver ion will result, 0.00532 M, and the very large value of $K_{f}$ ensures that essentially all the dissolved silver ion will be complexed by thiosulfate ion:

[Ag (S_{2} O_{3})_{2}^{3–}] = 0.00532 M

Rearranging the K expression for the combined equilibrium equations and solving for the concentration of thiosulfate ion yields

[S_{2} {O_{3}}^{2–}] = \frac{\sqrt{[Ag (S_{2} O_{3})_{2}^{3–}] [{Br}^{-}]}}{K} = \sqrt{\frac{(0.00532 M) (0.00532 M)}{24}} = 0.0011 M

Finally, the total mass of ${Na}_{2} S_{2} O_{3}$ required to provide enough thiosulfate to yield the concentrations cited above can be calculated.

Mass of ${Na}_{2} S_{2} O_{3}$ required to yield 0.00532 M $Ag (S_{2} O_{2})_{2}^{3–}$

0.00532 \frac{mol Ag (S_{2} O_{3})_{2}^{3–}}{1.00 L} \times \frac{2 mol S_{2} O_{3}^{2–}}{1 mol Ag (S_{2} O_{3})_{2}^{3–}} \times \frac{1 mol {Na}_{2} S_{2} O_{3}}{1 mol S_{2} O_{3}^{2–}} \times \frac{158.1 g {Na}_{2} S_{2} O_{3}}{1 mol {Na}_{2} S_{2} O_{3}} = 1.68 g

Mass of ${Na}_{2} S_{2} O_{3}$ required to yield 0.00110 M $S_{2} O_{3}^{2–}$

0.0011 \frac{mol S_{2} O_{3}^{2–}}{1.00 L} \times \frac{1 mol S_{2} O_{3}^{2–}}{1 mol {Na}_{2} S_{2} O_{3}} \times \frac{158.1 g {Na}_{2} S_{2} O_{3}}{1 mol {Na}_{2} S_{2} O_{3}} = 0.17 g

The mass of ${Na}_{2} S_{2} O_{3}$ required to dissolve 1.00 g of AgBr in 1.00 L of water is thus 1.68 g + 0.17 g = 1.85 g

Check Your Learning

AgCl(s), silver chloride, has a very low solubility:

AgCl (s) ⇌ {Ag}^{+} (a q) + {Cl}^{−} (a q),

K_sp = 1.6

\times

10^–10. Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed:

{Ag}^{+} (a q) + 2 {NH}_{3} (a q) ⇌ Ag {({NH}_{3})}_{2}^{+} (a q),

K_f = 1.7

\times

10⁷. What mass of NH₃ is required to prepare 1.00 L of solution that will dissolve 2.00 g of AgCl by formation of

Ag {({NH}_{3})}_{2}^{+} ?

Key Concepts and Summary

Systems involving two or more chemical equilibria that share one or more reactant or product are called coupled equilibria. Common examples of coupled equilibria include the increased solubility of some compounds in acidic solutions (coupled dissolution and neutralization equilibria) and in solutions containing ligands (coupled dissolution and complex formation). The equilibrium tools from other chapters may be applied to describe and perform calculations on these systems.

Chemistry End of Chapter Exercises

A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?

Calculate the equilibrium concentration of Ni²⁺ in a 1.0-M solution [Ni(NH₃)₆](NO₃)₂.

0.014 M

Calculate the equilibrium concentration of Zn²⁺ in a 0.30-M solution of $Zn {(CN)}_{4}^{2−} .$

Calculate the equilibrium concentration of Cu²⁺ in a solution initially with 0.050 M Cu²⁺ and 1.00 M NH₃.

7.2 $\times$ 10^–15 M

Calculate the equilibrium concentration of Zn²⁺ in a solution initially with 0.150 M Zn²⁺ and 2.50 M CN^–.

Calculate the Fe³⁺ equilibrium concentration when 0.0888 mole of K₃[Fe(CN)₆] is added to a solution with 0.00010 M CN^–.

4.4 $\times$ 10⁻²² M

Calculate the Co²⁺ equilibrium concentration when 0.010 mole of [Co(NH₃)₆](NO₃)₂ is added to a solution with 0.25 M NH₃. Assume the volume is 1.00 L.

Calculate the molar solubility of Sn(OH)₂ in a buffer solution containing equal concentrations of NH₃ and ${NH}_{4}^{+} .$

Calculate the molar solubility of Al(OH)₃ in a buffer solution with 0.100 M NH₃ and 0.400 M ${NH}_{4}^{+} .$

[OH⁻] = 4.5 $\times$ 10⁻⁶; [Al³⁺] = 2 $\times$ 10^–16 (molar solubility)

What is the molar solubility of CaF₂ in a 0.100-M solution of HF? K_a for HF = 6.4 $\times$ 10^–4.

What is the molar solubility of BaSO₄ in a 0.250-M solution of NaHSO₄? K_a for ${HSO}_{4}^{−}$ = 1.2 $\times$ 10^–2.

$K_{sp} = [{Ba}^{2+}] [{SO}_{4}^{2−}] = [{Ba}^{2+}] (0.049) = 1.08 \times 10^{–10} [{Ba}^{2+}] = 2.2 \times 10^{–9}$ ; (molar solubility)

What is the molar solubility of Tl(OH)₃ in a 0.10-M solution of NH₃?

What is the molar solubility of Pb(OH)₂ in a 0.138-M solution of CH₃NH₂?

[OH^–] = 7.6 $\times$ 10⁻³ M; [Pb²⁺] = 2.1 $\times$ 10^–11 (molar solubility)

A solution of 0.075 M CoBr₂ is saturated with H₂S ([H₂S] = 0.10 M). What is the minimum pH at which CoS begins to precipitate?

$CoS (s) ⇌ {Co}^{2+} (a q) + S^{2−} (a q) K_{sp} = 2.3 \times 10^{27}$

$H_{2} S (a q) + 2 H_{2} O (l) ⇌ 2 H_{3} O^{+} (a q) + S^{2−} (a q) K = 8.9 \times 10^{−27}$

A 0.125-M solution of Mn(NO₃)₂ is saturated with H₂S ([H₂S] = 0.10 M). At what pH does MnS begin to precipitate?

$MnS (s) ⇌ {Mn}^{2+} (a q) + S^{2−} (a q) K_{sp} = 2.3 \times 10^{−13}$

$H_{2} S (a q) + 2 H_{2} O (l) ⇌ 2 H_{3} O^{+} (a q) + S^{2−} (a q) K = 1.0 \times 10^{−26}$

7.66

Both AgCl and AgI dissolve in NH₃.

(a) What mass of AgI dissolves in 1.0 L of 1.0 M NH₃?

(b) What mass of AgCl dissolves in 1.0 L of 1.0 M NH₃?

The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service.

Solve the following problem:

${MgF}_{2} (s) ⇌ {Mg}^{2+} (a q) + 2 F^{−} (a q)$

In a saturated solution of MgF₂ at 18 °C, the concentration of Mg²⁺ is 1.21 $\times$ 10^–3 M. The equilibrium is represented by the preceding equation.

(a) Write the expression for the solubility-product constant, K_sp, and calculate its value at 18 °C.

(b) Calculate the equilibrium concentration of Mg²⁺ in 1.000 L of saturated MgF₂ solution at 18 °C to which 0.100 mol of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.

(c) Predict whether a precipitate of MgF₂ will form when 100.0 mL of a 3.00 $\times$ 10^–3-M solution of Mg(NO₃)₂ is mixed with 200.0 mL of a 2.00 $\times$ 10^–3-M solution of NaF at 18 °C. Show the calculations to support your prediction.

(d) At 27 °C the concentration of Mg²⁺ in a saturated solution of MgF₂ is 1.17 $\times$ 10^–3 M. Is the dissolving of MgF₂ in water an endothermic or an exothermic process? Give an explanation to support your conclusion.

(a) K_sp = [Mg²⁺][F^–]² = (1.21 $\times$ 10^–3)(2 $\times$ 1.21 $\times$ 10^–3)² = 7.09 $\times$ 10^–9
(b) 7.09 $\times$ 10^–7 M
(c) Determine the concentration of Mg²⁺ and F^– that will be present in the final volume. Compare the value of the ion product [Mg²⁺][F^–]² with K_sp. If this value is larger than K_sp, precipitation will occur.
0.1000 L $\times$ 3.00 $\times$ 10^–3 M Mg(NO₃)₂ = 0.3000 L $\times$ M Mg(NO₃)₂
M Mg(NO₃)₂ = 1.00 $\times$ 10^–3 M
0.2000 L $\times$ 2.00 $\times$ 10^–3 M NaF = 0.3000 L $\times$ M NaF
M NaF = 1.33 $\times$ 10^–3 M
ion product = (1.00 $\times$ 10^–3)(1.33 $\times$ 10^–3)² = 1.77 $\times$ 10^–9 This value is smaller than K_sp, so no precipitation will occur.
(d) MgF₂ is less soluble at 27 °C than at 18 °C. Because added heat acts like an added reagent, when it appears on the product side, the Le Châtelier’s principle states that the equilibrium will shift to the reactants’ side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic.

Which of the following compounds, when dissolved in a 0.01-M solution of HClO₄, has a solubility greater than in pure water: CuCl, CaCO₃, MnS, PbBr₂, CaF₂? Explain your answer.

Which of the following compounds, when dissolved in a 0.01-M solution of HClO₄, has a solubility greater than in pure water: AgBr, BaF₂, Ca₃(PO₄)₂, ZnS, PbI₂? Explain your answer.

BaF₂, Ca₃(PO₄)₂, ZnS; each is a salt of a weak acid, and the $[H_{3} O^{+}]$ from perchloric acid reduces the equilibrium concentration of the anion, thereby increasing the concentration of the cations

What is the effect on the amount of solid Mg(OH)₂ that dissolves and the concentrations of Mg²⁺ and OH^– when each of the following are added to a mixture of solid Mg(OH)₂ and water at equilibrium?

(a) MgCl₂

(b) KOH

(d) NaNO₃

(e) Mg(OH)₂

What is the effect on the amount of CaHPO₄ that dissolves and the concentrations of Ca²⁺ and ${HPO}_{4}^{2−}$ when each of the following are added to a mixture of solid CaHPO₄ and water at equilibrium?

(a) CaCl₂

(b) HCl

(d) NaOH

(e) CaHPO₄

Effect on amount of solid CaHPO₄, [Ca²⁺], [OH^–]: (a) increase, increase, decrease; (b) decrease, increase, decrease; (c) no effect, no effect, no effect; (d) decrease, increase, decrease; (e) increase, no effect, no effect

Identify all chemical species present in an aqueous solution of Ca₃(PO₄)₂ and list these species in decreasing order of their concentrations. (Hint: Remember that the ${PO}_{4}^{3−}$ ion is a weak base.)

Glossary

coupled equilibria: system characterized the simultaneous establishment of two or more equilibrium reactions sharing one or more reactant or product

Attribution. Chemistry 2e, OpenStax, Rice University, and the contributing authors credited in the book. Licensed under Creative Commons Attribution 4.0 International. Source revision 32c719622f of openstax/osbooks-chemistry-bundle (2026-02-04). CC BY 4.0. Original source ↗. Text edition of a pinned CC BY 4.0 source revision of the OpenStax book, converted from CNXML by the EvidenceHelix importer. Original prose, learning objectives, tables, exercises, glossary and captions are preserved; images are not copied (figures link to the original source and keep their credits). No new content was added. Later revisions of this book can carry different licence terms and are not used.