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1 Brønsted–Lowry Definition of Acids and Bases
742 Chapter 16 | Acids and Bases, A Molecular Look
In the gas phase, the proton is transferred directly from the HCl molecule to the NH3
n We are using “curved” arrows
to show how electrons shift and
rearrange during the formation of the
ions. The electron pair on the nitrogen
atom of the NH3 molecule binds to H+
as it is removed from the electron pair
in the H!Cl bond. The electron pair
in the H!Cl bond shifts entirely to
the Cl as the Cl- ion is formed.
Electron pair in the bond becomes
a lone pair on Cl as the chloride
ion is formed.
H 9 N9 H
Electron pair on N binds to
a proton (H+) as the proton separates from
the electron pair of the H9Cl bond.
n Although the positive charge is
shown on just one of the hydrogens
in NH4+, it is actually spread equally
over all four hydrogens.
As ammonium ions and chloride ions form, they attract each other, gather, and settle as
crystals of ammonium chloride.
Proton Transfer Reactions
Johannes Brønsted (1879–1947), a Danish chemist, and Thomas Lowry (1874–1936), a
British scientist, realized that the important event in most acid–base reactions is simply
the transfer of a proton from one particle to another. Therefore, they redefined acids as
substances that donate protons and bases as substances that accept protons. The heart of
the Brønsted–Lowry concept of acids and bases is that acid–base reactions are proton transfer
reactions. The definitions are therefore very simple.
Brønsted–Lowry Definitions of Acids and Bases2
An acid is a proton donor.
A base is a proton acceptor.
Accordingly, hydrogen chloride is an acid because when it reacts with ammonia, HCl
molecules donate protons to NH3 molecules. Similarly, ammonia is a base because NH3
molecules accept protons.
Even when water is the solvent, chemists use the Brønsted–Lowry definitions more
often than those of Arrhenius. Thus, the reaction between hydrogen chloride and water to
form hydronium ion (H3O+) and chloride ion (Cl-), which is another proton transfer
reaction, is clearly a Brønsted–Lowry acid–base reaction. Molecules of HCl are the acid in
this reaction, and water molecules are the base. HCl molecules collide with water molecules and protons transfer during the collisions.
n In H3O+, the positive charge is
actually spread equally over all three
Collision of molecules
permits proton transfer
Although Brønsted and Lowry are both credited with defining acids and bases in terms of proton transfer,
Brønsted carried the concepts further. For the sake of brevity, we will often use the terms Brønsted acid and
Brønsted base when referring to the substances involved in proton transfer reactions.
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16.1 | Brønsted–Lowry Definition of Acids and Bases
Conjugate Acids and Bases
Under the Brønsted–Lowry view, it is useful to consider any acid–base reaction as a chemical equilibrium, having both a forward and a reverse reaction. We first encountered a
chemical equilibrium in a solution of a weak acid on page 170. Let’s examine it again in
the light of the Brønsted–Lowry definitions, using formic acid, HCHO2, as an example.
Because formic acid is a weak acid, we represent its ionization as a chemical equilibrium
in which water is not just a solvent but also a chemical reactant, a proton acceptor.3
HCHO2(aq) + H2O �
� H3O+(aq) + CHO2-(aq)
−O 9 C 9 H
formate ion (CHO2−)
Conjugate acid–base pairs
formic acid (HCHO2)
Only the H in red is available in an
In the forward reaction, a formic acid molecule donates a proton to the water molecule
and changes to a formate ion, CHO2- (Figure 16.2a). Thus HCHO2 behaves as a Brønsted
acid, a proton donor. Because water accepts this proton from HCHO2, water behaves as a
Brønsted base, a proton acceptor.
Now let’s look at the reverse reaction (Figure 16.2b). In it, H3O+ behaves as a Brønsted
acid because it donates a proton to the CHO2- ion. The CHO2- ion behaves as a
Brønsted base by accepting the proton.
The equilibrium involving HCHO2, H2O, H3O+, and CHO2- is typical of proton
transfer equilibria in general, in that we can identify two acids (e.g., HCHO2 and H3O+)
and two bases (e.g., H2O and CHO2-). Notice that in the aqueous formic acid equilibrium, the acid on the right of the arrows (H3O+) is formed from the base on the left
(H2O), and the base on the right (CHO2-) is formed from the acid on the left (HCHO2).
Two substances that differ from each other only by one proton are referred to as a conjugate
acid–base pair. Thus, H3O+ and H2O are such a pair; they are alike except that H3O+ has
one more proton than H2O. One member of the pair is called the conjugate acid because
it is the proton donor of the two. The other member is the conjugate base because it is the
pair’s proton acceptor. We say that H3O+ is the conjugate acid of H2O, and H2O is the conjugate base of H3O+. Notice that the acid member of the pair has one more H+ than the base
The pair HCHO2 and CHO2- is the other conjugate acid–base pair in the aqueous
formic acid equilibrium. HCHO2 has one more H+ than CHO2-, so the conjugate acid
of CHO2- is HCHO2; the conjugate base of HCHO2 is CHO2-. One way to highlight
Figure 16.2 | Brønsted–Lowry
acids and bases in aqueous
formic acid. (a) Formic acid
transfers a proton to a water
molecule. HCHO2 is the acid and
H2O is the base. (b) When
hydronium ion transfers a proton
to the CHO2- ion, H3O+ is the
acid and CHO2- is the base.
Just as we have represented acetic acid as HC2H3O2, placing the ionizable or acidic hydrogen first (as we do
in HCl or HNO3), so we give the formula of formic acid as HCHO2 and the formula of the formate ion as
CHO2-. As the chemical structure shows, however, the acidic hydrogen in formic acid is bonded to O, not C.
Many chemists prefer to write formic acid as HCO2H.
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744 Chapter 16 | Acids and Bases, A Molecular Look
the two members of a conjugate acid–base pair in an equilibrium equation is to connect
them by a line.
HCHO2 + H2O
Conjugate acid–base equilibria
n Notice how the conjugate acid
always has one more H+ than the
In any Brønsted–Lowry acid–base equilibrium, there are invariably two conjugate
acid–base pairs. It is important that you learn how to pick them out of an equation by
inspection and to write them from formulas.
Determining the Formulas of Conjugate Acids and Bases
What is the conjugate base of nitric acid, HNO3, and what is the conjugate acid of the
hydrogen sulfate ion, HSO4-?
Any reference to a conjugate acid or a conjugate base signals that the problem involves the Brønsted–Lowry acid–base concept. The key to answering the questions
posed here is knowing the definitions, so be sure you’ve learned them.
the Tools: The tool required states that members of any conjugate acid–
base pair differ by one (and only one) H+, with the member having the greater number of
hydrogens being the acid.
We are asked to find the conjugate base of HNO3, which means HNO3 must
be the acid member of the pair. To find the formula of the base, we remove one H+ from
the acid. Removing one H+ (both the atom and the charge) from HNO3 leaves NO3-, so
the nitrate ion is the conjugate base of HNO3.
We are also asked to find the conjugate acid of HSO4-, so HSO4- must be the base
member of an acid–base pair. To find the formula of the acid, we add one H+ to the base,
which is equivalent to adding one hydrogen to the formula of the base and increasing its
positive charge (or decreasing its negative charge) by one unit. Adding an H+ to HSO4gives its conjugate acid, H2SO4.
the Answers Reasonable? As a check, we can quickly compare the two formulas
in each pair.
In each case, the formula on the right has one less H+ than the one on the left, so it is the
conjugate base. We’ve answered the question correctly.
16.1 | Which of the following are conjugate acid–base pairs? Describe why the others are
not true conjugate acid–base pairs. (a) H3PO4 and H2PO4-, (b) HI and H+, (c) NH2and NH3, (d) HNO2 and NH4+, (e) CO32- and CN-, (f ) HPO42- and H2PO4-. (Hint:
Recall that conjugate acid–base pairs must differ by one H+.)
16.2 | Write the formula of the conjugate base for each of the following Brønsted–Lowry
acids: (a) H2O, (b) HI, (c) HNO2, (d) H3PO4, (e) H2PO4-, (f) HPO42-, (g) H2S, (h) NH4+.
16.3 | Write the formula of the conjugate acid for each of the following Brønsted–
Lowry bases: (a) HO2-, (b) SO42-, (c) CO32-, (d) CN-, (e) NH2-, (f ) NH3, (g) H2PO4-,
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16.1 | Brønsted–Lowry Definition of Acids and Bases
Identifying Conjugate Acid–Base Pairs
in a Brønsted–Lowry Acid–Base Reaction
The anion of sodium hydrogen sulfate, HSO4-, reacts as follows with the phosphate ion,
HSO4-(aq) + PO43-(aq) → SO42-(aq) + HPO42-(aq)
Identify the two conjugate acid–base pairs.
n Sodium hydrogen sulfate (also
called sodium bisulfate) is used
to manufacture certain kinds
of cement and to clean oxide
coatings from metals.
There are two things to look for in identifying the conjugate acid–base pairs
in an equation. One is that the members of a conjugate pair are alike except for the number of hydrogens and the charge. The second is that the members of each pair must be on
opposite sides of the arrow, as in Equation 16.1, and we cannot have both acids (or both
bases) on the same side of the equation. In each pair, of course, the acid is the one with the
greater number of hydrogens.
n Assembling the Tools: We will use Equation 16.1 as a template to establish the relationships among the acid–base members of the two pairs.
Two of the formulas in the equation contain “PO4,” so they must belong
to the same conjugate pair. The one with the greater number of hydrogens, HPO42-,
must be the Brønsted–Lowry acid, and the other, PO43-, must be the Brønsted–Lowry
base. Therefore, one conjugate acid–base pair is HPO42- and PO43-. The other two ions,
HSO4- and SO42-, belong to the second conjugate acid–base pair; HSO4- is the conjugate acid and SO42- is the conjugate base.
HSO4−(aq) + PO43– (aq)
SO4 2– (aq) + HPO4 (aq)
the Answer Reasonable? A check assures us that we have fulfilled the requirements
that each conjugate pair has one member on one side of the arrow and the other member
on the opposite side of the arrow and that the members of each pair differ from each other
by one (and only one) H+.
16.4 | Sodium cyanide solution, when poured into excess hydrochloric acid, releases
hydrogen cyanide as a gas. The reaction is
NaCN(aq) + HCl(aq) → HCN( g ) + NaCl(aq)
Identify the conjugate acid–base pairs in this reaction. (Hint: It may be more obvious if
the spectator ions are removed.)
16.5 | One kind of baking powder contains sodium bicarbonate and calcium dihydrogen
phosphate. When water is added, a reaction occurs by the following net ionic equation.
HCO3-(aq) + H2PO4-(aq) → H2CO3(aq) + HPO42-(aq)
Identify the two Brønsted acids and the two Brønsted bases in this reaction. (The H2CO3
decomposes to release CO2, which causes the cake batter to rise.)
16.6 | When some of the strong cleaning agent “trisodium phosphate” is mixed with
household vinegar, which contains acetic acid, the following equilibrium is one of the
many that are established. (The position of equilibrium lies to the right.) Identify the pairs
of conjugate acids and bases.
PO43-(aq) + HC2H3O2(aq) �
� HPO42-(aq) + C2H3O2-(aq)
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746 Chapter 16 | Acids and Bases, A Molecular Look
Some molecules or ions are able to function either as an acid or as a base, depending on
the kind of substance mixed with them. For example, in its reaction with hydrogen chloride, water behaves as a base because it accepts a proton from the HCl molecule.
H2O + HCl( g ) → H3O+(aq) + Cl-(aq)
On the other hand, water behaves as an acid when it reacts with the weak base
H2O + NH3(aq) �
� NH4+(aq) + OH-(aq)
n From the Greek amphoteros, “partly
one and partly the other.”
Here, H2O donates a proton to NH3 in the forward reaction.
A substance that can be either an acid or a base depending on the other substance present is said to be amphoteric. Another term is amphiprotic, which stresses that the proton
donating or accepting ability is of central concern.
Amphoteric or amphiprotic substances may be either molecules or ions. For example, anions of acid salts, such as the bicarbonate ion of baking soda, are amphoteric.
The HCO3- ion can either donate a proton to a base or accept a proton from an acid.
Thus, toward the hydroxide ion, the bicarbonate ion is an acid; it donates its proton
HCO3-(aq) + OH-(aq) → CO32-(aq) + H2O
Toward hydronium ion, however, HCO3- is a base; it accepts a proton from H3O+.
HCO3-(aq) + H3O+(aq) → H2CO3(aq) + H2O
[Recall that H2CO3 (carbonic acid) almost entirely decomposes to CO2(g) and water as it
16.7 | Which of the following are amphoteric and which are not? Provide reasons for
your decisions. (a) H2PO4-, (b) HPO42-, (c) H2S, (d) H3PO4, (e) NH4+, (f ) H2O, (g) HI,
(h) HNO2. (Hint: Amphoteric substances must be able to provide an H+ and also react
with an H+.)
16.8 | The anion of sodium monohydrogen phosphate, Na2HPO4, is amphoteric. Using
H3O+ and OH-, write net ionic equations that illustrate this property.
16.2 | Strengths of Brønsted–Lowry
Acids and Bases
Brønsted–Lowry acids and bases have differing abilities to lose or gain protons. In this
section we examine how these abilities can be compared and how we can anticipate differences according to the locations of key elements in the periodic table.
Comparing Acids and Bases to a Relative Standard
When we speak of the strength of a Brønsted–Lowry acid, we are referring to its ability
to donate a proton to a base. We measure this by determining the extent to which the
reaction of the acid with the base proceeds toward completion—the more complete
the reaction, the stronger the acid. To compare the strengths of a series of acids, we have
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16.2 | Strengths of Brønsted–Lowry Acids and Bases
to select some reference base, and because we are interested most in reactions in aqueous media, that base is usually water (although other reference bases could also be
In Chapter 5 we discussed strong and weak acids from the Arrhenius point of view, and
much of what we said there applies when the same acids are studied using the Brønsted–
Lowry concept. Thus, acids such as HCl and HNO3 react completely with water to give
H3O+ because they are strong proton donors. Hence, we classify them as strong Brønsted–
Lowry acids. On the other hand, acids such as HNO2 (nitrous acid) and HC2H3O2 (acetic
acid) are much weaker proton donors. Their reactions with water are far from complete,
and we classify them as weak acids.
In a similar manner, the relative strengths of Brønsted bases are assigned according to
their abilities to accept and bind protons. Once again, to compare strengths, we have to
choose a standard acid. Because water is amphiprotic, it can serve as the standard acid as
well. Substances that are powerful proton acceptors, such as the oxide ion, react completely and are considered to be strong Brønsted–Lowry bases.
O2 − + H 2O
→ 2OH −
Weaker proton acceptors, such as ammonia, undergo incomplete reactions with water; we
classify them as weak bases.
Hydronium Ion and Hydroxide Ion in Water
Both HCl and HNO3 are very powerful proton donors. When placed in water they react
completely, losing their protons to water molecules to yield H3O+ ions. Representing
them by the general formula HA, we have
HA + H 2O
→ H3O+ + A −
Because both reactions go to completion, we really can’t tell which of the two, HCl or
HNO3, is actually the better proton donor (stronger acid). This would require a reference
base less willing than water to accept protons. In water, both HCl and HNO3 converted
quantitatively to another acid, H3O+. The conclusion, therefore, is that H3O+ is the strongest acid we will ever find in an aqueous solution, because stronger acids react completely
with water to give H3O+.
A similar conclusion is reached regarding hydroxide ion. We noted that the strong
Brønsted base O2- reacts completely with water to give OH-. Another very powerful
proton acceptor is the amide ion, NH2-, which also reacts completely with water.
Of course, we can also express this reaction in the form of a regular chemical equation,
NH 2− + H 2O 100
→ NH3 + OH−
Using water as the reference acid, we can’t tell which is the better proton acceptor, O2or NH2-, because both react completely, being replaced by another base, OH-. Therefore,
we can say that OH- is the strongest base we will ever find in an aqueous solution, because
stronger bases react completely with water to give OH-.
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