The
Brønsted-Lowry
Theory of Acids
Two important theories have been developed to explain the properties of acids, bases, and salts. These are
the Arrhenius Theory and the Bronsted-Lowry Theory. Here, we’ll look at the Bronsted-Lowry theory.
This Theory of Acids
was proposed independently in 1923 by
This Theory of Acids was proposed independently in 1923 by two people
This Theory of Acids
was proposed independently in 1923 by
Johannes Nicolaus Brønsted, a Danish
chemist, and
Johannes Nicolaus Brønsted, a Danish chemist, and
This Theory of Acids
was proposed independently in 1923 by
Johannes Nicolaus Brønsted, a Danish
chemist, and
Thomas Martin Lowry, an English chemist.
Thomas Martin Lowry, an English chemist. We’ll introduce the main points of this theory.
A hydrogen atom “H”
has 1 proton and 1 electron.
Before we do that, just a quick word here about a hydrogen atom. It’s atomic number is one, so
it has 1 proton.
A hydrogen atom “H”
has 1 proton and 1 electron.
A neutral hydrogen atom also has one electron.
A hydrogen atom “H”
has 1 proton and 1 electron.
Over 99.98% of the
hydrogen atoms on Earth
contain no neutrons.
It is known that over 99.98% of the hydrogen atoms on Earth contain no neutrons.
A hydrogen atom “H”
has 1 proton and 1 electron,
and no neutrons.
Over 99.98% of the
hydrogen atoms on Earth
contain no neutrons.
So we’ll state here that a hydrogen atom has no neutrons, which is nearly always the case.
A hydrogen atom “H”
has 1 proton and 1 electron, and no neutrons.
If 1 electron is removed from an H atom, it
forms an H+ ion.
If one electron is removed from a hydrogen atom, it forms an H + ion.
A hydrogen atom “H”
has 1 proton and 1 electron, and no neutrons.
If 1 electron is removed from an H atom, it
forms an H+ ion. H  H   e 
Here’s the equation showing that.
A hydrogen atom “H”
has 1 proton and 1 electron, and no neutrons.
If 1 electron is removed from an H atom, it
forms an H+ ion. H  H   e 
1p
+
H –
1e
A neutral H atom has 1 proton and 1 electron,
A hydrogen atom “H”
has 1 proton and 1 electron, and no neutrons.
If 1 electron is removed from an H atom, it
forms an H+ ion. H  H   e 
Removing 1 electron,
1p
+
H –
1e
So removing an electron
A hydrogen atom “H”
has 1 proton and 1 electron, and no neutrons.
If 1 electron is removed from an H atom, it
forms an H+ ion. H  H   e 
Removing 1 electron, leaves 1 proton and “0”
electrons.
+1p
+
H –
1e
will leave a charge of +1, 1 proton, and zero electrons.
A hydrogen atom “H”
has 1 proton and 1 electron, and no neutrons.
If 1 electron is removed from an H atom, it
forms an H+ ion. H  H  e 
Removing 1 electron, leaves 1 proton and “0”
electrons.
H
+1p
So an H+ ion contains no electrons, no neutrons, and one proton
A hydrogen atom “H”
has 1 proton and 1 electron, and no neutrons.
If 1 electron is removed from an H atom, it
forms an H+ ion. H  H  e 
Removing 1 electron, leaves 1 proton and “0”
electrons.
Therefore, an H+ ion is
H
+1p
the same as 1 proton.
Therefore, an H+ ion is the same thing as one proton.
A hydrogen atom “H”
has 1 proton and 1 electron, and no neutrons.
If 1 electron is removed from an H atom, it
forms an H+ ion. H  H  e 
Removing 1 electron, leaves 1 proton and “0”
electrons.
Therefore, an H+ ion is
the same as 1 proton.
H+ = 1 proton
Or we can say that H+ equals 1 proton. We use the terms “H+ ion” and “proton” interchangeably
in Chemistry 12.
According to the Brønsted-Lowry
Theory:
According to the Brønsted-Lowry Theory:
According to the Brønsted-Lowry
Theory:
An acid is any species that donates a
proton (H+) to another species.
An acid is any species that donates a proton, or H+ ion to another species. Let‘s look at an
example of this.
According to the Brønsted-Lowry
Theory:
An acid is any species that donates a proton (H+) to another
species.
Cl H
HCl
OH
H
H2O
Let’s start with a molecule of Hydrogen chloride, HCl, and a molecule of water, H2O. Here are
Lewis structures for these molecules.
Cl H
–
HCl
+
OH
H
H2O
HCl is a polar molecule. There is a partial negative charge on the chlorine atom and a partial
positive charge on the hydrogen atom, shown by the delta minus and delta plus.
Cl H
–
HCl
+
–
OH
+
H+
H2O
The water molecule is also polar: the oxygen atom has a partial negative charge and each
hydrogen atom has a partial positive charge.
Cl H
–
HCl
+
–
OH
+
H+
H2O
The negative charge on the oxygen pulls the partially positive hydrogen atom (click) away from
the chlorine. The hydrogen leaves its shared electron with the chlorine atom.
Cl
Cl–
H
–
OH
+
H+
H2O
The chlorine atom has gained an electron, so it acquires a negative charge and becomes a Cl
minus ion.
Cl
Cl–
+
H
–
OH
+
H+
H2O
The hydrogen atom lost an electron, so it acquires a positive charge, forming an H+ ion.
Cl
Cl–
Which is also called a proton.
+
H
–
OH
+
H+
H2O
Cl
Cl–
+
H
–
OH
+
H+
H2O
The proton moves to the water molecule and (click) attaches to one of its lone pairs.
+
Cl
Cl–
H
OH
H
Instead of staying with the hydrogen atom, the positive charge is considered as the charge of
the whole ion. So we’ll (click) move it over here.
Cl
Cl–
H
OH
H
+
H3O +
There are now 3 H atoms attached to one O, so the formula is H3O instead of H2O
Cl
Cl–
H
OH
H
We’ll draw square brackets around the H3O because it is an ion.
+
H3O +
Cl
Cl–
And write the positive charge here in the formula.
H
OH
H
+
H3O +
Cl
Cl–
Chemists call the H3O + ion, the hydronium ion.
H
OH
H
+
Hydronium ion
H3O +
So we can summarize the whole process here. We started with a molecule of HCl
+
And we added a molecule of water.
+
Which gave us
+
An H3O +, or hydronium ion
+
plus
+
+
a Cl minus, or chloride ion.
+
+
HCl (g)  H 2O(l ) 

H 3O(aq)

We can now write an equation to show this process. We write HCl gas

Cl (aq)
+
HCl (g)  H 2O(l ) 
Plus H2O liquid.
+

H 3O(aq)


Cl (aq)
+
HCl (g)  H 2O(l ) 
forms
+

H 3O(aq)


Cl (aq)
+
HCl (g)  H 2O(l ) 
H3O + aqueous
+

H 3O(aq)


Cl (aq)
+
HCl (g)  H 2O(l ) 
Plus Cl minus aqueous
+

H 3O(aq)


Cl (aq)
HCl (g)  H 2O(l ) 
We can also represent the process like this.

H 3O(aq)


Cl (aq)
H+
HCl (g)  H 2O(l ) 

H 3O(aq)


Cl (aq)
This shows that an H+ ion, or proton is being transferred from the HCl molecule to the water
molecule.
proton transfer
HCl (g)  H 2O(l ) 
So this can also be called a proton transfer.

H 3O(aq)


Cl (aq)
proton transfer
HCl (g)  H 2O(l ) 

H 3O(aq)
H2O has gained 1
proton, so it forms
an H3O+ ion
H2O has gained 1 proton, so it forms an H3O+ ion


Cl (aq)
proton transfer
HCl (g)  H 2O(l ) 

H 3O(aq)


Cl (aq)
HCl has lost 1
proton, so it forms
a Cl– ion
HCl has lost 1 proton, or H+ ion, so it forms a Cl– ion
According to the Brønsted-Lowry
Theory:
An acid is any species that donates a proton (H+) to another
species.
proton transfer
HCl (g)  H 2O(l ) 

H 3O(aq)


Cl (aq)
Brønsted-Lowry
Acid
Because the HCl is losing, or donating a proton, according to the Bronsted-Lowry theory, it is called
an acid. To indicate this, we often call it a Bronsted-Lowry acid, or Bronsted acid for short.
According to the Brønsted-Lowry
Theory:
An acid is any species that donates a proton (H+) to another
species.
A base is any species that accepts a
proton (H+) from another species.
BrønstedLowry
definition
of a BASE
According the Bronsted-Lowry theory, a base is defined as any species that gains or accepts a
proton, or H+ ion, from another species.
According to the Brønsted-Lowry
Theory:
An acid is any species that donates a proton (H+) to another
species.
A base is any species that accepts a proton (H+) from another
species.
proton transfer
HCl (g)  H 2O(l ) 

H 3O(aq)


Cl (aq)
Brønsted-Lowry
Base
Because the H2O is gaining, or accepting a proton, it can be defined as a Bronsted-Lowry Base,
or Bronsted base for short.
According to the Brønsted-Lowry
Theory:
An acid is any species that donates a proton (H+) to another
species.
proton transfer
HCl (g)  H 2O(l ) 
Acid
Base
So in this reaction, HCl is an acid,

H 3O(aq)


Cl (aq)
According to the Brønsted-Lowry
Theory:
An acid is any species that donates a proton (H+) to another
species.
A base is any species that accepts a proton (H+) from another
species.
proton transfer
HCl (g)  H 2O(l ) 
Acid
and water is a base.
Base

H 3O(aq)


Cl (aq)
proton transfer
HCl (g)  H 2O(l ) 
Not ions
Because the reactants HCl and H2O are not ions.

H 3O(aq)


Cl (aq)
proton transfer
HCl (g)  H 2O(l ) 
Not ions
But the products H3O+ and Cl minus ARE ions.

H 3O(aq)

Ions

Cl (aq)
HCl (g)  H 2O(l ) 

H 3O(aq)


Cl (aq)
of HCl
We can call this process the ionization of HCl.
Write the equations for the ionization of the
following acids when they are added to water.
Identify the acids and the bases on the reactant
side:
HNO3(aq)  H 2O(l ) 
HCN (g)  H 2O(l ) 

H 3O(aq)

H 3O(aq)



NO 3(aq)

CN (aq)
Here’s a question. We’re asked to write the equations for the ionization of the following acids
when they are added to water, and to identify the acids and the bases on the reactant side:
HNO3(aq)  H 2O(l ) 
HCN (g)  H 2O(l ) 

H 3O(aq)

H 3O(aq)



NO 3(aq)

CN (aq)
We’ll start the first reaction by adding liquid water as the other reactant.
HNO 3(aq)  H 2O(l ) 
HCN (g)  H 2O(l ) 

H 3O(aq)

H 3O(aq)



NO3(aq)

CN (aq)
The formula for this compound starts with an H, so we assume it acts as an acid.
HNO3(aq)  H 2O(l ) 

H 3O(aq)


NO 3(aq)
acid
HCN (g)  H 2O(l ) 
And we’ll label it as an acid.

H 3O(aq)


CN (aq)
HNO3(aq)  H 2O(l ) 
acid

H 3O(aq)


NO3(aq)
base
HCN (g)  H 2O(l ) 

H 3O(aq)
So the other reactant, water, must act as a base in this case.


CN (aq)
HNO3(aq)  H 2O(l ) 
acid
Now an acid is a proton donor


NO3(aq)
base
HCN (g)  H 2O(l ) 
proton
donor

H 3O(aq)

H 3O(aq)


CN (aq)
HNO3(aq)  H 2O(l ) 
acid


NO3(aq)
base
HCN (g)  H O
proton
donor

H 3O(aq)
proton
acceptor
2 (l )
And a base is a proton acceptor.


H 3O(aq)


CN (aq)
H+
HNO3(aq)  H 2O(l ) 
acid


NO3(aq)
base
HCN (g)  H O
proton
donor

H 3O(aq)
proton
acceptor
2 (l )


H 3O(aq)


CN (aq)
So that means a proton, or H+ ion will be transferred (click) from the acid, HNO3, to the base,
H2O.
H+
HNO3(aq)  H 2O(l ) 
acid

H 3O(aq)


NO3(aq)
base
Will gain an H+ ,

which is one H
3 one
(aq)
and
+ charge
HCN (g)  H 2O(l )  H O


CN (aq)
So this means that water will gain an H+, which is one H and one + charge
H+
HNO3(aq)  H 2O(l ) 
acid

H 3O(aq)


NO3(aq)
base
The 
hydronium
3 (aq)
ion
HCN (g)  H 2O(l )  H O


CN (aq)
Adding an H and one + charge to water, gives us H3O+, the hydronium ion.
H+
HNO3(aq)  H 2O(l ) 
acid


NO3(aq)
base
Has lost one H+,
which is one H and
(g)
2 (l )
one + charge
HCN

H 3O(aq)
 HO


H 3O(aq)


CN (aq)
The acid HNO3, has lost one H+, which means it has lost one H and one + charge
H+
HNO3(aq)  H 2O(l ) 
acid

H 3O(aq)


NO3(aq)
base
HCN (g)  H 2O(l ) 


H 3OSo
 CN
it will
form (aq)
(aq)
–
NO3 , the nitrate
ion
We remove one H atom from HNO3 giving us NO3, and taking away one + charge is the same as adding
one negative charge, so we have NO3 minus, the nitrate ion. Both of the new ions formed are aqueous.
HNO3(aq)  H 2O(l ) 
acid

H 3O(aq)


NO3(aq)
base
HCN (g)  H 2O(l ) 

H 3O(aq)


CN (aq)
of HNO3
So this is the equation for the ionization of HNO3. HNO3 is the acid on the left side and water is
the base on the left side.
HNO3(aq)  H 2O(l ) 
acid

H 3O(aq)


NO3(aq)
base
HCN (g)  H 2O(l )

H 3O(aq)
Now, we’ll see what we get if we add HCN to water.


CN (aq)
HNO3(aq)  H 2O(l ) 
acid

H 3O(aq)


NO3(aq)
base
HCN (g)  H 2O(l )

H 3O(aq)
acid
Because HCN starts with H, we’ll treat it as an acid.


CN (aq)
HNO3(aq)  H 2O(l ) 
acid
And H2O as a base


NO3(aq)
base
HCN (g)  H 2O(l )
acid

H 3O(aq)
base

H 3O(aq)


CN (aq)
HNO3(aq)  H 2O(l ) 
acid
H+


NO3(aq)
base
HCN(g)  H2O(l )
acid

H 3O(aq)

H3O(aq)
base
So there will be a (click) proton transfer from HCN to H2O.


CN(aq)
HNO3(aq)  H 2O(l ) 
acid
H+


NO3(aq)
base
HCN (g)  H 2O(l )
acid

H 3O(aq)

H 3O(aq)

base
The H2O will gain a proton, or H+ and form H3O +, or hydronium

CN (aq)
HNO3(aq)  H 2O(l ) 
acid
H+


NO3(aq)
base
HCN (g)  H 2O(l )
acid

H 3O(aq)

H 3O(aq)
base
And the HCN will lose a proton, or H+, and form CN minus


CN (aq)

H 3O(aq)
HNO3(aq)  H 2O(l )  of HCN

acid
base
HCN (g)  H 2O(l )
acid

NO3(aq)

H 3O(aq)


CN (aq)
base
So this is the equation for the ionization of HCN. HCN is the acid on the left and water is the
base. A double arrow is used here because the ionization of HCN does not go to completion.

H 3O(aq)
HNO3(aq)  H 2O(l )  of HCN

acid
base
HCN (g)  H 2O(l )
acid

NO3(aq)

H 3O(aq)


CN (aq)
base
In a solution of HCN, only a few molecules are ionized. You’ll be shown how you can tell
whether to use a single arrow or double arrow later in the course.
Amphiprotic
Species
Consider the reaction:
NH 3(g)  H 2O(l )

NH4(aq)


OH(aq)
Now, lets consider this reaction. Again, the double arrow here just tells us that this reaction
does not go to completion. Instead, an equilibrium exists here.
NH 3(g)  H 2O(l )

NH4(aq)
We see that the NH3 has been converted to NH4 +.


OH(aq)
NH 3(g)  H 2O(l )

NH4(aq)


OH(aq)
In doing so, it gains one H and one + charge, therefore it gains one H+, or one proton.
NH 3(g)  H 2O(l )

NH4(aq)


OH(aq)
base
Because it gains a proton in this case, the NH3 is classified as a base.
NH 3(g)  H 2O(l )

NH4(aq)


OH(aq)
base
Looking at the water, we see that is has been converted to OH minus.
NH 3(g)  H 2O(l )

NH4(aq)


OH(aq)
base
One less
H+ than
H2O
OH minus has one less H and one less + than H2O, therefore it has one less H+ than H2O.
NH 3(g)  H 2O(l )

NH4(aq)


OH(aq)
base
therefore when H2O converts to OH minus, it loses an H+, or proton.
NH 3(g)  H 2O(l )
base

NH4(aq)
acid
For that reason, we identify H2O as an acid in this case.


OH(aq)
NH 3(g)  H 2O(l )
base


OH(aq)


NO3(aq)
acid
HNO3(aq)  H 2O(l ) 
acid

NH4(aq)

H 3O(aq)
base
Now, we’ll look at a previous reaction we had in which HNO3 reacts with water.
NH 3(g)  H 2O(l )
base


OH(aq)


NO3(aq)
acid
HNO3(aq)  H 2O(l ) 
acid

NH4(aq)
base
In this reaction, water acted as a base.

H 3O(aq)
NH 3(g)  H 2O(l )
base


OH(aq)


NO3(aq)
acid
HNO3(aq)  H 2O(l ) 
acid

NH4(aq)

H 3O(aq)
base
So we can see that, depending on what it’s reacting with, water can play the role of an acid or
the role of a base.
NH 3(g)  H 2O(l )
base


OH(aq)


NO3(aq)
acid
HNO3(aq)  H 2O(l ) 
acid

NH4(aq)
base
Amphiprotic
Such a species is said to be amphiprotic.

H 3O(aq)
NH 3(g)  H 2O(l )
base


OH(aq)


NO3(aq)
acid
HNO3(aq)  H 2O(l ) 
acid

NH4(aq)

H 3O(aq)
base
Amphiprotic
can act
either as an
acid or as a
base
An amphiprotic species is one that can act either as an acid or as a base, depending on what it is
with. Water is one amphiprotic substance. There are many more, as we shall see later in this unit.
Monoprotic, Diprotic,
Triprotic, and
Polyprotic Acids
HCl (g)  H 2O(l ) 
acid

H 3O(aq)
base
HNO3(aq)  H 2O(l ) 
acid


Cl (aq)
base
Consider these two acids, HCl, and HNO3.

H 3O(aq)


NO3(aq)
HCl (g)  H 2O(l ) 
acid

H 3O(aq)
base
HNO3(aq)  H 2O(l ) 
acid


Cl (aq)
base
Both of these are able to lose one proton only.

H 3O(aq)


NO3(aq)
Monoprotic
HCl (g)  H 2O(l ) 
acid
Monoprotic

H 3O(aq)
base
HNO3(aq)  H 2O(l ) 
acid


Cl (aq)

H 3O(aq)


NO3(aq)
base
An acid that has one proton available to donate is called a monoprotic acid. So both HCl and
HNO3 are monoprotic acids.
H 2SO4(aq)  H 2O(l ) 
Now, consider this acid, H2SO4. It’s called

H 3O(aq)


HSO4(aq)
H 2SO4(aq)  H 2O(l ) 
sulphuric acid
sulphuric acid

H 3O(aq)


HSO4(aq)
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)
sulphuric acid
Notice it has 2 H atoms at the beginning of the formula.


HSO4(aq)
Diprotic acid
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
sulphuric acid
Acids that have two protons they can donate, are said to be diprotic. So H2SO4 is a diprotic
acid.
Diprotic acid
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
sulphuric acid
However, when diprotic acids like H2SO4, are added to water, they do not lose both of their
protons at once. They do it in steps, losing one proton at a time.
H+
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
sulphuric acid
In the first step of the reaction of H2SO4 in water, the H2SO4 loses one proton to water.
H+
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)

sulphuric acid
So the water is converted to H3O +, or hydronium. (end of statement)

HSO4(aq)
H+
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
Will lose one
H and one +
charge
sulphuric acid
Because the H2SO4 is losing an H+, it means it’s losing one H atom and one + charge.
H+
H 2SO4(aq)  H 2O(l ) 
Will lose one
H and one +
charge
sulphuric acid
This gives us HSO4 with a minus charge.

H 3O(aq)


HSO4(aq)
H+
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
sulphuric acid
Called the
hydrogen sulphate
or bisulphate ion
HSO4 minus is called the hydrogen sulphate, or bisulphate ion.
H+
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)
sulphuric acid
Notice that HSO4 minus has one hydrogen it can lose.


HSO4(aq)
Has one H that
it can lose
H+
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
sulphuric acid

HSO4(aq)
 H2O(l )

H3O(aq)

2
SO4(aq )
So in the second step of the ionization of sulphuric acid, the HSO4 minus will react with
H+
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
sulphuric acid

HSO4(aq)
Water.
 H2O(l )

H3O(aq)

2
SO4(aq )
H+
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
sulphuric acid
H+

HSO4(aq)
 H 2 O (l )
And donate its proton to water

H 3O(aq)

2
SO4(aq )
H+
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
sulphuric acid
H+

HSO4(aq)
 H 2 O (l )

H 3O(aq)
Which would produce another hydronium ion. (end of statement)

2
SO4(aq )
H+
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
sulphuric acid
H+

HSO4(aq)
 H 2 O (l )

H 3O(aq)

2
SO4(aq )
Loses one H
and one +
charge
The HSO4 minus ion loses a proton, so it loses one hydrogen and one positive charge.
H+
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
sulphuric acid
H+

HSO4(aq)
 H 2 O (l )
Loses one H
and one +
charge
So it will be left as SO4.

H 3O(aq)

2
SO4(aq )
H+
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
sulphuric acid

HSO4(aq)
 H 2 O (l )

H 3O(aq)

2
SO4(aq )
hydrogen
sulphate ion
And losing one positive charge will cause its charge to go down one, from negative 1 to
negative 2, or 2 minus.
H+
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
sulphuric acid

HSO4(aq)
 H 2 O (l )

H 3O(aq)

2
SO4(aq)
hydrogen
sulphate ion
The product SO4 2minus, or sulphate ion, does not have any hydrogens to donate, so this is the
last step in the ionization of sulphuric acid
H+
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
sulphuric acid

HSO4(aq)
 H 2 O (l )

H 3O(aq)

2
SO4(aq )
hydrogen
sulphate ion
The double arrow here implies there is an equilibrium. This reaction does not go to completion.
H 2SO4(aq)  H 2O(l ) 

H 3O(aq)


HSO4(aq)
sulphuric acid

HSO4(aq)
 H 2 O (l )

H 3O(aq)

2
SO4(aq)
hydrogen
sulphate ion
Whereas in the first step of the ionization of sulphuric acid, there is a single arrow, which means this step
goes to completion. 100% of the sulphuric acid is converted to hydronium and hydrogen sulphate ions.
H 3PO4(aq)  H 2O(l )
phosphoric acid
Now we’ll look at phosphoric acid.

H 3O(aq)


H 2PO4(aq)
H 3PO4(aq)  H 2O(l )

H 3O(aq)


H 2PO4(aq)
phosphoric acid
We see that it has 3 H’s at the front of its formula, which means it has three protons it can
lose.
Triprotic acid
H 3PO4(aq)  H 2O(l )

H 3O(aq)
phosphoric acid
For that reason, phosphoric acid is called a triprotic acid.


H 2PO4(aq)
Triprotic acid
H 3PO4(aq)  H 2O(l )

H 3O(aq)


H 2PO4(aq)
phosphoric acid
When it’s combined with water, it doesn’t lose all three protons at once. Just one at a time.
H+
H 3PO4(aq)  H 2O(l )

H 3O(aq)


H 2PO4(aq)
phosphoric acid
In the first step, one proton is transferred to a water molecule, so it produces one hydronium
ion. (end of statement)
H+
H 3PO4(aq)  H 2O(l )
Loses one H
and one +
charge
because it loses one H and one positive charge,

H 3O(aq)


H 2PO4(aq)
H+
H 3PO4(aq)  H 2O(l )

H 3O(aq)
Loses one H
and one +
charge
the other product would have 2 H’s and a charge of negative 1.


H 2PO4(aq)
H+
H 3PO4(aq)  H 2O(l )

H 3O(aq)


H 2PO4(aq)
called the
dihydrogen
phosphate ion
H2PO4 minus is called the dihydrogen phosphate ion. We see that it still has two protons it can
donate.
H+
H 3PO4(aq)  H 2O(l )

H 3O(aq)


H 2PO4(aq)
H+

H 2PO4(aq)
 H 2 O (l )

H 3O(aq)

2
HPO4(aq )
In the second step, the H2PO4 minus ion loses one of its protons to water
H+
H 3PO4(aq)  H 2O(l )

H 3O(aq)


H 2PO4(aq)
H+

H 2PO4(aq)
 H 2 O (l )
so it produces another hydronium ion

H 3O(aq)

2
HPO4(aq )
H+
H 3PO4(aq)  H 2O(l )

H 3O(aq)


H 2PO4(aq)
H+

H 2PO4(aq)
 H 2 O (l )

H 3O(aq)
Loses one H
and one +
charge
And since this loses one H and one positive charge,

2
HPO4(aq )
H+
H 3PO4(aq)  H 2O(l )

H 3O(aq)


H 2PO4(aq)
H+

H 2PO4(aq)
 H 2 O (l )
Loses one H
and one +
charge
The other product would be HPO4 2 minus.

H 3O(aq)

2
HPO4(aq)
H+
H 3PO4(aq)  H 2O(l )

H 3O(aq)


H 2PO4(aq)
H+

H 2PO4(aq)
 H 2 O (l )

H 3O(aq)

2
HPO4(aq )
Called the
monohydrogen
phosphate ion
this product is called the monohydrogen phosphate ion. We can see that it has one proton
available to donate.
H+
H 3PO4(aq)  H 2O(l )

H 3O(aq)


H 2PO4(aq)
H+

H 2PO4(aq)
 H 2 O (l )

H 3O(aq)
 H 2 O (l )

H 3O(aq)

2
HPO4(aq)

3
PO4(aq )
H+
2
HPO4(aq )
So in the last step, the HPO4 2minus donates its single proton to water, producing another
hydronium ion. (end of statement)
H+
H 3PO4(aq)  H 2O(l )

H 3O(aq)


H 2PO4(aq)
H+

H 2PO4(aq)
 H 2 O (l )

H 3O(aq)
 H 2 O (l )

H 3O(aq)

2
HPO4(aq)

3
PO4(aq )
H+
2
HPO4(aq )
Loses one H and
one + charge
And because it loses one H and one positive charge,
H+
H 3PO4(aq)  H 2O(l )

H 3O(aq)


H 2PO4(aq)
H+

H 2PO4(aq)
 H 2 O (l )

H 3O(aq)
 H 2 O (l )

H 3O(aq)

2
HPO4(aq)

3
PO4(aq)
H+
2
HPO4(aq )
Loses one H and
one + charge
the final ion that it forms is just PO4 3 minus.
H+
H 3PO4(aq)  H 2O(l )

H 3O(aq)


H 2PO4(aq)
H+

H 2PO4(aq)
 H 2 O (l )

H 3O(aq)
 H 2 O (l )

H 3O(aq)

2
HPO4(aq)

3
PO4(aq)
H+
2
HPO4(aq )
The phosphate ion
This is simply called the phosphate ion
Polyprotic Acid
Polyprotic acid
Polyprotic Acid
—just a general term for any
acid that can donate more than
one proton.
is just a general term for any acid that can donate more than one proton.
Polyprotic Acid
—just a general term for any
acid that can donate more than
one proton.
e.g.) H2CO3 (carbonic acid-diprotic)
Examples could be carbonic acid, which is diprotic
Polyprotic Acid
—just a general term for any
acid that can donate more than
one proton.
e.g.) H2CO3 (carbonic acid-diprotic)
H3PO4 (phosphoric acid-triprotic)
phosphoric acid, which is triprotic
Polyprotic Acid
—just a general term for any
acid that can donate more than
one proton.
e.g.) H2CO3 (carbonic acid-diprotic)
H3PO4 (phosphoric acid-triprotic)
H4P2O7 (pyrophosphoric acid)
and pyrophosphoric acid, which has 4 protons it can donate.