Acids and Bases

The earliest definition was given by Arrhenius:
◦ An acid contains a hydrogen atom and dissolved in
water for form a hydrogen ion, H+
HCl(g)
acid
H+(aq)
Cl−(aq)
+
◦ A base contains hydroxide and dissolved in water to
form OH-
NaOH(s)
base
Na+(aq)
+
OH− (aq)



The Arrhenius definition correctly predicts
the behavior of many acids and bases
However, this definition is limited and
sometimes inaccurate
For example, H+ does not exist in water.
Instead, it reacts with water to form the
hydronium ion, H3O+
H+(aq)
+
hydrogen ion:
does not really exist
in solution
H2O(l)
H3O+(aq)
hydronium ion:
actually present in
aqueous solution


A Brønsted–Lowry acid is a proton (H+) donor
A Brønsted–Lowry base is a proton (H+) acceptor
This proton is donated.
HCl(g) + H2O(l)


H3O+(aq) + Cl−(aq)
HCl is a Brønsted–Lowry acid because it donates
a proton to the solvent water
H2O is a Brønsted–Lowry base because it accepts
a proton from HCl


A Brønsted–Lowry acid must contain a
hydrogen atom
Common Brønsted–Lowry acids (HA):
HCl
hydrochloric acid
HBr
hydrobromic acid
HNO3
nitric acid
H2SO4
sulfuric acid
H
O
acidic H
H C C
atom
O H
H
acetic acid

A monoprotic acid contains one acidic
proton
◦ HCl

A diprotic acid contains two acidic protons
◦ H2SO4

A triprotic acid contains three acidic protons
◦ H3PO4

A Brønsted–Lowry acid may be neutral or it
may carry a net positive or negative charge
◦ HCl, H3O+, HSO4−


A B-L base is a proton acceptor, so it must be
able to form a bond to a proton
A base must contain a lone pair of electrons that
can be used to form a new bond to the proton
This e− pair forms a new
bond to a H from H2O
H
N
H
+
H
Brønsted–Lowry
base
H2O(l)
+
H
H
N
H
H
+ OH− (aq)

Common B-L bases:
Lone pairs make these
neutral compounds bases.
NH3
ammonia
H2O
water
The OH− is the base in
each metal salt.
NaOH
sodium hydroxide
KOH
potassium hydroxide
Mg(OH)2
magnesium hydroxide
Ca(OH)2
calcium hydroxide
This e− pair
stays on A
H A
acid
This e− pair forms
a new bond to H+.
gain of H+
+
B
base
loss of H+
A
−
+
H
B+
gain of H+
H A
acid
+
B
base
A −
+
H B+
conjugate conjugate
acid
base
loss of H+


The product formed by loss of a proton from an
acid is called its conjugate base
The product formed by gain of a proton by a base
is called its conjugate acid



HBr and Br- are a conjugate acid-base pair
H20 and H30+ are conjugate acid-base pair
The net charge must be the same on both
sides of the equation
gain of H+
H Br
acid
+ H2O
base
loss of H+
Br− + H3O+
conjugate conjugate
acid
base

When a species gains a proton (H+), it gains a
+1 charge
H2O
base
zero charge

add H+
H3O+
+1 charge
When a species loses a proton (H+), it
effectively gains a -1 charge
HBr
acid
zero charge
lose H+
Br−
−1 charge

Amphoteric compound: A compound that
contains both a hydrogen atom and a lone pair
of e−; it can be either an acid or a base
+
H
H
O
H
add H+
H2O as a base
H
O
H
H2O as an acid
H
O
H
conjugate acid
remove H+
−
H
O
conjugate base


When a covalent acid dissolves in water, the
proton transfer that forms H3O+ is called
dissociation
When a strong acid dissolves in water, 100%
of the acid dissociates into ions
HCl(g) + H2O(l)


H3O+(aq) + Cl−(aq)
A single reaction arrow is used, because the
product is greatly favored at equilibrium
Common strong acids are HI, HBr, HCl, H2SO4,
and HNO3


When a weak acid dissolves in water, only a
small fraction of the acid dissociates into ions
Unequal reaction arrows are used, because
the reactants are usually favored at
equilibrium
CH3COOH(l) + H2O(l)

H3O+(aq) + CH3COO−(aq)
Common weak acids are H3PO4, HF, H2CO3,
and HCN

When a strong base dissolves in water, 100%
of the base dissociates into ions
NaOH(s) + H2O(l)


Na+(aq) +
−OH(aq)
Common strong bases are NaOH and KOH
When a weak base dissolves in water, only a
small fraction of the base dissociates into
ions
NH3(g) + H2O(l)
NH4+(aq) + −OH(aq)

A strong acid readily donates a proton,
forming a weak conjugate base
HCl
strong acid

Cl−
weak conjugate base
A strong base readily accepts a proton,
forming a weak conjugate acid
OH−
strong base
H2O
weak conjugate acid

A Brønsted–Lowry acid–base reaction
represents an equilibrium
H A
acid


+
B
base
A − +
H B+
conjugate conjugate
acid
base
The position of the equilibrium depends upon
the strengths of the acids and bases
The stronger acid reacts with the stronger
base to form the weaker acid and the weaker
base

When the stronger acid and base are the
reactants on the left side, the reaction readily
occurs and the reaction proceeds to the right
H A + B
stronger stronger
acid
base

A − + H B+
weaker
weaker
base
acid
A larger forward arrow means that the
products are favored

When the stronger acid and base are the
products on the right side, equilibrium favors
the reactants and little product forms
H A + B
weaker weaker
acid
base

A − + H B+
stronger stronger
base
acid
A larger reverse arrow means that the
reactants are favored
HOW TO Predict the Direction of Equilibrium in an
Acid–Base Reaction
Are the reactants or products favored in
Example
the following acid–base reaction?
gain of H+
HCN(g)
acid
+
−OH(aq)
base
−CN(aq)
+ H2O(l)
conjugate conjugate
acid
base
loss of H+
Step [1]
Identify the acid in the reactants and the
conjugate acid in the products.
HOW TO Predict the Direction of Equilibrium in an
Acid–Base Reaction
Step [2]
Determine the relative strength of the acid
and the conjugate acid.
•From Table 9.1, HCN is a stronger acid than H2O.
Step [3]
HCN(g)
stronger
acid
Equilibrium favors the formation of the
weaker acid.
+
−OH(aq)
−CN(aq)
+ H2O(l)
weaker
acid
Products are favored.
HA(aq) + H2O(l)
K=
H3O+(aq) + A
− (aq)
[H3O+][A − ]
[H2O] [HA]
acid dissociation Ka = K [H2O]
constant


Ka =
[H3O+][A − ]
[HA]
The stronger the acid, the larger the Ka value
Equilibrium favors formation of the weaker
acid, the acid with the smaller Ka value


Water can behave as both a Brønsted–Lowry
acid and a Brønsted–Lowry base
Thus, two water molecules can react together
in an acid–base reaction:
loss of H+
H O H +H O H
acid
base
−
H
O H
+
conjugate
base
gain of H+
H O H
conjugate
acid
+

From the reaction of two water molecules, the
following equilibrium constant expression
can be written:
[H3O+][OH−]
K =
[H2O]2
ion-product Kw = K[H20]2 = [H3O+][OH−]
constant

Kw is a constant for all aqueous solutions at
25 oC
Kw
Kw
=
=
[H3O+][OH−]
(1.0 x 10−7) x (1.0 x 10−7)
Kw
=
1.0 x 10−14

To calculate [OH−]
when [H3O+] is known:
Kw = [H3O+][OH−]

To calculate [H3O+]
when [OH−] is known:
Kw = [H3O+][OH−]




The lower the pH, the higher the concentration
of H3O+
Acidic solution: pH < 7  [H3O+] >1 x 10−7
Neutral solution: pH = 7  [H3O+] = 1 x 10−7
Basic solution: pH > 7  [H3O+] < 1 x 10−7

If the [H3O+] in a cup of coffee is 1.0 x 10−5
M, then what is the value of [OH−]?
Kw
−14
1.0
x
10
−9 M
[OH−] =
=
=
1.0
x
10
1.0 x 10−5
[H3O+]



In this cup of coffee, [H3O+] > [OH–]
Is the solution is acidic or basic overall?
Acidic

A logarithm has the same number of places
after the decimal as there are digits in the
original number
If [H3O+] = 1.2 x 10–5 M for a solution, what is its pH?
pH = –log [H3O+]
pH = –log (1.2 x 10–5) 2 digits
pH = –(–4.92) 2 decimal places
pH = 4.92
The solution is acidic because the pH < 7
If the pH of a solution is 8.50, what is the [H3O+]?
pH = −log [H3O+]
8.50 = −log [H3O+]
−8.50 = log [H3O+]
antilog (−8.50 ) = [H3O+] 2 decimal places
[H3O+] = 3.2 x 10−9 M 2 digits
The solution is basic because [H3O+] > 1 x 10–7 M.

Neutralization reaction: An acid-base reaction
that produces a salt and water as products
HA(aq) + MOH(aq)
acid
base


H
OH(l) + MA(aq)
water
salt
The acid HA donates a proton (H+) to the OH−
base to form H2O
The anion A− from the acid combines with the
cation M+ from the base to form the salt MA
HOW TO Draw a Balanced Equation for a Neutralization
Reaction Between HA and MOH
Example
Write a balanced equation for the reaction
of Mg(OH)2 with HCl.
Step [1]
Identify the acid and base in the reactants
and draw H2O as one product.
HCl(aq) + Mg(OH)2(aq)
acid
base
H2O(l) + salt
water
HOW TO Draw a Balanced Equation for a Neutralization
Reaction between HA and MOH
Step [2]
Determine the structure of the salt.
•The salt is formed from the parts of the acid and
base that are not used to form H2O.
HCl
Mg(OH)2
H+
Cl−
reacts to used to
form H2O form salt
Mg2+
2 OH−
used to
react to
form salt form water
Mg2+ and Cl− combine to form MgCl2.
HOW TO Draw a Balanced Equation for a Neutralization
Reaction between HA and MOH
Step [3] Balance the equation.
Place a 2 to
balance O and H
2 HCl(aq) + Mg(OH)2(aq)
acid
base
Place a 2 to
balance Cl
2 H2O(l) + MgCl2(aq)
water
salt

A net ionic equation contains only the species
involved in a reaction
HCl(aq) + NaOH(aq)

H—OH(l) + NaCl(aq)
Written as individual ions:
H+(aq) + Cl−(aq) + Na+(aq) + OH− (aq)
H—OH(l) + Na+(aq) + Cl−(aq)


Omit the spectator ions, Na+ and Cl–
What remains is the net ionic equation:
H+(aq) + OH− (aq)
H—OH(l )

A bicarbonate base, HCO3−, reacts with one
H+ to form carbonic acid, H2CO3
H+(aq) + HCO3−(aq)
H2CO3(aq)
H2O(l) + CO2(g)

Carbonic acid then decomposes into H2O and
CO2
◦ For example:
HCl(aq) + NaHCO3(aq)
NaCl(aq) + H2CO3(aq)
H2O(l) + CO2(g)

A carbonate base, CO32–, reacts with two H+
to form carbonic acid, H2CO3
2 H+(aq) + CO32–(aq)
H2CO3(aq)
H2O(l) + CO2(g)

For example:
2 HCl(aq) + Na2CO3(aq)
2 NaCl(aq) + H2CO3(aq)
H2O(l) + CO2(g)


A salt can form an acidic, basic, or neutral
solution depending on whether its cation and
anion are derived from a strong or weak acid
and base
For the salt M+A◦ The cation M+ comes form the base
◦ The anion comes from the acid HA
NaCl
Na+
from NaOH
strong base
Cl−
from HCl
strong acid
A salt derived from a strong acid and strong base
forms a neutral solution (pH = 7).
NaHCO3
Na+
from NaOH
strong base
HCO3−
from H2CO3
weak acid
A salt derived from a strong base and a weak acid
forms a basic solution (pH > 7).
NH4Cl
NH4+
from NH3
weak base
Cl−
from HCl
strong acid
A salt derived from a weak base and a strong acid
forms an acidic solution (pH < 7)

The ion derived from the stronger acid or
base determines whether the solution is
acidic or basic



To determine the concentration of an acid or
base in a solution, we carry out a titration
If we want to know the concentration of an
acid solution, a base of known concentration
is added slowly until the acid is neutralized
When the acid is neutralized:
◦ # of moles of acid = # of moles of base
◦ This is called the end point of the titration.
mole–mole
conversion
factor
M (mol/L)
conversion
factor
Moles of
base
[1]
Volume of
base
[2]
Moles of
acid
[3]
Volume of
acid
M (mol/L)
conversion
factor
HOW TO Determine the Molarity of an Acid Solution
from Titration
Example What is the molarity of an HCl solution if
22.5 mL of a 0.100 M NaOH solution are
needed to titrate a 25.0 mL sample of the
acid?
volume of base (NaOH)
22.5 mL
conc. of base (NaOH)
0.100 M
volume of acid (HCl)
25.0 mL
conc. of acid (HCl)
?
HOW TO Determine the Molarity of an Acid Solution
from Titration
Step [1]
Determine the number of moles of base
used to neutralize the acid.
M (mol/L)
conversion
factor
22.5 mL NaOH x
1L
x 0.100 mol NaOH
1000 mL
1L
=
0.00225 mol NaOH
Moles of
base
HOW TO Determine the Molarity of an Acid Solution
from Titration
Step [2] Determine the number of moles of acid
that react from the balanced chemical
equation.
HCl(aq) + NaOH(aq)
Moles of
base
0.00225 mol NaOH x
H2O(l) + NaCl(aq)
mole–mole
conversion
factor
Moles of
acid
1 mol HCl = 0.00225 mol HCl
1 mol NaOH
HOW TO Determine the Molarity of an Acid Solution
from Titration
Step [3]
M
=
=
Determine the molarity of the acid from
the number of moles and the known
volume.
mol
L
=
0.00225 mol HCl x 1000 mL
1L
25.0 mL solution
0.0900 M HCl


A buffer is a solution whose pH changes
very little when acid or base is added
Most buffers are solutions composed of
roughly equal amounts of:
◦ A weak acid
◦ The salt of its conjugate base

The buffer resists change in pH because
◦ Added base, OH−, reacts with the weak acid
◦ Added acid, H3O+, reacts with the conjugate base

If an acid is added to the following buffer
equilibrium, then the excess acid reacts with
the conjugate base, so the overall pH does
not change much

If a base is added to the following buffer
equilibrium, then the excess base reacts with
the conjugate acid, so the overall pH does not
change much

The effective pH range of a buffer depends on
its Ka
H3O+(aq) + A −(aq)
HA(aq) + H2O(l)
Ka =

[H3O+][ A −]
[HA]
Rearranging this expression to solve for
[H3O+]:
[HA]
+
[H3O ] =
determines the
buffer pH
Ka
x
[ A −]


Normal blood pH is between 7.35 and 7.45
The principle buffer in the blood is carbonic
acid/ bicarbonate (H2CO3/HCO3−)
CO2(g) + H2O(l)
H2CO3(aq)
H2O
H3O+(aq) + HCO3−(aq)


CO2 is constantly produced by metabolic
processes in the body
The amount of CO2 is related to the pH of the
blood

Respiratory acidosis results when the body
fails to eliminate enough CO2, due to lung
disease or failure

Respiratory alkalosis is caused by
hyperventilating; very little CO2 is produced
by the body