Acids and
Bases
What do lemon juice, lime
juice, sauerkraut, vinegar
and sour cream have in
common?
Acids

Sour

Electrolytes

Change the color of some indicators


litmus paper changes from blue to red
phenolphthalein is

React with metals to produce hydrogen

React with bases to produce water and a
salt
Naming acids

Binary





two elements
hydro-element-ic
ex: hydrochloric  HCl
hydrosulfuric  H2S
Ternary



three or more elements
oxyacids  polyatomic ions
ex: sulfuric  H2SO4
chloric  HClO3
Bases




Bases
Bases
Bases
Bases
have a bitter taste.
are slippery.
are electrolytes.
change the color of some indicators:
 litmus paper changes from red to blue
 phenolphthalein is
 Bases react with acids to form water and a salt.
 An example of a base is soap
 Bases are called alkaline
Naming Bases

Name positive ion – hydroxide

ex: NaOH  sodium hydroxide

NH4OH  ammonium hydroxide
Naming acids

Stop and practice by doing worksheet 1
Theories of Acids and Bases
Arrhenius
Bronsted-Lowry
Lewis
Arrhenius Acids


Oldest, most common, traditional
Defines an acid as any compound
containing hydrogen that ionizes to
yield [H3O+ ] in aqueous solution.


Standard acids that begin with an -H
Ex: HCl, HI, H2SO4
Arrhenius Acids

Frequently ionizes more that one
hydrogen

Called polyprotic (why?)
Monoprotic ionizes 1 hydronium
Diprotic ionizes 2 H+
Triprotic ionizes 3 H+

What is acetic acid?



HNO3
H2SO4
H3PO4
Arrhenius Bases


Oldest, most common, traditional
Defines a base as any compound
containing hydroxide that ionizes to
yield [OH- ] in aqueous solution.


Standard bases that end with an -OH
Ex: NaOH, KOH
Arrhenius Bases


Families I and II on the periodic
table readily form bases when mixed
with water.
Alkali/alkaline means basic



Family I/II + H2O  metal(OH) + H2
Na
+ H2O  NaOH
+ H2
Note the base quickly dissociates.
Acid/Base/Salt

Acid – hydrogen ions

Base –hydroxide ions

Salt – metal-nonmetal

ionic compound
Bronsted-Lowry acid/base

Includes all Arrhenius and more.

Acid = hydrogen donor


It gives away hydrogen ions
Base = hydrogen acceptor

It takes hydrogen
Bronsted-Lowry acid/base

NH3 + H2O  NH4+ + OHBase
Acid
Although we are mixing an acid and a base the solution
is not neutral. It is basic, since there is more –OH- on
the product side.
Bronsted-Lowry acid/base

HF + H2O  H3O+ + FAcid
Base
Although we are mixing an acid and a base the solution
is not neutral. It is acidic, since there is “extra” H3O+
on the product side.
Bronsted-Lowry acid/base
Notice the reverse of this equation.

HF + H2O  H3O+ + FAcid
Base
In the reverse equation the acid and base changes
places, this is called conjugate acid and conjugate
base pairs.
Bronsted-Lowry acid/base

HF + H2O  H3O+ + FAcid
Base
conj. Acid
conj. Base
Conjugate acid – formed when a base acquires a proton
H+ to form the acid.
Conjugate base – the particle that remains after the
proton has been released by the acid.
Bronsted-Lowry acid/base


The Bronsted-Lowry theory is needed to explain
how things such as NH3 and NaHCO3 (ammonia and
sodium bicarbonate) are bases.
Although not all Bronsted-Lowry acid/bases are
Arrhenius, all Arrhenius acid/bases are BronstedLowry.
Bronsted-Lowry acid/base

Identify the acid, base, conj. acid,
conj. base for the following reaction.
HNO3 +
NaOH  NaNO3 + H2O
Bronsted-Lowry acid/base

Identify the acid, base, conj. acid,
conj. base for the following reaction.
NaHCO3 + HCl  NaCl + H2CO3
Amphoteric


A substance that can be both an acid
and a base.
Water has a split personality. It can
be both an acid and base.
Lewis acid/base

Acid – electron pair acceptor.

Base – electron pair donor.
Note: this theory does not need H+
for an acid or base.
Lewis acid/base
BF3
+
F
l
F-B
NH3 
H
l
+
:N-H 
l
F
l
H
Acid
Base
F
l
H
l
N–H
l
l
F H
F-B
..
Lewis acid/base
F
l
F-B
l
F
H
l
+
:N-H 
l
H
F
l
H
l
F-B . . N – H
l
F
l
H
Acid
Base
Ammonia donates electrons to Boron, ammonia is the base.
Boron wants eight electrons, so it will accept the electrons
so it is a acid.
Summary of Acid/Base
Theories
Acid
Arrhenius
releases H+ ions releases OHin water.
ions in water.
donates a
Bronsted-Lowry
proton (H+)
Lewis
Bases
accepts an
electron pair
accepts
protons. (H+)
donates an
electron pair
Summary of Acid/Base
Theories
Arrhenius
Bronsted-Lowry
Lewis Acid & Bases
Hydrogen Ions from Water

At times water collides with so much
force that one water molecule looses
a H+ to another water molecule – this
is called self ionization
..
H–O: +
l
H
..
H–O:
l
H

..
+
H–O–H
l
H
+
.. H–O:
..
Hydrogen Ions from Water
Simplified

Water decomposes into hydrogen ion
and hydroxide.
..
H–O:
l
H

H
+
+
.. H–O:
..
•In reality there is so much water that H+ never
occurs it is always H3O+ (hydronium)
Pure water



In pure water the [OH-] = [H3O+]
Neutral - any aqueous solution where
[OH-] = [H3O+].
[H3O+] = 1.0 E -7 mol/L
Pure water

[OH-] and [H3O+] are inversely
related.

Therefore, [OH-] [H3O+] = Kw

Kw = 1.0E-14 (mol/L)2

Ion-product constant for water
Acidic solutions
HCl + H2O  H+ + Cl- + H+ + OH2H+ and 1 OHRemember there is so much water in an
aqueous solution that the H+s are
really H3O+.
Therefore, [H3O+] > [OH-]
[OH-] vs. [H3O+]

[H3O+] > [OH-]  acid

[H3O+] < [OH-]  basic


Ex: NaOH + H2O  Na+ + OH- + H+ + OH-
[H3O+] = [OH-]  neutral
+
O]
[H3
vs.
problems
[OH ]
Calculate [OH-] if:
a)
[H3O+] = 1E-1 M
b)
[H3O+] = 1E-12 M
Calculate [H3O+] if:
a)
[OH-] = 1E-3 M
b)
[OH-] = 1E-10 M
practice
How strong
is your acid?
How strong is your acid?

A strong acid
completely ionizes
(dissociates) in
aqueous solution.
HCl
water
+
-
H + Cl
How strong is your acid?

A weak acid slightly
ionizes (dissociates)
in aqueous solution.
HC2H3O2
water
+
H +
C2H3O2
Qualitative vs. Quantitative
“strength”

“strong” and “weak” are qualitative in
chemistry we must also have a
quantitative description.
Acid dissociation constant



Acid dissociation constant (Ka) – the
ratio of the concentration of the
dissociated form of the acid to the
undissociated.
Ex: HX  H+ + XKa = [H+] [X-]
[HX]
Acid dissociation constant




If Ka is large then a lot of the acid
dissociates.
If Ka is small then a little of the acid
dissociates.
Ka is big = strong acid
Ka is small = weak acid
Strong and weak bases

Strong bases – dissociate
“completely” into metal ions and
hydroxide ions in aqueous solutions.


NaOH
Na+ + OH-
Weak bases – do NOT dissociate
“completely” in an aqueous solution.

NH3
+
H2O  NH4OH
NH4+ + OH-
Neutralization reactions
Neutralization reactions


A reaction in which an acid and a base
(Arrhenius) react in aqueous solution
to produce water and a salt.
Double displacement reaction
HCl + NaOH  H2O + NaCl
H2SO4 + 2KOH  2H2O + K2SO4
Relative
Strengths
of Acids
and Bases
(Downhill
rule)
Downhill Rule
Simple way to predict BronstedLowry reactions.

1.
2.
3.
4.
5.
Write the reaction
Decide which is the acid/base (ignore all
spectators)
Locate the acid in the left hand column and
the base in the right.
If the line connecting the two goes down
left-to-right the rxn will occur.
Write the products. (opposite sides with
spectators)
Neutralization practice problem

Will a neutralization reaction occur
between:
HNO3 + KOH  ?
acid
base (OH-)
K+ spectator
Neutralization practice problem

It is “downhill”
Opposite side
Opposite side
-
HNO3 + KOH  NO3 + H2O + K
acid
base (OH-) conj.
K+ spectator
Base
conj.
Acid
+
spectator
Neutralization practice problem

Join the two independent ions.
In water they are dissociated, but
they are written as a compound.
HNO3 + KOH
acid

KNO3 + H2O
base (OH-)
conj.
K+ spectator
Base
conj.
Acid
Neutralization practice problem

Will a neutralization reaction occur
between:
NH4+ + NaOH  ?
acid
base (OH-)
Na+ spectator
Neutralization practice problem

It is “downhill”
NH4Cl +
acid
Cl- spectator
NaOH  NH3 + H2O + NaCl
base (OH-)
conj.base
Na+ spectator
conj. Acid
spectators
Neutralization practice problem


Will a neutralization reaction occur
between:
HF + H2O  ?
Neutralization practice problem

It is “uphill”

HF + H2O  No rxn
pH and pOH
pH and
pOH
pH



Indicates the hydronium ion
concentration.
The negative logarithm of the
hydronium ion concnetration.
pH = - log [H3O+]
pH scale

Always use pH to determine acid/base/neutral
0 1 2 3 4 5 6 7 8 9 1011 12 13 14
strong
acid
neutral
strong
base
pH
What is the pH of pure water?
U: pH
K: [H3O+] = 1.0E-7M
P: pH = - log [H3O+]
S: pH = - log [1.0E-7M]
=
7

pH
Calculate the pH of a solution with a
hydronium ion concentration of 5.0E6M?
U: pH
K: [H3O+] = 5.0E-6M
P: pH = - log [H3O+]
S: pH = - log [5.0E-6M]
5.30

pOH




Indicates the hydroxide ion concentration.
The negative logarithm of the hydroxide
ion concnetration.
pOH = - log [OH-]
never use pOH to determine
acid/base/neutral!! (ZERO CREDIT!!)
Equations
Kw = 1.0E-14 = [H3O+][OH-]
 pH = -log[H3O+]
 pOH = -log[OH-]
 pH + pOH = 14

pH and pOH problems
Determine the pH and pOH of a solution
which contains 0.0035 mole of H3O+ per
liter. Is this an acid/base or neutral
solution?
P: pH = -log [H3O+] ; pH + pOH = 14
S: pH = - log [0.0035M]
= 2.46
 pOH = 14 – 2.46 = 11.54
 acid

pH and pOH Problems

What is the pOH of a solution
containing 0.042M KOH? Is this a/b/n?
since this is a strong base  100% ionizes
Therefore: KOH
 K+
+
OHbefore 4.2E-2
0
0
after
0
4.2E-2M
4.2E-2M
pH and pOH Problems (cont’d)

What is the pOH of a solution containing
0.042M KOH? Is this a/b/n?
u: pOH
 k: [OH-] = 4.2E-2M
 p: pOH= -log [OH ]
 s: pOH = -log [4.2E-2M]
pOH = 1.38
pH = 14 – 1.38 = 12.62  base

pH and pOH Problems

What is the pH of a solution containing
0.035M H2SO4? Is this a/b/n?
since this is a strong acid  100% ionizes
Therefore: H2SO4  2H+
+
SO4-2
before 3.5E-2
0
0
after
0
7.0E-2M
3.5E-2M
pH and pOH Problems (cont’d)





What is the pH of a solution containing
0.035M H2SO4? Is this a/b/n?
u: pH
k: [H+] = 7.0E-2M
p: pH= -log [H+]
s: pH = -log [7.0E-2M]
pH = 1.15  acid
pH and pOH Problems





Calculate the [H3O+] of a solution that
has a pH of 3.70.
u: [H3O+]
K: pH = 3.70
p: pH = -log [H3O+] 
s: [H3O+] = 10-3.70
= 2.0E-4M
10-pH
Titration
Titration

an experimental procedure in which
the unknown concentration of a known
volume of solution is determined by
measuring the volume of a solution of
known concentration required to
react completely with it.
Titration



an analytical method in which a
standard solution is used to
determine the concentration of
another solution.
Standard solution – one of known
concentration.
End point – the point at which
neutralization is achieved.
Steps for Titration
1.
2.
3.
A measured amount of an acid/base of
unknown concentration is added to a
flask.
An appropriate indicator
(phenolphthalein, BB, or methyl orange) is
added to the solution.
A measured amount of base/acid (known
concentration) is mixed into the first
solution.
Titration Experiment
Preview


A ring stand, a
clamp, a burette, a
250mL flask, and a
white sheet of
paper are required
materials
The materials
should be
assembled as the
picture shows.
Titration Experiment
Preview




Condition the burette.
Add a small amount of
NaOH to the burette
(about 5 mL).
Place the burette
almost horizontal and
turn.
Empty contents through
stopcock into sink.
Titration Experiment
Preview



Be sure the stopcock of
the burette is closed!
Using the funnel/stirring
rod at the top, fill the
burette with the NaOH
solution.
In the flask place the
measured HCl solution,
and 5 drops of the
indicator.
Titration Experiment
Preview


Begin the titration
by opening the
stopcock to make the
NaOH solution flow
dropwise into the
flask.
Be sure to swirl the
flask constantly so
the solutions mix
thoroughly.
Titration Experiment Preview


Continue to add NaOH
solution to the flask
until a color change
begins (colorless to
pink)
At this point, slow the
flow even more. (It is
over titrated when the
solution turns fuchsia.)
Titration curve

Graphical depiction of pH of solution
as it is neutralized by a given volume
of titrate added.
Titration curve:
strong vs. strong

Titration curve:
strong acid vs. weak base

Titration curve:
weak acid vs. strong base

Titration curve:
weak vs. weak
Titration curve summary
Titration problem

1.
2.
This is a stoichiometric problem.
First write the balanced equation.
Convert:
Molaritymolemolemolarity.
Titration prob. (method II)

1.
2.
3.
4.
This is a stoichiometric problem.
First write the balanced equation.
Convert from Molarity and volume to
moles of known solution (base)
Convert: mole base mole acid.
Use volume of acid to convert to
Molarity.
Titration problem sample 1

In a laboratory experiment, 20.0 mL
of NH3(aq) solution is titrated to the
methyl orange endpoint using 15.65mL
of a 0.200M HCl solution. What is the
concentration of the aqueous
solution?
Titration problem sample 1
Method I
NH3 + HCl  NH4Cl
U: [NH3]
K: [HCl] = .200M; vHCl= 15.65 mL HCl; VNH3=
20.0mL NH3
p: M HClmole HCl  mole NH3  M NH3
S: .200 mole HCl 0.01565L HCl 1mole NH3= 0.157 M NH3
1 L HCl soln
1 mole HCl
0.0200 L NH3 soln
Titration problem sample 1
Method II
NH3 + HCl  NH4Cl
U: [NH3]
K: [HCl] = .200M; vHCl= 15.65 mL
HCl; VNH3= 20.00mL NH3
p: M HCl= mol HCl/L HCl soln
mole HCl  mole NH3
M NH3 = mole NH3/L NH3 soln
Titration problem sample 1
Method II
NH3 + HCl  NH4Cl
S: .200 mole HCl = x moles HCl
1 L HCl soln
0.01565 L HCl
X mole HCl = 3.13E-3 moles HCl
3.13E-3 mole HCl
1mole NH3 =
3.13E-3 mol NH3
1 mole HCl
M NH3 = 3.13E-3 mol NH3
0.0200 L
= 0.157 M NH3
Titration problem sample 2

If 20.0 mL of a 0.300M solution of
NaOH is required to neutralize
30.0mL of sulfuric acid solution, what
is the molarity of the sulfuric acid
solution?
Titration problem sample 1
Method I
2NaOH + H2SO4  Na2SO4 + 2H2O
U: [H2SO4]
K: [NaOH ] = 0.300M; VNaOH= 20.0 mL
NaOH; VH2SO4= 30.00mL H2SO4
p: M NaOHmole NaOH  mole H2SO4 M H2SO4
S: .300 mole NaOH 0.0200L NaOH 1mole H2SO4= .100 M H2SO4
1 L NaOH soln
2 mole NaOH
.0300 L H2SO4 soln
Titration problem sample 1
Method II
2NaOH + H2SO4  Na2SO4 + 2H2O
U: [H2SO4]
K: [NaOH ] = 0.300M; VNaOH= 20.0 mL
NaOH; VH2SO4= 30.00mL NH3
p: M NaOH= mol NaOH/L NaOH soln
mole NaOH  mole H2SO4
M H2SO4 = mole H2SO4 /L H2SO4 soln
Titration problem sample 2
Method II
2NaOH + H2SO4  Na2SO4 + 2H2O
S: .300 mole NaOH
=
1 L NaOH soln
x moles NaOH
0.0200 L NaOH
X mole NaOH = 6.00E-3 moles NaOH
6.00E-3 mole NaOH
1mole H2SO4 = 3.00E-3 mol H2SO4
2 mole NaOH
M NaOH = 6.00E-3 mol H2SO4 =
0.0300 L
0.100 M H2SO4
Titration problem sample 3

How many liters of a 1.50M sulfuric
acid solution are needed to neutralize
completely 120.0 grams NaOH?
Titration problem sample 3
Method I
2NaOH + H2SO4  Na2SO4 + 2H2O
U: L H2SO4
K: [H2SO4] = 1.500M; 120.0g NaOH
p: g NaOHmole NaOH  mole H2SO4 L H2SO4
S: 120.0g NaOH 1 mole NaOH 1mole H SO 1 L H SO = 1.00L H2SO4
2
40.0 gNaOH
2mol NaOH
4
2
4
1.50 mol H2SO4
Titration problem sample 1
Method II
2NaOH + H2SO4  Na2SO4 + 2H2O
U: L H2SO4
K: [H2SO4] = 1.500M; 120.0g NaOH
p: g NaOH mol NaOH
mole NaOH  mole H2SO4
M H2SO4 = mole H2SO4 /L H2SO4 soln
Titration problem sample 1
Method II
2NaOH + H2SO4  Na2SO4 + 2H2O
S:
120.0g NaOH 1 mole NaOH =
40.0 gNaOH
3.00 mole NaOH
3.00 mol NaOH
1mole H2SO4 = 1.50 mol H2SO4
2 mole NaOH
1.50mol NaOH = 1.50 mol H2SO4
1 L NaOH soln
x L H2SO4 soln
X L H2SO4 soln = 1.00 L H2SO4 soln