Chapter 19
Acids, Bases, and Salts
Anything in black letters = write it in
your notes (‘knowts’)
19.1 – Acid-Base Theories
Acids
Taste sour
Dissolve active metals to produce hydrogen gas
Turns litmus paper RED
Bases
Taste bitter
Feels slippery on skin (dissolves oils on skin)
Turns litmus paper BLUE
Have you seen the litmus paper yet??
These are experimental definitions, they do not explain
(theory) how an acid is different from a base.
Arrhenius defined an acid
and base theoretically.
Svante Arrhenius
(1857 – 1927)
First, a vocab word…
Dissociate to split or separate from another
Arrhenius Definition (~1887)
ACID – substance that dissociates in water
to form hydrogen ions (H+).
HCl (aq)  H+ (aq) + Cl- (aq)
BASE – substance that dissociates in water
to form hydroxide ions (OH-).
NaOH (aq)  Na+ (aq) + OH- (aq)
When an acid is placed in water, H+ ions are
produced.
Hydrogen ions can also be thought of as H3O+ ions.
H3O+ = hydronium ion
HCl (aq)  H+ (aq) + Cl- (aq)
or equivalently,
HCl
+
H 2O  H 3O +
+
Cl-
Brønsted-Lowry Definition (~1923)
ACID – donates H+
BASE – accepts H+
Johannes Bronstad
(1879 – 1947)
Thomas Lowry
(1874 – 1936)
B-L definition covers more examples than
the Arrhenius definition.
ammonia
ammonium ion
water donates a H+ and so is a B-L acid
ammonia accepts a H+ and so is a B-L base
Conjugate Acid – formed when a base
accepts a H+
Conjugate Base – formed when an acid
donates a H+
EXAMPLES…
ACIDS donate H+,
BASES accept H+
label each as acid, base, c. acid, c. base
HCO3-
+
acid
HF
+
+
acid +
OH-
+
base +
H 2O
base


HCO3- 
base
+
c. base
F-
H 3O +
+ c. acid
+
H2CO3
 c. base + c. acid
HCO3- 
acid
CO3-2
H2O
+
CO3-2
 c. acid + c. base
Amphoteric – substance that can be an acid or a
base – depending on what it reacts with.
Water is amphoteric
ACIDS donate H+, BASES accept H+
label each reactant as an acid and base,
label the products as conjugate acids or conjugate bases.
HNO3 + H2O  H3O+ + NO3-
CH3COOH + H2O  H3O+ + CH3COONH3 + H2O  NH4+ + OHH2O + CH3COO-  CH3COOH + OH-
Lewis Acids and Bases (not covered)
Acid/Base Indicators
Litmus
Acid – red, Base – blue, Neutral - colorless
Phenolphthalein
Acid – colorless, Base – pink, Neutral - colorless
Cabbage
Acid – red/pink, Base – yellow/green, Neutral – blue/purple
Assignment:
Chapter 19 Worksheet #1 (FRONT)
19.2 – Hydrogen Ions and Acidity
Molarity (M) – unit used to express the concentration of a
solution
mol solute (mol)
Molarity =
liters of soln (L)
anything in [brackets] means the concentration in molarity
[H+] = ‘the hydrogen ion concentration’
[OH-] = ‘the hydroxide ion concentration’
Self-Ionization of Water
Water ionizes to produce a small amount of H+ and
OH- ions.
H2O
 H+ +
OH-
In pure water at 25̊C
[H+] = [OH-] = 1 x 10-7 M
Ion-product constant for water (Kw)
Kw = [H+][OH-] = 1.0 x 10-14
remember…anything in [brackets] represents the concentration
in molarity
A solution is acidic if [H+] > 1.0 x 10-7 M
…or if the pH of the solution is below 7
Just as the mole was used to simplify large numbers of atoms,
pH is used to simplify small concentration numbers
pH = ‘power of the hydrogen ion’
pH = -log[H+]
Instead of writing out numbers like these…
[H+] = 1 x 10-7 M
[H+] = 2.4 x 10-4 M
pH = 7.00
[H+] = 7.3 x 10-10 M
pH = 9.14
pH = 3.62
we can write number like these
Instead of saying “This solution has a hydronium
ion concentration of 2.4 x 10-4 M”.
We can just say “This solution has a pH of 3.62”.
Not only is pH an easier number to talk about, pH
is understood by most people, whereas molarity is
not.
The pH scale is used to describe how acidic or
basic (alkaline) a substance is.
Examples
Pure water has [H+] = 1.00 x 10-7 M
The pH of water would be
pH = -log[H+]
pH = - log [1.00 x 10-7]
pH = 7
Examples
[H+] = 2.3 x 10-5 M. Calculate the pH.
pH = - log [H+]
pH = - log [2.3 x 10-5]
pH = 4.64
Examples
[H+] = 1.0 x 10-5 M. Calculate the pH.
pH = - log [H+]
pH = - log [1.0 x 10-5]
pH = 5.0
Examples
pH = 4.2. Calculate [H+]
pH = - log [H+]
-4.2 = log [H+]
10-4.2 = 10log [H+]
10-4.2 = 6.31 x 10-5 M = [H+]
Summary of pH
Acidic
Neutral
[H+]
100 M
10-7 M
pH
0
7
14
10-7 M
100 M
[OH-] 10-14 M
pH = - log [H+]
Basic
10-14 M
[H+] = 10-pH
Kw = [H+][OH-] = 1.0 x 10-14
19.3 – Strengths of Acids and Bases
Strong acids & bases completely dissociate (split
into ions) in water;
Weak acids & bases partly dissociate and only form
a small amount of ions.
HCl is a strong acid,
HCl (aq)  H+ (aq) + Cl- (aq)
it completely ionizes
Concentration
Initial
at Equilibrium
[HCl]
[H3O+]
[Cl-]
0.10 M
0
0
0
0.10 M
0.10 M
CH3COOH is a weak acid,
CH3COOH (aq)
CH3COO- (aq) + H+ (aq)
it forms a small amount of ions
Concentration
Initial
at Equilibrium
[CH3COOH]
0.10 M
0.0987 M
[CH3COO-]
[H3O+]
0
0
1.34 x 10-3 M 1.34 x 10-3 M
0.00134 M
0.00134 M
Chemical Equilibrium – occurs when forward
rxn rate equals reverse rxn rate; dynamic
CH3COOH + H2O
CH3COO-
+ H3O+
Le Chatelier’s Principle – at equilibrium, a rxn
will shift forward or backward in response to any
change in conditions (temp, pressure,
concentration)
Increase [CH3COOH], rxn shifts the rxn to the right.
Increase [CH3COO-], rxn hifts rxn to the left.
Le Chatelier’s Principle Example 1
CH3COOH + H2O
CH3COO-
+ H3O+
1. Add methyl orange to acetic acid; divide into 3’s
red pH 4.0, orange pH 5.0, yellow pH 6.0
2. Add Na+CH3COO- to acetic acid
Doing this, increases [CH3COO-] and causes a shift in
the rxn to the left; increasing the pH.
3. Add NaOH to the acetic acid
Doing this, decreases [H3O+] causing a shift in the rxn
to the left; increasing the pH.
Le Chatelier’s Principle Example 2
NaCl (aq)

Na+
+
Cl-
What will happen when drops of HCl (aq)
are added to a saturated solution of NaCl?
HCl (aq)

H+
+
Cl-
the rxn will shift to the left because the
[Cl-] concentration increased.
Le Chatelier’s Principle Example 3
Co(H2O)6+2 (aq) + 4Cl- (aq) + heat  CoCl4-2 (aq) + 6H2O
1. Add ~ 3 mL of conc. HCl to about 2 mL of 0.1 M CoCl2
2. Add water to reverse the rxn
3. Add ~ 2 mL of 0.1M AgNO3(aq).
Ag+ (aq)
+
Cl- (aq)  AgCl (s)
Buffer – solution that maintains a fairly
stable pH when small amounts of acid or
base are added.
Buffer solutions consist of a
weak acid and its conjugate base
or
a weak base and its conjugate acid.
Acetic Acid/Acetate Ion Buffer
CH3COO-
CH3COOH
The acetate ion reacts with any added acid.
CH3COO-
+
H+
CH3COOH
The acetic acid reacts with any added base.
CH3COOH
+
OH-
 CH3COO-
+ H2O
Universal pH Indicator Color Chart
Blood is buffered at pH b/w 7.35 - 7.45.
This is done mainly by the carbonic acid/bicarbonate buffer.
Carbonic acid (H2CO3) neutralizes added bases.
Bicarbonate ion (HCO3-) neutralizes added acids.
H2CO3 (aq)  H2O + CO2
Hold your breath, CO2 level builds up in blood,
causing [H2CO3] ↑, pH ↓
Hyperventilate, CO2 level drops in blood,
causing [H2CO3] ↓, pH ↑
Questions
What determines if an acid or base is strong or weak?
What occurs during chemical equilibrium?
Describe Le Chateliers Principle.
CH3COO-
+
H+
CH3COOH
Which direction will the rxn shift if [CH3COO-] increases?
What would happen to the pH of the solution?
What is a buffer?
What two things must a buffer contain?
How does a buffer work?
Le Chatelier’s Principle and Reversible Reactions ANALYSIS
Part 1
1. (a) H2SO4 caused the solution to turn from yellow to orange.
(b) Adding H2SO4 caused the reaction to shift to the right.
(c) Adding NaOH caused the reaction to shift to the left.
Part 2
1. (a) Fe+3 (ferric or iron(III)) & NO3-1 (nitrate)
(b) K+1 (potassium) & SCN-1 (thiocyanate)
(c) The Fe(SCN)+2 caused the solution to turn dark red.
(d) Adding Fe3(SO4)2 increased the concentration of Fe(SCN)+2.
(e) Adding Fe(NO3)3 increased the concentration of Fe(SCN)+2.
(f) Adding NaOH decreased [Fe(SCN)+2]
(g) Reducing [SCN-] caused the rxn to shift to the left (yellow)
19.4 – Neutralization Reactions
Neutralization Rxn – complete rxn of a strong base
with a strong acid
A neutralization rxn will produce a salt and water.
Acid
HCl
+
Base

+ NaOH 
Salt + H2O
NaCl + H2O
Titration – determining the concentration of an
unknown solution using a solution whose
concentration is known.
Standard – solution that
has a known concentration.
Equivalence Point –
point where the amount of
acid equals the amount of
base
End Point – point where
the indicator changes
color
ASSIGNMENT:
Read Section 19.4 (p. 672 – 675)
Answer Questions #35-43
EXAMPLE
10.0 mL of 0.5 M HCl solution is added to 20.0
mL of NaOH of unknown concentration. What
is the concentration of the NaOH?
HCl
+ NaOH 
0.5 M
10.0 mL
xM
20.0 mL
NaCl + H2O
Since the reaction of HCl and NaOH is 1:1 and
twice the volume of NaOH was used, the NaOH
must half as strong as HCl; [0.25 M].
EXAMPLE
What volume of 0.10 M KOH is required to
neutralize 20.0 mL of 0.20 M H2SO4 solution?
H2SO4 + 2KOH 
0.20 M
0.10 M
20.0 mL
x mL
K2SO4 + 2H2O
Since KOH requires twice as many moles as H2SO4,
you should double your answer.
19.5 – Salts in Solution
not covered…
Write the balanced chemical equation for each
neutralization reaction
Sulfuric acid + magnesium hydroxide
Phosphoric acid + calcium hydroxide
Nitric acid + ammonium hydroxide
Chapter 19 Quiz #2
1. What color will litmus paper be in an acidic solution?
2. What color will phenolphthalein indicator be in an basic
solution?
3. What does [H+] mean?
4. What two products are always formed in an acid-base
neutralization reaction?
5. Explain the difference between a strong acid and a
weak acid.
6. Explain why the pH of pure water is 7.00
7. What is a buffer?
8. How is the molarity of a solution calculated?
9. A student titrated 10.0 mL of an HCl solution. The
titration required 23.3 mL of 0.24M NaOH solution.
a. Which solution was the standard?
b. Which solution was more concentrated?
c. Convert both volumes to liters
d. Calculate the number of moles of NaOH that reacted.
e. Calculate the number of moles of HCl that reacted.
f.
Calculate the molarity of the HCl solution.
10. Calculate the pH of solutions with the following
hydrogen ion concentrations.
a. [H+] = 1.23 x 10-4M
b. [H+] = 3.42 x 10-7M
11. Calculate the hydrogen ion concentrations of
solutions with the given pH.
a. pH = 3.14
b. pH = 9.2