ACIDS and BASES
Unit 10, Chapter 19
pH indicators


pH indicators are valuable tool for
determining if a substance is an acid or
a base.
The indicator will change colors in
solution.
Things to use…



pH meter will indicate the numeric
value of acid or base based on the pH
range
Chemical indicators: phenolphthalein,
universal indicator…
Natural indicators: poinsettia, red
cabbage juice…
Properties of Acids and Bases

ACIDS





Have a sour taste
Change the color of
many indicators
Are corrosive (react
with metals)
Neutralize bases
Conduct an electric
current

BASES





Have a bitter taste
Change the color of
many indicators
Have a slippery
feeling
Neutralize acids
Conduct an electric
current
The Arrhenius Theory
of Acids and Bases
Arrhenius Theory of Acids and Bases:
an acid contains hydrogen and ionizes in
solutions to produce H+ ions:
HCl  H+(aq) + Cl-(aq)
Arrhenius Theory of Acids and Bases:
a base contains an OH- group and ionizes
in solutions to produce OH- ions:
NaOH  Na+(aq) + OH-(aq)
Neutralization

Neutralization: the combination of H+
with OH- to form water.
H+(aq) + OH-(aq)  H2O (l)

Hydrogen ions (H+) in solution form
hydronium ions (H3O+)
In Reality…
H+ + H2O  H3O+
Hydronium Ion
(Can be used
interchangeably with H+)
Commentary on Arrhenius Theory…
One problem with the Arrhenius
theory is that it’s not comprehensive
enough. Some compounds act like acids
and bases that don’t fit the standard
definition.
Bronsted-Lowry Theory
of Acids & Bases
Bronsted-Lowry Theory of Acids & Bases:

An acid is a proton (H+) donor

A base is a proton (H+) acceptor
for example…
Proton transfer
HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
Base
Acid
another
Water is a proton
donor, and thus an
example… acid.
ACID
CONJUGATE
BASE
NH3(aq) + H2O(l)  NH4+ (aq) + OH- (aq)
BASE
Ammonia is a proton
acceptor, and thus a
base
CONJUGATE
ACID
Amphoteric Substances
A substance that can act as both an
acid and a base (depending on what it is
reacting with) is termed amphoteric.
Water is a prime example.
Conjugate acid-base pairs

Conjugate acid-base pairs differ by
one proton (H+)
A conjugate acid is the particle formed
when a base gains a proton.
A conjugate base is the particle that
remains when an acid gives off a proton.
Examples: In the following reactions,
label the conjugate acid-base pairs:


H3PO4 + NO2-  HNO2 + H2PO4acid
base
c. acid
c. base
CN- + HCO3-  HCN + CO32base
acid
c. acid c. base

HCN + SO32-  HSO3- + CNacid
base
c. acid c. base

H2O + HF  F- + H3O+
c. base c. acid
base acid
SUMMARY OF ACID-BASE THEORIES
Theory
Acid Definition
Base Definition
Arrhenius
Theory
Any substance which
releases H+ ions in
water solution.
Any substance which
releases OH- ions in
water solution
BrǿnstedLowry Theory
Any substance which
donates a proton.
Any substance which
accepts a proton.
Strength of Acids and Bases

A strong acid dissociates completely in sol’n:


A weak acid dissociates only partly in sol’n:


HNO2  H+(aq) + NO2-(aq)
A strong base dissociates completely in sol’n:


HCl  H+(aq) + Cl-(aq)
NaOH  Na+(aq) + OH-(aq)
A weak base dissociates only partly in sol’n:

NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
Acid-Base Reactions


Neutralization reactions: reactions
between acids and metal hydroxide
bases which produce a salt and water.
H+ ions and OH- ions combine to form
water molecules:

H+(aq) + OH-(aq)  H2O(l)
Buffered Solutions
A solution of a weak acid
and a common ion is called a
buffered solution.
Thus, the solution maintains
it’s pH in spite of added acid or
base.
Pg. 620 fig 19.27
Demo: buffered solution


Demo: tap water vs. dH2O


Both waters have Universal indicator in them
(= pH indicator (changes color in the
presence of ions), which is a type of weak
acids)
The water will change pH, and therefore
COLOR (which helps us determine if a
solution is acidic or basic) with the addition of
HCl (acid) and NaOH (base)
Universal Indicator Color Chart
PAGE 602 fig 19.13
pH scale
0
Acid
7
Neutral
14
Base


Why does it take more drops of acid or
base to make the tap water change
color than it does for the distilled
water?
What is distilled water made of? What is
tap water made of?
pH and pOH
Pg. 596 (in section 19.2)
Ionization of water
Experiments have shown that pure water ionizes very
slightly:
+
 2H2O  H3O + OH
 Measurements show that:
[H3O+] = [OH-]=1 x 10-7 M
+
 Pure water contains equal concentrations of H3O +
OH-, so it is neutral.

pH

pH is a measure of the
concentration of hydronium
ions in a solution.

pH = -log [H3O+]
or

pH = -log [H+]
Example: What is the pH of a solution
where [H3O+] = 1 x 10-7 M?
pH = -log [H3O+]
-7
 pH = -log(1 x 10 )
 pH = 7

Example: What is the pH of a solution
where [H3O+] = 1 x 10-5 M?
pH = -log [H3O+]
-5
 pH = -log(1 x 10 )
 pH = 5


When acid is added to water, the [H3O+]
increases, and the pH decreases.
Example: What is the pH of a solution
where [H3O+] = 1 x 10-10 M?
pH = -log [H3O+]
-10)
 pH = -log(1 x 10
 pH = 10


When base is added to water, the [H3O+]
decreases, and the pH increases.
The pH Scale
PAGE 598 Table 19.5 & fig 19.10
*You must use pH to determine if something
is acidic, basic or netural (not pOH)
0
Acid
7
Neutral
14
Base
pOH

pOH is a measure of the
concentration of hydroxide
ions in a solution.

pOH = -log [OH-]
Example: What is the pOH of a solution
where [OH-] = 1 x 10-5 M?
pOH = -log
-5
 pOH = -log(1 x 10 )
 pOH = 5

[OH ]
How are pH and pOH related?

At every pH, the following relationships hold true:

[H3O+] • [OH-] = 1 x 10-14 M

pH + pOH = 14
Example 1: What is the pH of a solution
where [H+] = 3.4 x 10-5 M?
pH = -log [H+]
-5
 pH = -log(3.4 x 10
M)
 pH = 4.5

Example 2: What is the pH of a solution
where [H+] = 5.4 x 10-6 M?
pH = -log [H+]
-6
 pH = -log(5.4 x 10 )
 pH = 5.3

Example 3: What is the [OH-] and pOH
for the solution in example #2?





[H3O+][OH-]= 1 x 10-14
(5.4 x 10-6)[OH-] = 1 x 10-14
[OH-] = 1.9 x 10-9 M
pH + pOH = 14
pOH = 14 – 5.3 = 8.7
Example #4

Classify each solution as acidic, basic, or neutral
***MUST SOLVE FOR pH and use the
pH scale
a. [H+] = 6.0 x 10-10 M
b. [OH-] = 3.0 x 10-2 M
c. [H+] = 2.0 x 10-7 M
d. [OH-] = 1.0 x 10-7 M
basic
basic
acidic
neutral
Example #5


Which is the MOST basic from
question #4?
B.
Acids and bases: Titrations


The amount of acid or base in a
solution is determined by carrying out a
neutralization reaction;
an appropriate acid-base indicator
(changes color in specific pH range)
must be used to show when the
neutralization is completed.
This process
is called a
titration: the
addition of a
known amount
of solution to
determine
the volume or
concentration
of another
solution.
Buret
Solution with
Indicator
Read a
buret
volume to
2 decimal
places
Textbook page 615


Figure 19.22 a-c
End point: the point at which the
indicator changes color
(show lab in demo form…)
3 Steps to do a titration (pg. 615):
1.
2.
3.
Add a measured amount of an acid of
unknown concentration to a flask.
Add an appropriate indicator to the
flask
Add measured amounts of a base of
known concentration using a buret.
Continue until the indicator shows that
neutralization has occurred. This is
called the end point of the titration
4 steps to a titration CALC:




1) balanced equation
2) calculate the number of moles of
acid or base in known solution
3) calculate the number of moles in
unknown solution used during the
titration
4) determine molarity of unknown
solution and the pH
Example:


In a titration, 27.4 mL of 0.0154 M
Ba(OH)2 is added to a 20.0 mL sample
of HCl solution of unknown
concentration. What is the molarity and
pH of the acid solution?
Equation: (Step 1)
Ba(OH)2 + 2 HCl  BaCl2 + 2 H2O
 (steps)
Step 2

Calculate the number of moles of
known solution (Ba(OH)2)
Calculate moles of known
solution:
Mol Ba(OH)2 = 27.4mLx 0.0154mol x
1L
1L

1000mL
4.22 x 10-4
mol Ba(OH)2
Step 3


Calculate moles of unknown solution
Use stoichiometry and the balanced
equation:
Ba(OH)2 + 2 HCl  BaCl2 + 2 H2O

Calculate moles of unknown solution:
Mol HCl = 4.22 x 10-4 mol Ba(OH)2 x 2 mol HCl=
8.44 x 10-4 mol HCl
1 mol Ba(OH)2
Ba(OH)2 + 2 HCl  BaCl2 + 2 H2O
Use coefficients
from bal. eq to get
molar ratio
Step 4

Determine molarity and pH
Calculate the M and pH

Molarity =
8.44 x 10-4 mol HCl = 4.22 x 10-2 M HCl
0.0200 L

pH: HCl dissociates into H+ and Cl- ions
4.22 x 10-2 M HCl = 4.22 x 10-2 M H+ = 4.22 x 10-2 M Cl
pH = -log[4.22 x 10-2 M H+] = 1.375 = 1.38
Extra Calculation (same steps as
the one we just did)


**** DO ON THE BACK OF YOUR
NOTETAKERS OR CONT. IN NOTES
A 25 mL solution of H2SO4 is completely
neutralized by 18 mL of 1.0 M NaOH.
What is the concentration of the H2SO4
solution?

Follow steps #1-4
Titration Curve (pg. 615)


A graph showing how the pH changes
as a function of the amount of added
titrant in a titration.
Data for the graph is obtained by
titrating a solution and measuring the
pH after EVERY drop of added titrant.
Equivalence point =


The point on the curve where the
moles of acid equal the moles of
base
the midpoint of the steepest part of
the curve is a good approximation of
the equivalence point.


Knowledge of the equivalence point can
then be used to choose a suitable
indicator for a given titration;
the indicator must change color (end
point) at a pH that corresponds to the
equivalence point.
pg. 602 figure 19.12
HANDOUT: titration curve WS