ACIDS AND
BASES
WHAT IS AN ACID?
 An
acidic solution contains more
hydrogen ions (H+) than hydroxide (OH-)
ions
 They must be in aqueous solutions
 Example:
H2O  H+ + OHMore of these
WHAT ARE THE PROPERTIES OF
AN ACID?
 Acidic
solutions taste sour
 Citric acid in lemons and limes,
acetic acid in vinegar
 Able to conduct electricity
 That’s why there is battery acid
 Cause litmus paper to turn red
 Able to neutralize bases
WHAT IS A BASE?
A
basic solution contains more hydroxide
ions (OH-) than hydrogen (H+) ions
 They must be in aqueous solutions
 Example:
H2O  H+ + OHMore of these
WHAT ARE THE PROPERTIES OF
A BASE?
 Basic
solutions taste bitter
 Basic solutions feel slippery
 Bases are one of the ingredients
of soaps and detergents
 Bases also conduct electricity
 Turns litmus paper blue
 Able to neutralize acids
INDICATORS
Scientists
have a number of
ways to measure if something is
acidic or basic
These are called indicators
ARRHENIUS ACID
 The
first person to determine how water
became acidic or basic
Svante
 Arrhenius
Arrhenius
acid: a compound that
donates a hydrogen ion (H+)
 Arrhenius base: a compound that
donates a hydroxide ion (OH-)
EXAMLES
Acids:
HCl,
HF, HI, H2SO4, HNO3
Bases
NaOH,
KOH, Ca(OH)2,
HOW IT WORKS
 The
actual molecule that makes
aqueous solutions acidic is called
the hydronium ion (H3O+)
 H2O
+ HCl  H3O+ + Cl- (acid solution)
 Abbreviated:
 NaOH
 NOTE:
HCl  H+ + Cl-
 Na+ + OH- (basic solution)
Acids and bases only donate 1
H+ or 1 OH- at a time
TRY THESE
 For
1.
2.
3.
4.
each of the following, draw the
reaction for the acid or base. Indicate
whether it is an acid or base:
HCl

?
KOH

?
H2SO4

?
Mg(OH)2 
?
ANSWERS
1.
2.
3.
4.
HCl  H+ + ClKOH  K+ + OHH2SO4  H+ + HSO4Mg(OH)2  MgOH+ + OH-
WATER: THE SPECIAL CASE
 Most
compounds either donate a
hydrogen ion (H+) or a hydroxide
ion (OH-).
 Water is an example of a
compound that can do both
 H2O + H2O  H3O+ + OH Since water can act as an acid or a
base it is called: amphoteric
HOW DO WE MEASURE ACID?
 We
have a scale that measures how
acidic or basic a substance is:
 pH scale
 Scale goes from 0-14
 The smaller the number the more acidic
 The bigger the number the more basic
 pH of 7 is neutral
 Each pH number is 10x stronger than the
previous one
 pH 4 is 10X stronger than pH 5
pH
Scale
HOW DO WE FIGURE OUT THE
CONCENTRATION OF AN
ACID?
 So
far we have a scale that
measures how acidic or basic: pH
scale
 Yet how does this relate to our
previous measurement: the mole
 Molarity: the number of moles of
solute dissolved per liter of solution,
also known as molar concentration
 Has the unit of molarity (M)
FURTHER DEFINITION
 Solution:
a uniform mixture of solids,
liquids or gases; also called
homogenous
 Solute: one or more substances
dissolved in a solution
 The stuff you put into the mixture
 Solvent: the substance that
dissolves a solute to form a solution
 The liquid that you add the stuff to
EXAMPLE
 Molarity
of an acid is always based
on the concentration of the H+ ion
 Also written as [H+]
 REMINDER: Acids and bases
always donate 1 H+ or 1 OH- at a
time
 If you have 2 moles of HCl and you
add it to 1L of water, what is the
molarity of H+ in the solution?
EXAMPLE
 HCl
 H+ + Cl 2 moles of HCl = 2 moles of H+
 2moles = 2M solution
1 Liter
 NOTE: To get molarity(M), you must have
the solution in moles and Liters (L)
 Molarity will also be a conversion
(moles/L)
LET’S TRY SOME
1.
2.
3.
If you have 25.0g of sodium hydroxide in
250mL of water, what is the molarity(M)
of the hydroxide ion?
If you add 155 g of hydrochloric acid in
500mL of water, what is the molarity(M)
of the hydronium ion?
You add 330g of chromic acid to 100mL
of water. What is the molarity(M) of the
hydronium ion?
ANSWERS
1.2.5M
2.8.5M
3.28M
HOW DO WE DO DILUTIONS?
A
lot of times, when working
with acids, you want to
reduce the concentrations
so they are not so strong.
HOW DO WE DO THIS?
EXAMPLE
You
have 500mL of a 6M
solution of HCl. How do you
dilute this solution to a 0.5M
solution?
Here’s
the formula
(V1)(M1) = (V2)(M2)
ANSWER
 (500mL)(6M)
= (V2)(0.5M)
 3000mL•M = V2(0.5M)
0.5M
0.5M
 6000mL = V2
 Therefore,
you have to add enough water
to bring the volume to 6000mL
TRY THESE
1.
2.
3.
If you have 100mL of a 12M solution
of NaOH. What is the final volume
needed to make a 0.25M solution?
You start with 50mL of a 3.5M solution
of HF. What is the final volume
needed to make a 1.0M solution?
You put 22.8g of HCl in 250mL of
water. How much water do you
need to make it a 1.0M solution?
ANSWERS
1.
4800mL
2.
175mL
3.
625mL
HOW DO WE CONVERT
BETWEEN pH AND MOLARITY
 We
have two scales to measure the
amount of H+ in a solution
 pH
 Molarity (M)
 There
has to be a way to convert
between pH and molarity
pH FORMULA
 The
formula to convert between pH and
concentration is:
pH =
 Let’s
+
-log[H ]
examine what these mean
EXAMINING THE FORMULA
 pH:
the scale of how acidic or basic
 log: is a math conversion using the
base power of 10 (we will examine
this on your calculator)
 [H+]: the molar concentration of
hydronium ions in solution
LET’S TRY AN EXAMPLE
 If
you have a solution that has a 1.0x10-3 M
of H+, what is the pH?



Step 1: Type 1.0x10-3 into your calculator
Step 2: Press the log button
Step 3: Take the negative of the answer
 You
should get pH = 3
LET’S GO THE OTHER WAY


If you have a solution that has a pH of 3.5, what is
the concentration of H+?
 Step 1: Take the inverse log of the negative of
the number
 TI-30X: Type 2nd LOG and then enter the
number
 Casio: Type in the number and type shift log
 Step 2: Make sure you enter the negative of the
pH
[H+] = 3.16x10-4 M
TRY THESE
1.
2.
3.
4.
A solution has a pH of 8.25. What is the
molar concentration of H+?
A solution has a pH of 1.44. What is the
molar concentration of H+?
If you have [H+] = 2.45x10-9 M, what is the
pH?
If you have [H+] = 1.5x10-2 M, what is the
pH?
ANSWERS
1.
5.62X10-9 M
2.
3.63x10-2 M
3.
8.61
4.
1.04
USING pH IN A DILUTION
You
have a 3250mL of a
solution with a pH of 3.55.
What final volume do you need
to make the concentration of
the solution 2.5x10-5 M?
ANSWER
 V1=
3250mL
 M1= pH 3.55 = 2.82 x 10-4 M
 V2= ?
 M2= 2.5x10-5 M
 V1M1 = V2M2
 (3250mL)(2.82 x10-4 M) = V2
(2.5x10-5 M)
 V2 = 3.7x104 mL or 37L
USING pH IN A DILUTION
 You
have 2.75L of a sample with a
pH of 5.83. What final volume do
you need to create a solution with
a neutral pH?
ANSWER
 V1=
2.75L
 M1= pH 5.83 = 1.48 x 10-6 M
 V2= ?
 M2= pH 7.00 = 1.00x10-7 M



V1M1 = V2M2
(2.75L)(1.48 x10-6 M) = V2 (1.00x10-7 M)
V2 = 40.7L
NEUTRALIZATION REACTIONS
 So
1.
2.

far . . .
An acid produces H+
A base produces OHWhat do you get when you combine an
acid with a base?
NEUTRALIZATION REACTIONS

Answer: You get water and a salt.

Example:
HCl + NaOH  H2O + NaCl


When you mix an acid and a base, this is
called a neutralization reaction.
Reminder: Pure water is pH 7
YOU DON’T ALWAYS FULLY
NEUTRALIZE
 When
you mix an acid and a base,
you don’t produce all water and a
salt.
 For example, if you mix 250mL of
1.5M HCl with 250mL of 1.0M NaOH,
will this reaction be fully
neutralized? What is the new
concentration?
ANSWER: NO
 Since
the acid is more
concentrated than the base, only
part of the acid is neutralized.
 How do we find out the new
concentration of the solution after
we mix the acid and base?
 Step
NEUTRALIZATION
1. Find out the moles of HCl
 250mL of 1.5M HCl
 0.250L | 1.5moles = 0.375 moles HCl
1L
 Step 2. Find out the moles of NaOH
 250mL of 1.0M NaOH
 0.250L | 1.0moles = 0.250moles NaOH
1L
NEUTRALIZATION
 Step
3. Figure how much acid or base
you have left by subtracting


0.375moles H+ – 0.250 moles OH0.125 moles of HCl were not neutralized
 Step


4. Figure out the new molarity
0.125 moles is now in 500mL (250mL acid +
250mL of base)
0.125moles/0.500L = 0.25M HCl
TRY THIS ONE
You
mix 300mL of 2.5M HCl with
150mL of 0.75M KOH. What is
the acid concentration of the
solution?
ANSWER
 Total
moles of HCl:

0.300L| 2.5 moles = 0.75moles H+
1L
0.150L| 0.75 moles = 0.1125moles OH1L
0.75moles – 0.1125 moles = 0.6375 moles

0.6375 moles =


0.450L
1.42M
TRY THIS ONE
You
have 320mL of pH 4.5
solution. If you add 220mL
of a 0.0032M solution of
sodium hydroxide. What is
the new concentration?
ANSWER
-3
1.29x10
M
FULL NEUTRALIZATION
 Not
only can you discover the new
concentration, you can also figure out
how much of an acid or a base you need
to fully neutralize a sample.
 Remember to fully neutralize a sample,
you need to make it pH 7.

This means all the H+ reacts with all of the
OH-
EXAMPLE
 You
have 25.00mL of 3.2M HCl. How
much 1.0M NaOH do you have to add to
make the solution fully neutral?
 Step 1. Find out how much H+ you have


0.025L | 3.2moles = 0.080moles H+
1L
Therefore you need 0.080 moles OH- to fully
neutralize
ANSWER
 Step
2. How much NaOH do you need to
add?

0.080 moles |
 Final
1L
= .080L
1 mole
answer: You need to add 80mL of
1.0M NaOH to fully neutralize 25.00mL of
3.2M HCl.
SHORTCUT
 Since
you are doing a full neutralization,
you can also use the dilution formula:
V1M1 = V2M2
 Try
out the previous problem to see if you
get the same answer.
 NOTE: This only works on FULL
neutralization
TRY THIS
You
start with 250mL of a 0.50M
HCl solution. How much 5.0M
NaOH do you have to add to
the solution to completely
neutralize it?
LAST THING
 How
do we calculate an unknown
concentration of an acid or base
by using a known concentration
 To do this, you need to find the
point where the acids and bases
completely cancel out
 You can do this using pH indicators
DEFINITIONS
 Titration:
process in which an acid-base
neutralization reaction is used to
determine the concentration of a solution
of unknown concentration.
 Equivalence point: the point at which the
moles of H+ from an acid equals the moles
of OH- from a base.
 End point: point in a titration when an
indicator changes color
EXAMPLE
 You
start with 350mL of HCl. If you titrate
this with 235mL of 2.0M NaOH, what is the
concentration of your starting HCl?
 Step1: Find the moles of OH- used.

.235L | 2.0moles = 0.47moles OH1L
EXAMPLE
 Step
2: Determine the
concentration of acid
 0.47 moles OH- = 0.47moles H+
 0.47
moles H+ = 1.3M HCl
0.350L
SHORTCUT

A titration is another example of a full
neutralization

You can again use the dilution formula to solve
V1M1
Try
= V2M2
the previous problem to
see if you get the same
answer
BRONSTED-LOWRY ACIDS
 With
our current definition of acids and
bases, we have some problems
 For example:


NH3 (ammonia) and Na2CO3 (sodium
carbonate) both can act as bases
Yet, neither can donate a hydroxide (OH-)
ion
 How
can we explain this?
BRONSTED-LOWRY ACIDS
A
Danish chemist (Johannes Bronsted)
and an English chemist (Thomas Lowry)
came up with a more inclusive definition
 Bronsted-Lowry acid: a compound that is
able to donate a hydrogen ion (H+)
 Bronsted-Lowry base: a compound that is
able to accept a hydrogen ion (H+)
EXAMPLE
H2O
+ H2O  H3O+ + OH-
This water will
accept the H+.
Therefore it is the
base.
This water will donate
the H+. Therefore it
is the acid.
LET’S LOOK AT OUR ORIGINAL
EXCEPTIONS
 NH3
(ammonia) and Na2CO3 (sodium
carbonate) both act as bases

Yet, neither can donate a hydroxide (OH-)
ion
 NH3
+ H2O  NH4+ + OH-
 Na2CO3
+ H2O  NaHCO3 + OH-
OBSERVATION
 You
may have also observed that
(by this definition) if you start with an
acid, you make a base and vice
versa.
 Example
 HF + H2O  H3O+ + FAcid
Base
OBSERVATION
 You
may also notice that if you have an
acid on one side of the reaction, you
must also have a base.
 Example
HF
Acid
+ H2O  H3O+ + FBase
Acid
Base
CONJUGATE ACID-BASE PAIRS
 Chemical
reactions are reversible so
either side can be an acid or base.
 We call these conjugate acid-base pairs
 Conjugate
acid-base pairs: two
substances related to each other by the
donating and accepting of a single
hydrogen ion
GENERAL FORMAT
Conjugate
base
Acid
HX + H2O  H3O+ + XBase
Conjugate
acid