2.
Acid-Base Reactions
• Acid + Base
Water + Salt
• Properties related to every
day:
– antacids depend on
neutralization
– farmers use it to control soil
pH
– formation of cave stalactites
– human body kidney stones
from insoluble salts
3.
Acid-Base Reactions
• Neutralization Reaction - a reaction in
which an acid and a base react in an
aqueous solution to produce a salt and
water:
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2 H2O(l)
– Table 21.1, page 614 lists some salts
4.
Titration
• Titration is the process of adding a
known amount of solution of known
concentration to determine the
concentration of another solution
• Remember? - a balanced
equation has a mole ratio
5.
Titrations
BuretBuret - volumetric glassware
used for titrations.
It allows you to add a known
amount of your titrant to the
solution you are testing.
If a pH meter is used, the
equivalence point can be
measured.
An indicator will give you the
endpoint.
6.
Titrations
• Note
the color
change
which
indicates
that the
‘endpoint’
has been
reached.
Start End
8.
Titration
• The concentration of acid (or base) in
solution can be determined by
performing a neutralization reaction
–An indicator is used to show when
neutralization has occurred
–Often we use phenolphthalein-
because it is colorless in neutral and
acid; turns pink in base
9.
Steps - Neutralization reaction
#1. A measured volume of acid of
unknown concentration is added to a
flask
#2. Several drops of indicator added
#3. A base of known concentration is
slowly added, until the indicator
changes color; measure the volume
– Figure 21.4, page 617
– ASAP Sim26
10.
Neutralization
• The solution of known concentration
is called the standard solution
– added by using a buret
• Continue adding until the indicator
changes color
– called the “end point” of the titration
– Sample Problem 21.2, page 618
11.
Equivalents
• An equivalent is the amount of acid or base
that will give 1 mole of hydrogen ions, or in the
case of a base, hydroxide ions ( remember not
all acids and bases are the same strength and
so, a 1M solution of an acid may or may not
give 1 mole of hydrogen ions)
Example; 1 mole of HCl is 1 equivalent and will
neutralize 1 mole of NaOH, which is also 1
equivalent
BUT, 1 mole of H2SO4 has 2 equivalents and will
neutralize 2 moles of NaOH
12.
Equivalents
• The mass of one equivalent of
a substance is called the gram
equivalent mass
• The gram equivalent mass
for HCl is 36.5 g/mol
• The gram equivalent mass
for H2SO4 is 49g/mol (1 mole
H2SO4 is 2 equivalents, so 1
equivalent is ½ mole)
13.
Equivalent Problem
• What is the mass of 1 equivalent of
Ca(OH)2?
Molar mass is 74g/mol
1mole Ca(OH)2 has 2 equivalents
74g/mol = 37g/eq
2 eq./mol
14.
Equivalent problem
• How many equivalents are in
14.6 g of sulfuric acid (H2SO4)
1mole H2SO4 has 2 equivalents
Molar mass is 98.1g/mol
So… 98.1g/2eq = 49.1g/eq
14.6 g x 1eq/49.1g=.297 eq
15.
Normality
• Normality (N)=equivalent/L
• IF 1 mole of acid or base gives 1
equiv. then Molarity (M) will
equal Normality (N)
• If 1 mole of acid =2 equiv,, then
N=2M; if 1 mole of acid =3 equiv,
then N=3M
• If you know the normality you
can find the # of equiv. by.
Equiv= V (in L) x N
16.
Normality problem
• How many equiv. are in 2.5L
of 0.60N H2SO4?
Equiv=V x N
Equiv=2.5L x 0.60N=1.5 equiv.
17.
Normality & Titrations
• If you are diluting a solution of
known normality, or in the
case of titration you will find
the following formula useful:
V1N1=V2N2
In the case of titrations
VaNa=VbNb
18.
Problem Using V1N1=V2N2
• You need to make 250mL of
0.10N sodium hydroxide from
a stock solution of NaOH that
is 2.0N. How many mL of the
stock solution do you need?
V1N1=V2N2
250mLx 0.10N=V2x2.0N
250mLx 0.10N = V2
2.0N
13mL=V2
19.
• If a 35.0mL of a 0.20N HCl are
needed to neutralize 25.0mL
of an unknown base, what is
the normality of the base?
VaNa=VbNb
35.0mL x 0.20N =25.0mLxNb
35.0mL x 0.20N=Nb
25.0mL
0.28N=Nb
Problem Using V1N1=V2N2
20.
• How many mL of 0.500N
sulfuric acid are needed to
neutralize 50.0mL of 0.200N
potassium hydroxide?
VaNa=VbNb
Va x 0.500N = 50.0mL x 0.200N
Va= 50.0mL x 0.200N
0.500N
Va=20.0mL
Problem Using V1N1=V2N2
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