And Acid/Base dilution
Mr. Shields
Regents Chemistry
U15 L05
1
Neutralization
Remember we said previously that one of the
Property’s of acids and bases is that they react together
Do you remember what this reaction is called?
It’s neutralization. So what exactly is neutralization?
Neutralization occurs when acids and bases react. This
results in 2 products – a salt and water. For example:
HNO3 + LiOH  LiNO3 + H20
2
Titration
Titrations are carefully “Controlled” Neutralizations.
A titration is used to determine the amount of acid or
Base present in an unknown.
- for example: you might want to know how
acidic is in a sample of collected lake water.
3
Acid-Base Titration

Requires a standard solution an acid-base
indicator and a graduated burette


Typically the indicator used is Phenolphthalein
when titrating strong acids and bases
A Standard solution is an acid or base of known
molar concentration.


If titrating an acid we need a standardized base.
If titrating a base we need a standardized acid
4
Recall Molarity (M) = # moles/liter of solution
For example: If we add 0.5 moles of KOH to
750 ml of water what is the Molarity?
M = 0.5 mol / 0.75L = 0.67M
5
Titration



The Standard solution is slowly added
to the unknown solution drop by drop.
As the solutions mix, a neutralization
reaction occurs.
Eventually, enough standard solution is
added to neutralize the unknown
solution. This is the Equivalence point.
Example: 1 mol H2SO4 + 2 mol KOH  1 mol K2SO4 + 2 mol H20
6
Reading
A
Buret
Initial reading
2.4 ml
Final Reading 4.8 ml
1 ml
1 ml
2
2
3
3
4
4
5
5
6
6
Volume used = 2.4 ml
7
Equivalence point


Total number of moles of H+ ions donated
by acid = total number of moles of OHdonated by the base.
Total moles H+ = total moles OH-
Ex: 50ml 0.5M HCL + 50ml 0.5M KOH have equal
Amounts of H+ and OHHow many moles of HCl and KOH are there?
0.025 mol ea.
8
Titration

End-point = point at which indicator changes color.

In titrating an unk. acid phenolphthalein changes
from colorless to pink
(what would the color change be if we were titrating a
base with an acid?)

If the indicator is chosen correctly, the end-point is
very close to the equivalence point.
9
Titration of a strong acid with a
strong base
14-
Phenolphthalein
Color change:
8.2

pH
Equivalence Pt 
7-
0-

0 ml
End point
Between pH of 4 and 10, only
a few drops of base are added.

Volume of 0.100 M NaOH added (ml)

40ml
10
Calculation of Concentration
Think back to our discussion of solutions. Do you
Recall the equation for determining Molarity by
Dilution?
Calculating Molarity after dilution is accomplished very
simply by using the equation M1V1 = M2V2
For example if M1 = 2M and V1=100ml and it is then
Diluted with 100ml of H20 what is the new Molarity?
2M x 100ml = M2 x 200ml
M2= 200/200 = 1M
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Neutralization Equation
In neutralization we use a very similar equation
#H+ (MH+VH+ )= #OH- (MOH-VOH-)




MH+ = molarity of H+
MOH- = molarity of OHVH+ = volume of H+
VOH- = volume of OH-
#
= Subscript in the
acid or base for
H+ or OH-
12
and MaVa = MbVb

True for strong monoprotic acids and
monohydroxy bases Since both H+ and
OH- (S#) =1.


Or strong diprotic and dihydroxy acids &
bases since both H+ and OH- (S#) = 2
Like HCl & KOH Or H2SO4 & Ca(OH)2
13
Titration Problem #1


In a titration of 40.0 mL of a nitric acid solution,
the end point is reached when 35.0 mL of 0.100
M NaOH is added.
Calculate the concentration of the nitric acid
solution.
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Neutralization Reaction
 HNO3
+ NaOH  H2O + NaNO3
 HNO3
is a strong monoprotic acid.
 NaOH is a strong monohydroxy
base
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Variables
 Ma
=?
 Va = 40.0 mL
 Mb = 0.100 M
 Vb = 35.0 mL
 # = 1 and 1
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Plug and Chug
 MA
(40.0 mL) = (0.100 M )(35.0
mL)
X
= Molarity (M) = .0875 M
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Titration Problem #2

What is the concentration of a sulfuric acid
solution if 50.0 mL of a 0.250 M KOH
solution are needed to neutralize 20.0 mL
of the H2SO4 solution of unknown
concentration?
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Neutralization Reaction
 KOH
+ H2SO4  H2O + K2SO4
 H2SO4
is a strong Diprotic acid.
 KOH is a strong monohydroxy
base
 #H+ = 2, #OH- = 1)
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Variables
 Ma
=?
 Va = 20.0 mL
 Mb = 0.250 M
 Vb = 50.0 mL
 SA =2 & SB =1
20
Plug and Chug
 #AxMAxVA
= #BxMBxVB

2xMA (20.0 mL) = 1x(0.250 M) (50.0 mL)

MA= 0.312 M
21