Experiment #7 Acid Base Titrations

Experiment #7
Acid Base Titrations
What are we doing in
this experiment?
Determine the concentration of acetic acid in
vinegar.
To follow the ionization of phosphoric acid
with the addition of NaOH and plot a titration
curve.
What is acid-base titration?
A TITRATION WHICH DEALS WITH A
REACTION INVOLVING ACID AND A
BASE.
What is a titration?
The act of adding standard solution in small
quantities to the test solution till the reaction
is complete is termed titration.
What is a standard solution?
A standard solution is one whose
concentration is precisely known.
What is a test solution?
A test solution is one whose
concentration is to be estimated
Titration of Vinegar
against NaOH
Vinegar is an acetic acid solution of certain
concentration
So a titration of vinegar against NaOH actually
means, a reaction between acetic acid and NaOH.
HC2H3O2(aq)
Acetic acid
Acid
+
NaOH(aq) 
NaC2H3O2 + H2O(l)
Sodium
hydroxide
Sodium
Acetate
Base
salt
water
water
Titration of Vinegar
against NaOH
HC2H3O2(aq)
+
CH3COOH(aq)
NaOH(aq) 
NaC2H3O2 + H2O(l)
+ NaOH(aq)  CH3COONa + H2O(l)
1 molecule
1 molecule
1 molecule
1 mole
1 mole
1 mole
1 mole CH3COOH ≡ 1 mole NaOH
1 molecule
1 mole
Titration of Vinegar
against NaOH
In this experiment, we are trying to find the
[CH3COOH] in vinegar. So we have to know the
[NaOH] accurately first, before finding the
[CH3COOH] in Vinegar.
Finding the [NaOH] should be pretty easy right?!!
Ya! Kind of …..
How hard is it to make 500 mL of
0.1 M NaOH?
Should not be that hard right?!!
Calculate the weight of NaOH (2.0 g)
Weigh out the NaOH on the balance
Dissolve it in 500 mL of water in a volumetric flask
Hold on! We have a problem here!!
So what is your problem?
I don’t have a problem but Mr. NaOH
seems to have a problem here
NaOH is hygroscopic..
What do you mean by hygroscopic?
It absorbs moisture. NaOH absorbs moisture from air.
What if NaOH is hygroscopic?
Let us say, for the problem at hand,
we need 2.0 g NaOH to make a 0.1 M
NaOH solution. By the time we weigh
out the NaOH for our solution, it
would have absorbed moisture. So the
total weight of 2.0 g is not all due to
NaOH. It has some contribution from
the water that the NaOH absorbed.
What if NaOH is hygroscopic?
Since the total weight of 2.0 g is not all due to
NaOH, the concentration of the 0.1 M NaOH
that we are trying to make is not going to be
What we expect it to be.
We will not know the accurate concentration
of the NaOH solution, just by dissolving 2.0 g
of NaOH in 500 mL of water.
But the requirement for a titration is that we
know the concentration of at least one of the
Solutions very precisely.
How do we find the [NaOH] precisely?
Through standardization
What is standardization?
It is just a technical term for doing
a titration using a primary standard
to find the precise concentration of a
secondary standard.
What is a primary standard?
A primary standard should possess the following qualities:
(i) It must be available in very pure form
(ii) It should not be affected by exposure to moisture or air.
(iii) It should maintain its purity during storage.
(iv) The reactions involving the primary standard should
be stoichiometric and fast.
(v) It should have high molecular weight.
Which primary standard are we
going to use?
Potassium hydrogen phthalate, abbreviated as KHP.
Remember!! KHP is not the molecular formula for
Potassium hyrogen phthalate. It is just an abbreviation.
So when calculating the molecular weight of KHP,
do not add up the atomic weights of K, H and P.
C-H
HC
COOH
HC
COOK
CH
Potassium Hydrogen Phthalate, KHC8H4O4
Standardization
KHC8H4O4(aq) + NaOH(aq)  KNaC8H4O4 + H2O(l)
Acid
Base
salt
1 molecule
1 molecule
1 molecule
1 mole
1 mole
1 mole
1 mole KHC8H4O4 ≡ 1 mole NaOH
water
1 molecule
1 mole
Standardization
Vfinal- Vinital= Vused (in mL)
Vinitial
moles of KHP = Moles of NaOH
“ 0.1 M
NaOH ”
moles of KHP
= MNaOH × VNaOH
moles of KHP
= MNaOH × Vused
Vfinal
M
NaOH

moles of KHP
V
used
250mL
250mL
250mL
KHP + H2O+ 2-3 drops of phenolphthalein
End point:
Pale Permanent
Pink color
Titration of Vinegar vs. NaOH
Vfinal- Vinital= Vused (in mL
Vinitial
moles of acetic acid
= Moles of NaOH
“ 0.1 M
NaOH ”
moles of acetic acid
= MNaOH × VNaOH
moles of acetic acid
= MNaOH × Vused
Vfinal
M
acetic acid

moles of NaOH
V (L)
vinegar
250mL
250mL
250mL
vinegar + H2O+ 2-3 drops of phenolphthalein
End point:
Pale Permanent
Pink color
Hydrolysis of salts formed from
strong bases and weak acids
NaCH3COO (CH3COO-)
NaOH + CH3COOH
SB
WA
HA(aq) + OH-(l)
WA
SB
H2O + A-(aq)
conjugate conjugate
acid
base
Hydrolysis of salts formed from
strong bases and weak acids
A-(aq) + H2O (l)
Kb 
HA
HA OH 

eqb
A 
eqb

eqb
HA OH 

Kb 
A 

+ OH-(aq)
Hydrolysis of salts formed from
strong bases and weak acids
HA OH 

Kb 
A 

Multiplying and dividing the numerator and denominator by [H3O+]
HA OH  H O 



Kb
3
A 
H O 


3
Rearranging the equation
Kb 
HA

3

H O OH 


A  H O 


3
1

(1)
Hydrolysis of salts formed from
strong bases and weak acids
Remember!!
H3O+ (aq)
H2O (l) + H2O(l)
H+ (aq)
H2O(l)
+
OH-(aq)
+
Ionic product of water , Kw  H 3O  OH

Kw
(25 C )  H O  OH   110


3

OH-(aq)
14


(2)
Hydrolysis of salts formed from
strong bases and weak acids
Remember!!
For a monoprotic weak acid (HA) dissolved in water,
H3O+(aq) + A-(aq)
conjugate conjugate
acid
base
HA(aq) + H2O(l)
acid
base
H O  A 

Ka 
3

eqb
eqb
HA
eqb
H O  A 

Ka 
3
HA

HA
1





K a H 3O A 
(3)
Hydrolysis of salts formed from
strong bases and weak acids
HA

H O OH 
(1)
K 

A  H O 
1


3
b


3
Ionic product of water , Kw  H 3O  OH

HA
1



K a H 3O  A 


(3)
Substituting 3 and 2 in 1
1 Kw Kw
Kb 


Ka 1 Ka
Kb  K a  K w
(2)
pH at the equivalence point
HA(aq) + OH- (l)
A-
A-(aq) + H2O (l)
Kb 
+ H2O(aq)
HA
HA OH 

eqb
A 
HA OH 

A 
eqb

eqb

Kb

At equivalence point, [HA] = [OH-]

OH 

A 
 2
Kb

+ OH-(aq)
pH at the equivalence point

OH 
K 
A 
 2

b
Now we make 2 substitutions in the above equations
Kw
OH  

H 3O 

2
Kw
Kw
 
 2
K a A  H 3O 
H O 
 2
3
1
Kw Ka
 
A 
Kw
Kb 
Ka
pH at the equivalence point
H O 
 2
3
1
Kw Ka
 
A 
Kw Ka
H 3O  

A 

At equivalence point, moles of A-  moles of HA
moles of HA
A  
 CHA
Vsolution at equivalence ( L)

KwKa
H 3O  
CHA

pH at the equivalence point
KwKa
H 3O  
CHA

Taking log on both sides and multiplying both sides of the
equation by -1
KwKa
 Log H 3O    Log
C1 HA
 Kw Ka 2
pH   Log 

 CHA 
 KwKa 
1
pH   Log 

2
 CHA 

Titration curve of phosphoric acid, H3PO4
Phosporic acid is a triprotic acid
Titration curve of phosphoric acid, H3PO4