Chemistry Acid Base

1. Acid Base Titrations
Experiment #7

2. 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.

3. 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.

4. 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

5. 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) + NaOH(aq)  NaC2H3O2 + H2O(l)
Acetic acid Sodium
hydroxide
Sodium
Acetate
water
Acid Base salt water

6. Titration of Vinegar
against NaOH
HC2H3O2(aq) + NaOH(aq)  NaC2H3O2 + H2O(l)
CH3COOH(aq) + NaOH(aq)  CH3COONa + H2O(l)
1 molecule 1 molecule 1 molecule 1 molecule
1 mole 1 mole 1 mole 1 mole
1 mole CH3COOH ≡ 1 mole NaOH

7. 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 …..

8. 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!!

9. 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.

10. 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.

11. 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.

12. 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.

13. 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.

14. 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.
COOH
C-H
COOK
CH
HC
HC
Potassium Hydrogen Phthalate, KHC8H4O4

15. KHC8H4O4(aq) + NaOH(aq)  KNaC8H4O4 + H2O(l)
Standardization
Acid Base salt water
1 molecule 1 molecule 1 molecule 1 molecule
1 mole 1 mole 1 mole 1 mole
1 mole KHC8H4O4 ≡ 1 mole NaOH

16. 250mL 250mL 250mL
Vinitial
Vfinal
End point:
Pale Permanent
Pink color
“ 0.1 M
NaOH ”
Vfinal- Vinital= Vused (in mL)
Standardization
KHP + H2O+ 2-3 drops of phenolphthalein
moles of KHP = Moles of NaOH
moles of KHP
= MNaOH × VNaOH
moles of KHP
= MNaOH × Vused
used
NaOH
V
KHPofmoles
M =

17. 250mL 250mL 250mL
Vinitial
Vfinal
End point:
Pale Permanent
Pink color
“ 0.1 M
NaOH ”
Vfinal- Vinital= Vused (in mL
Titration of Vinegar vs. NaOH
vinegar + H O+ 2-3 drops of phenolphthalein
moles of acetic acid
= Moles of NaOH
moles of acetic acid
= MNaOH × VNaOH
moles of acetic acid
= MNaOH × Vused
)(LV
NaOHofmoles
M
vinegar
acidacetic
=

18. Hydrolysis of salts formed from
strong bases and weak acids
NaCH3COO (CH3COO-
)
NaOH + CH3COOH
SB WA
HA(aq) + OH-
(l) H2O + A-
(aq)
WA conjugate
acidSB
conjugate
base

19. Hydrolysis of salts formed from
strong bases and weak acids
A-
(aq) + H2O (l) HA + OH-
(aq)
[ ] [ ]
[ ]eqb
eqbeqb
b
A
OHHA
K −
−
=
[ ] [ ]
[ ]−
−
=
A
OHHA
Kb

20. Hydrolysis of salts formed from
strong bases and weak acids
[ ] [ ]
[ ]−
−
=
A
OHHA
Kb
Multiplying and dividing the numerator and denominator by [H3O+
]
[ ] [ ]
[ ]
[ ]
[ ]+
+
−
−
×=
OH
OH
A
OHHA
Kb
3
3
Rearranging the equation
[ ]
[ ] [ ]
[ ][ ]
1
3
3
−+
+−
×=
OHOH
OHA
HA
Kb
(1)

21. Hydrolysis of salts formed from
strong bases and weak acids
Remember!!
H2O (l) + H2O(l) H3O+
(aq) + OH-
(aq)
H2O(l) H+
(aq) + OH-
(aq)
[ ] [ ]−+
×= OHOHKwwaterofproductIonic 3
,
[ ] [ ] 14
3
101)25( −−+
×=×= OHOHCKw

(2)

22. Hydrolysis of salts formed from
strong bases and weak acids
Remember!!
For a monoprotic weak acid (HA) dissolved in water,
HA(aq) + H2O(l) H3O+
(aq) + A-
(aq)
acid conjugate
acidbase
conjugate
base
[ ] [ ]
[ ]eqb
eqbeqb
a
HA
AOH
K
−+
= 3
[ ] [ ]
[ ]HA
AOH
Ka
−+
= 3
[ ]
[ ][ ]−+
=
AOH
HA
Ka 3
1
(3)

23. Hydrolysis of salts formed from
strong bases and weak acids
[ ]
[ ] [ ]
[ ][ ]
1
3
3
−+
+−
×=
OHOH
OHA
HA
Kb
(1)
[ ] [ ]−+
×= OHOHKwwaterofproductIonic 3
, (2)
[ ]
[ ][ ]−+
=
AOH
HA
Ka 3
1
(3)
Substituting 3 and 2 in 1
a
ww
a
b
K
KK
K
K =×=
1
1
wab
KKK =×

24. pH at the equivalence point
HA(aq) + OH-
(l) A-
+ H2O(aq)
A-
(aq) + H2O (l) HA + OH-
(aq)
[ ] [ ]
[ ]eqb
eqbeqb
b
A
OHHA
K −
−
=
[ ] [ ]
[ ]−
−
=
A
OHHA
Kb
At equivalence point, [HA] = [OH-
]
[ ]
[ ]−
−
=
A
OH
Kb
2

25. [ ]
[ ]−
−
=
A
OH
Kb
2
pH at the equivalence point
Now we make 2 substitutions in the above equations
[ ]
[ ]+
−
=
OH
K
OH w
3 a
w
b
K
K
K =
[ ] [ ]2
3
2
+−
=
OHA
K
K
K w
a
w
[ ]
[ ]−
+
=
A
KKOH aw
1
2
3

26. pH at the equivalence point
[ ]
[ ]−
+
=
A
KKOH aw
1
2
3
[ ]
[ ]−
+
=
A
KK
OH aw
3
At equivalence point, moles of A-
≈ moles of HA
[ ] HA
eequivalencatsolution
C
LV
HAofmoles
A ==−
)(
[ ]
HA
aw
C
KK
OH =+
3

27. pH at the equivalence point
[ ]
HA
aw
C
KK
OH =+
3
Taking log on both sides and multiplying both sides of the
equation by -1
[ ]
HA
aw
C
KK
LogOHLog −=− +
3
2
1






−=
HA
aw
C
KK
LogpH






−=
HA
aw
C
KK
LogpH
2
1

28. Titration curve of phosphoric acid, H3PO4
Phosporic acid is a triprotic acid

29. Titration curve of phosphoric acid, H3PO4