Titrimetrey as analytical tool

1. Acid – Base Titrations
Pabitra Kumar Mani ,
Assoc. Professor,
ACSS,
BCKV

2. Bronsted-Lowry acid-base theory- An acid is a proton donor
and a base is a proton acceptor
The Bronsted-Lowry theory concentrates on proton
donation and acceptance.
•water molecules undergoing dissociation or self-ionisation
because of the reaction
2H2O(l) ⇋ [H3O+(aq)] + [OH-(aq)]
BUT, in this reaction, water acts as both acid and base
Therefore water is an amphoteric oxide
•e.g. water acting as a base - proton acceptor with a stronger acid like
the hydrogen chloride gas
HCl(g) + H2O(l) ⇋ H3O+(aq) + Cl-(aq)
This is how hydrochloric acid is formed which you write simply as
HCl. e.g. water acting as an acid - proton donor with a weak BUT
stronger base like ammonia gas
NH3(aq) + H2O(l) ⇋ NH4+(aq) + OH-(aq)
This is why ammonium solution is alkaline - sometimes wrongly called
'ammonium hydroxide' instead of aqueous ammonia

3. Lewis acid-base electron pair theory :
Acid is an electron pair acceptor but base is an electron pair
donor.
e.g. a non B-L, but a Lewis acid-base interaction is boron trifluoride
(Lewis-acid, electron pair acceptor) reacting with ammonia (Lewisbase, electron pair donor).
F3B + :NH3 ⇋ F3B-NH3
Acid
base
Note: In organic chemistry mechanisms, nucleophiles are
Lewis bases and electrophiles are Lewis acids and they may
fit into the Bronsted-Lowry definition too e.g. protonation of
alcohols and alkenes via acid.
In Transition Metal chemistry, ligands like water, can donate a
pair of non-bonding electrons (lone pair) into a vacant orbital of a
central metal ion and so dative covalent (co-ordinate) bonds
hold a complex together.
The central metal ion with vacant bonding orbitals can act as a
Lewis acid by accepting an electron pair to form a dative

4. HCl gas:
HCl(g) + H2O(l) ⇋ H3O+(aq)
+
Cl-(aq)
acid
base
conjugate acid. conjugate base.
The resulting solution is called hydrochloric acid
Ammonia:
NH3(aq) + H2O(l) ⇋ NH4+(aq)
base
acid
conjugate acid
+ OH-(aq)
conjugate base
HCO3-, can act as an acid with a base or act as a base with an acid, such
behaviour is described as amphoteric.
HCO3- + H3O+(aq) ==> 2H2O(l) + CO2(aq)
HCO3- acting as a base, accepting a proton from an acid.
HCO3- + OH-(aq) ==> H2O(l) + CO32-(aq)
HCO3- acting as an acid, donating a proton to the hydroxide ion base.

5. Volumetric analysis (Titrimetry)
(i) Acid-Base and (ii) Displacement titrimetry
Titrant: the reagent of known concn,
Titrand: the substance being titrated
H+ + OH- = H2O.......neutralisation rn
Acid
base
MX + HA ⇋ HX + MA
Salt
Acid1
Acid2
Salt (acid2 weak than acid1)
MX + BOH ⇋ BX + MOH
base1
base2
(base2 weaker tahn base1)
Primary standard..... (It must be easy to obtain, to purify, to dry and to
preserve in pure state)
H2C2O4.2H2O (Crystalline) primary std acids and reductant,
Na2CO3(anhy), KH(IO3), Phthalate, Borax (Na2B4O7.10H2O), Potassium
tetra borate , 36.39% H2SO4 –non-hygroscopic used a s primary std.
Constant boiling HCl can be used to prepare standard HCl soln
Benzoic acid –primary standard
Secondary standard..... NaOH,HCl, H2SO4, CuSO4.5H2O

6. Equivalence point is determined by using an indicator (W.Ostald)
(pH-indicator)
HIn ⇋
H+ +
Acidic
Indicator(mol.form)
InOH
In-
HIn and In- are differently coloured
ionic form
⇋ In+ + OH-
InOH and In+ are differently coloured
Basic
ionic form
Indicator(mol.form)
⇋
Tautomeric transformation
HIn
Hin*
⇋ H+ + In-
HIn and In- are differently coloured
HIn ⇋ H+ + In[ H ][ In ]
kIn-a =
+
[ HIn]
[H ]
+
= kIn-a
pH = - logkIn-a
−
[ HIn]
[ In ] [ HIn]
−
- log [ In ]
−
pH = pKIn-a + log [ In ]
−
[ HIn]
Alkaline colour intensity
acidic colour intensity

7. [ In ]
When [ HIn] = 10, pH= pKIn-a + 1
[ In ]
and When [ HIn] = 1/10, pH= pK In-a - 1
−
−
Operationally,
pH= pKIn-a ± 1
[ In ] 10 times concn of HIn i.e., when the color due to [ In ] is dominant
−
−
During titration, if pH at equivalence point, lies in the range pK In-a ± 1,
that indicator can be used to detect the e.p. In the particular titration
InOH
⇋ In+ + OH-
[ In ][ OH ]
=
[ InOH ]
+
KIn-b
−
[ InOH ]
∴ [OH ] = kIn-b [ In ]
-
[ In ]
+
-log[OH ] or, pOH = pKIn-b + log10
-
[ InOH ]
[ In ]
pH = 14 - pKIn-b -log10 [ InOH ]
+
[ In ]
+
pH = pKIn-a -log10 [ InOH ]
+
pH + pOH =pKw=14
pH -14 = - pOH
HA ⇋ H+ + ApKa+pKb= pKw

8. Indicator
pH range
Colour
Acid
Change
Alkaline
pKIn-a
Bromophenol
Blue
2.8-4.6
yellow
blue
4.0
Methyl orange
3.1-4.4
O
Y
3.7
Methyl red
4.2-6.3
R
Y
5.1
Bromothymol
Blue
6.0-7.6
Y
Blue
7.0
Phenolphthalein
8.3-10.0
COLOUR
LESS
R
9.6
Thymolphthalein
8.3-10.5
C
blue
9.2

9. 1. Strong acid –strong base titrations
1(N) HCl ------- 1(N) NaOH
ml of
alkali
added
[H+] g-ion/L
pH
0
1
0
50
(50x 1/150)x 1
0.48
50 ml of 1N=150 ml of S(N), S= (50/150) (N)
99
(1/199)x 1
2.3
99.9
0.1/199.9
3.3
100
As in water
7
100.1
(0.1/200.1)
10.7
[OH-]= 0.1/200.1
101
p(OH)=3.3,
pH=10.7
110
[OH-]= 1/201
p(OH) = 2.3
[OH-]= 10/210
pH=11.7
12.7
150
[OH-]= 50/250
13.3
Volume of alkali added (NaOH)

11. All the indicators, M.O. M.R. PhTh have indicator range pH lies in
the reflexion region. Hence , all these can be used.
When 0.1 N HCl is titrated with 0.1 N NaOH, the inflexion range
becomes narrower by two pH units(4.3-9.7)
With more dil strong acid and strong base the inflexion region becomes
narrower. In this case the choice of indicator becomes narrower. M.Orange
becomes unsuitable to detect the e.p. also, PhTh can’t be used.
BPB is suitable indicator for titration of very dilute strong acid.
Methyl red is better than Methyl orange because of its higher indicator pH
range for titration of strong dilute acid.
Presence of CO2 dissolved in water interferes with the indicator of
high pH ranges like PhTh, but acidic indicators like M.Orange is however,
remains unaffected by the presence of dissolved CO2.
Distilled water having pH 7.0 shows a pH of 5.7 when it is in
equilibrium with air containing 0.03% CO2 by volume. When satd. With CO2
distilled water shows pH of 3.7 at 25°C

12. Dependence of the magnitude of end –point break on concentration.
Curve1: 100 ml of 0.1M HCl versus 0.1 M NaOH.
Curve 2: 100 ml of 0.01M HCl versus 0.01 M NaOH
Curve 3: 100 ml of 0.001M HCl versus 0.001 M NaOH

13. 2. Weak
acid - strong base titration
CH3COOH (HAc)
0.1 N, 100 ml
- NaOH
0.1N
ml of
alkali
added
[H+] g-ion/L
pH
0
√(1.75 x10-5 )x0.1
pH=4.8+log50/50
2.9
50
4.8(pKa)
HAc ⇋ H+ + Ac99
pH=4.8 +log 99/1
6.8
99.9
0.1/199.9
7.8
100
pH=7+2.4+½ log(0.01)
8.9
100.1
pH due to 0.1 ml NaOH 10.7
in 200.1ml soln
[OH-]= 0.1/200.1
101
p(OH)=3.3,
[OH-]= 1/201
p(OH) = 2.3
pH=10.7
pH=11.7
[ H ][ Ac ]
=
+
K
a
[ HAc]
−
=

+ 2
H





[ HAc]
Since [H+]= [Ac-] only for pure HAc
and [HAc] = c
∴[H+] = √(Ka . c)
(i) pH=½ pKa - ½log c
=½x4.8 -½log(0.1)=2.9

14. [ H ][ Ac ]
=
+
K
a
[ HAc]
−
[H ]
+
Ka[ HAc]
=
Ac -
[ ]
log Ac −




2. ∴pH = pK a +
[ HAc]
This form of ionisation constant equation is called the Henderson- Hasselbalch
equation. It is useful for calculating the pH of a weak acid soln containing its salt.
Salt of weak acid and strong base
2
HAc + OH- = Ac- + H2O
−
OH − 
[ HAc] [ OH ] 

K =

At equi.p. Ac undergoes hydrolysis
−
=
h
[ Ac ]
c
[HAc] = [ OH-]
Ac- + H2O ⇋HAc + OH2
OH − = K w / K a .c
[ HAc] OH −  H + K




w


K =
∴OH −  = Kw.C / Ka
x  + =
h
−


Ac


K
H 


a


1
1


1


[
]
∴ pOH = 2 pK w −
3.
pH =
1
1
1
pK w + pK a + log C
2
2
2
pK a − log c
2
1
1


∴ pH = pKw −  1 pK w − pK a − log c 
2
2
2


2
3. General eqn for calculating e.p. When an weak acid is titrated with a strong
base and also when any salt of weak acid and strong base is dissolved in water

15. Inflection region narrower
and on the alkaline pH side
As the titration curve
indicates, methyl orange
and methyl red can’t be
used to detect the E.P but
alkaline range indicator
like PhTh, ThTh, Thymol
blue can be used as
indicator. For the acids
with Ka. As low as 10-7,
the e.p. Is at pH=10. i.e.,
at a still higher pH, but
the rate of change in pH
in the neighbourhood of
the theoretical or
stoichiometric e.p. is very
little pronounced due to
considerable hydrolysis.
Dependence of the titration curve of weak acid on concentration. Curve1:
100 ml of 0.1M HOAc versus 0.1 M NaOH. Curve 2: 100 ml of 0.01M HOAc l
versus 0.01 M NaOH, Curve 3: 100 ml of 0.001M HOAc versus 0.001 M NaOH

19. Salt of weak
acid, strong
electrolyte Bronsted Base
Sodium acetate is a weak
base(conjugate base of weak
acid).The weaker the
conjugate acid , the
stronger the conjugate base,i.e., the more
strongly the salt will combine with a proton as
from the water

21. What is the pH when 25.00 mL of 0.20 M CH3COOH has been
titrated with 40.0 mL of 0.10 M NaOH?
Solution:
1)Determine moles of acid and base before reaction:
CH3COOH: (0.20 mol/L) (0.02500 L) = 0.0050 mol
NaOH: (0.10 mol/L) (0.04000 L) = 0.0040 mol
2) Determine moles of acid and salt after reaction:
CH3COOH: 0.0050 mol - 0.0040 mol = 0.0010 mol
CH3COONa: 0.0040 mol
3) Use Henderson-Hasselbalch Equation to determine pH:
pH = 4.572 + log (0.0040/0.0010)
pH = 4.572 + 0.602
pH = 5.354

22. 3. Titration of Weak base with Strong acid
100 ml 0.1 N NH4OH vs.HCl(0.1N)
ml of acid [H+] g-ion/L
added
pH
0
14-2.37-0.5=11.1
11.1
50
14 - 4.8 – 0 = 9.2 (pKa of
conjugate acid of the
weak base)
9.2
99
14-4.8-log(99/1) = 7.2
7.2
99.9
14-4.8- log(99.9/1) = 6.2
6.2
100
7- 2.4 - ½log(0.05) = 5.25
5.25
100.1
S= (0.01/200.1)=3.8
3.8
NH4OH +HCl⇋ NH4Cl + H2O
NH4OH ⇋ NH4+ + OH-
[
Since [NH4+]= [OH-] only for free
base and [NH4OH] = c
∴OH −  = Kb [ NH 4OH ]




pOH=½ pKb - ½log c
(i) pH=pKw -½ pKb + ½log c
0.1 mlx0.1N= S x 200.1 , S= 0.01/200.1
NH +  OH − 




4 
K =
b
NH OH 




4
Kb NH 4OH − 




OH -  =




NH 4+
∴
[
pH + pOH = pKw
pOH = pkw-pH
]
NH +  OH − 
−2


 = OH


4 
K =
[NH 4OH]
b
NH OH 




4
]
pOH = pK b + log
∴pH = pK w − pK b − log
[NH4+ ]
[ NH4OH ]
[NH4+ ]
[ NH4OH ]

23. At equivalent point only NH4Cl is present, but NH4+ undergoes hydrolysis(h)
NH4+ + H2O ⇋NH4OH + H+
K
h
=
H +










C
H +


 X 


 NH + 
H +
4 







[ NH 4 OH ] H + 


2
=
=
K
w
K
b
NH4Cl : a weak Bronsted acid will
follow hydrolysis
Kw
∴H +  = K w .C / K b


K


b
1
1
1
pH = pK w − pK b − log C
2
2
2
It is clear from the titration curve neither PhPh nor ThPh i.e., the alkaline
range indicator can be used to detect the e.p in the titration of 0.1N NH 4OH
agt a strong acid. Indicators with pH range on the slightly acidic side
such as M.O, M.R or BCG etc can be used for the purpose. BCG is
suitable for titration of all weak bases with Kb greater than10-6 .

25. 4. Titration of weak acid - weak base
100 ml 0.1N
CH3COOH
VS.
NH4OH
(0.1N) aq NH3
At e.p. The pH will depend upon the relative strength of acid and base. In this
case both the weak acid anion and the weak base cation undergo hydrolysis.
Ac- + H2O ⇋ HAc + OH NH 4 + + H 2 O ⇋ NH 4 OH + H +
The hydrolytic equilibrium of a salt of weak acid and weak base is expressed by the eq.
[ΜΟΗ ][ΗΑ ]
+
Κ =
M + A +H2O ⇋ MOH +HA
h
Μ +Α − ....eqn 1






H + 
[ΜΟΗ ][ΗΑ ] x   x


Κ =
h
Μ + Α − H + 


 



 

Kw
Kh =
....eqn 2
K a Kb
OH −




[ ΜΟΗ ][ ΗΑ ] x
OH −
Κh =




Μ + OH − 






H +  OH − 

 


 

x
H + 
A − 








If x is the hydrolysis of 1 mol of the salt dissolved in V L of solution then
the individual concns are:
[MOH]= [HA]=x/V; [M+]= [A-] = (1-x)/V from eqn 1,
 x  x 
V V 
x2
   =
Κh =
1 - x 1 - x 
(1 - x) 2
 V  V 



∴ Kh =
x
(1 - x)
....eqn 3

26. HA ⇋ H+ + A-
Ka
H +   A − 
  
=   
[ HA ]

∴ H + = K a .




∴ H +  = K a [ HA ] = K a ( x ) = Ka . K


A - 
h


(1 − x)
 
 
Kw
=
K a Kb
K a .K w
Kb
No conc. term for the salt
produced appears
If the ionisation constant of the acid and the base
1
1
1
pH = pK w + pK a − pK b are equal, that is Ka=Kb, pH= ½pKw =7.0,
2
2
2
pH of Ammonium acetate is 7.0+2.7 - 2.7 = 7.0
Hence, during titration of weak acid agt a weak base pH at e.p. is independent of
the concn. of acid and base and depends on only the relative strength of acid
and base.
Accordingly, no sharp colour change obtd. at e.p with any simple pH
indicator. However, a mixed indicator of suitable composition which exhibits
a sharp colour change over a very limited pH range may sometimes be used for
the detection of e.p.Thus, HAc-NH4OH titration, neutral red + methylene blue
has been used with some success.
On the whole it is better to avoid such titrations by indicator method. Usually
potentiometric and conductometric titrations are preferable in such cases.

27. There is no sharp change in the pH during the titration. Hence, no
sharp equivalence point can be obtained with common indicators.
However, a mixed indicator, which shows a sharp colour change
over limited pH range may be used.

29. Titration of Poly basic acids
For a dibasic acid if the diff.between primary and secondary dissociation
constant is very large, K1/K2 ≥104, the solution of acid behaves like a
mixture of two mono-basic acids having independent dissociation constant
and during titration of these acids agt. a strong base, we can have two
seperate e.p. Thus for H2SO3(K1= 1.7x 10-2 , K2=1x 10-7) there will be a
sharp change of pH at the first e.p. But the 2nd e.p satge is less
pronounced because K2 is very low. However, ThPh indicator the 2nd e.p.
Can be well detected.
H2CO3 (K1= 4.3x 10-7 , K2=5.6x 10-11) it can be seen from K1 and K2 value
that 2nd dissociation stage is extremely weak and no suitable indicator is
available for direct titration of H2CO3 as a dibasic acid.
Provided the 1st dissociation stages of the acid is weak and K1 can be
neglected in comparison with concn C we can have
H +  =




K1K 2
pH=½pK1 +½pK2
H +  =




K1K 2C
Usual expression
K1 + C
For H2CO3 , pH (e.p.)= 8.2 With a knowledge of the pH at the stoichiometric
e.p and also by knowing the course of neutralisation curve we select an
appropriate indicator for the titration of any dibasic acid for which K 1/K2 is at
4

32. H3PO4 (K1= 7.5x10-3, K2=6.2 x10-8, K3=5x10-13
H3PO4 in soln will behave as the mixture of 3 monobasic acids because
K1/K2 and K2/K3 ratios are greater than 104.Neutralisation with a strong
base can proceed almost completely to the end of the 1st stage before the
2nd dissociation stage is affected and the 2nd stage can be neutralised
completely before the tertiary stage is apparent. The pH–change during
titration of H3PO4 agt.caustic soda soln( NaOH) can be shown as ;
H3PO4 is a good example of a titration
where the first two equivalence points,
corresponding to base reaction with the
first and second protons, respectively, are
clearly visible.(sharp change in pH at the
equivalence point)
The acid dissociation constants for H3PO4
are quite different from each other with pKa's
of 2.15, 7.2, and 12.15 Because the pKa are
so different, the protons are reacted at
different pH's. This is illustrated in the plot of
the relative fraction as a fn of pH

33. (i)With 50% of 1st acidity (1st replaceable H+ half neutralised)
[H PO ]
−
pH = pK1 +log10
(ii) At 1st e.p.
2
4
[ H 3PO 4 ]
= pK1 = 2.1
pH = ½pK1 + ½pK2 = ½(2.1+7.2)= 4.65
(iii) With 1.5 equivalence of alkali added,
pH = pK2 +log10
[HPO ]
[H PO ]
4
2
−2
4
−
= pK2 = 7.2
(iv) With 2nd equivalence point,
pH = ½pK2 + ½pK3 = ½(7.2+12.3)= 9.75
(v) With 2.5 equivalence of alkali added,
pH = pK3 +log10
(vi) With 3rd equivalence point,
PO − 3 
 4 


HPO − 2 

4 


= pK1 = 2.1
pH = ½pKw + ½pK3 +½log C = ½(7+6.5-0.5)= 12.65

34. The pH at points where the relative fraction of two species are equal, e.g.,
where two relative concentration lines cross, have a simple relationship to
the acid equilibration constants. For example, the first crossing occurs for
[H3PO4] = [ H2PO4-]. The relationship to pH is most easily found by
recognizing that all principle species are given in the first proton ionization
equation

35. Figure 1. Titration curve for titration of phosphoric acid with sodium hydroxide.
The point marked "a" is the initial point. The pH of the solution equals pKa1, pKa2,
and pKa3 at points b, d, and f, respectively. Points c, e, and g are the first,
second, and third equivalence points, respectively. The points between a and c,
between c and e, and between e and g are the first, second, and third buffer
zones, respectively.