2- Weak acid - Strong base titration :eg. CH
3
COOH (pKa = 4.74) and NaOH
We have sample of 100 ml 0.1 N HAc and tirate against 0.1 N NaOH a.Before the titration pH is due to the sample i.e. HAc (weak acid) therfore pH = 1/2pKa + 1/2 pCa
= 2.37 + 0.5 = 2.87
b- During titration eg. after adding 40 ml NaOH
HAc + NaOH NaAc + H
2
O
HAc will be present ( not completely neutralized) and NaAc i.e. weak acid and its salt, therfore pH is calculated from eq. of acidic buffer pH = pKa + log salt/ acid = 4.74 + log 40/60 = 4.53
c- At the equivalence point (end point) all HAc is neutralized and only NaAc is present , therfore pH is calculated from: pH = 1/2 pKw + 1/2 pKa -1/2 pCs (salt of weak acid and strong base)
= 7 + 2.37 -1/2 (- log 100 x 0.1 /200) = 8.72
d- After end point the pH is calculated from excess titrant i.e. NaOH.
N.B.
The end point is at the alkaline side and abrupt change in the curve is from pH 7 to pH 11. Therfore M.O and M.R. indicator are not suitable.
Therfore use Ph.Ph. ind or any ind of pH range on the alkaline side.

3- Strong acid - weak base titration :eg. NH
4
OH (pKb = 4.74)and HCl
We have sample 100 ml 0.1 N NH
4
OH and titrate against 0.1 NHCl a- Before titration pH is due to the sample NH
4 pH is calculated from eq. of weak base.
OH pH = pKw - 1/2 pKb -1/2 pCb
= 14 – 2.37 0.5 =11.13
b During titration eg. After adding 90 ml HCl
NH
4
OH + HCl NH
4
Cl + H
2
O
NH
4
OH will be present(not completely neutralized) and NH4Cl i.e. weak baseand its salt.therfore pH is calculated from eq. of basic buffer pH = pKw -pKb - log salt/ base
= 14 - 4.74 - log 90/10 = 8.31
c- At the equivalence point (end point) all NH
4 only present , therfore pH is calculated from :
OH is neutralized and NH
4
Cl is pH = 1/2 pKw- 1/2 pKb + 1/2 pCs( salt of strong acid and weak base)
= 7 - 2.37 + 1/2 (- log 100 x 0.1 /200) = 5.13
d-After the end point the pH is calculated from excess titrant i.e. HCl
N.B.
The end point is at the acidic side = 5.13 .The abrupt change is from pH 6 to 4 ,
Therfore Ph.Ph. indicator cannot be used. Use M.O. or M.R. or any indicator of pH range at acidic side.

Applications
1- Direct titration methods :-
Direct titration is useful for :-
A- Strong acid B- Strong base
C- Weak acid or base if Ka and Kb not less than 10-7
Determination of acids :-
1- Strong acids can be titrated against strong alkali using Ph.Ph. or M.O.
On titrating weak acid only Ph.Ph. is suitable.
2- Acids like benzoic acid , salicylic acid which are not insoluble in water are dissolved in neutral ethanol then add water and titrate against NaOH using Ph.Ph. as indicator.
3- Boric acid: weak acid is a monobasic acid i.e. release 1 H+ , can be titrated against NaOH only after potentiation by adding any poly hydroxy compound eg. glycerol using Ph.Ph. as indicator.
Determination of bases :-
1- Strong base can be titrated against strong acids using M.O. or Ph.Ph .
For weak bases we use M.O. indicator or any indicator of pH range on the acidic side.

2- Displacement titration
It is used for easily hydrolysable salts :-
A- Salt of strong base and weak acid eg. borax, Na
2
CO
3
B- Salt of weak base and strong acid eg. FeCl
3
, Al
2
(SO
4
)
3
N.B.
Always titrate the strong part of the salt.
eg.1. KCN We titrate KOH by standard acid eg. HCl eg.2. Borax Na
2
B
4
O
7
Borax hydrolyze in water to give:
Na
2
B
4
O
7
+ 7 H
2
O 4H
3
BO
3
+ 2 NaOH v.weak acid tit.

 titrate

HCl using M.O indicator.(let reading = x)
2 HCl using M.O
eg.3. Na
2
CO
3 sodium carbonate .
Na
2
CO HCl NaHCO
3
HCl NaCl + H
3 pH = 8.3 3.8
2
O +CO
2
Ph.Ph. M.O.

Na
2
CO
3 can be determined by titration against HCl by 2 methods : a- Using M.O. as indicator it will give total CO b- Using Ph.Ph. as indicator it will give 1/2 CO
3
-2
3
-2 and considered as half neutralization step.
But care that the 1st step to NaHCO
3 takes place on 2 separate steps;
Na
2
CO
3
+ 2HCl 2NaCl + CO
2
+H
2
O
Na
2
CO
3
+CO
2
+ H
2
O 2 NaHCO
3
2Na
2
CO
3
+2HCl 2NaHCO
3
+ 2 NaCl i.e.
For 1/2 neutralization, we must prevent the escape of CO
2 by :
1- cooling 2- dilution 3-stirring
4- dipping the nozzle of burrete under the surface of the soln.

3- Indirect or back (residual titration)
- In it we add Known excess of standard to the sample and titrate the excess unreacted standard.
Conc. of sample A = Known excess standard - b
When do we use back titration ?
1- When sample is volatile .eg. NH
3
, formic acid.
2- When sample is insoluble eg. ZnO, CaO, CaCO
3
, BaCO
3
3- When reaction require heat of standard solution.
4- When reaction proceed only in presence of excess reagent eg. with lactic acid.
Determination of inorganic ammonium salts :
NH
4
Cl + NaOH NaCl + NH
3
+ H
Add known excess NaOH , then boil to remove
2
O
NH
3 and titrate excess NaOH using HCl and M.R. as indicator.
Determination of nitrogen in organic compounds : ( by Kjeldahl's method) :
Organic cpd presence of K
2
SO
4
Conc.H2SO4
Δ
NH
3
-Nitrogen of organic cpd is reduced to NH
3 or Na
2
SO
4
(NH
4
)
2
SO
4 by digestion with conc. H
(to raise the boiling point of acid) and
2
SO
4 in the
CuSO
4 or HgO as a catalyst.

-The organic cpd. is oxidised to CO to NH
3
2 and the acid is reduced to SO
2 which is fixed with excess acid as (NH
4
)
2
SO
4 and nitrogen then add known excess NaOH and titrate excess unreacted alkali against HCl as before.
4- Other indirect Titrations :
Determination of esters :-
Esters are hydrolysed by reflux with known excess of NaOH, cool and the excess unreacted NaOH is titrated against standard HCl using Ph.Ph. indicator.
Determination of ammonium salt and amino acids (Formol titration)
Another method for determination of ammonium salts is formol titration .
When formaldehyde is added to the sample , hexamethylene tetramine (hexamine) which is neutral is formed and equivalent amount of acid which can be titrated against NaOH.
4NH
4
Cl + 6HCHO (CH
2
)
6
N
4
+ 4HCl + 6H
2
O
4(NH
4
)
2
SO
4
+ 6HCHO (CH
2
)
6
N
4
+ 2H
2
SO
4
+ 6H
2
O
N.B.
HCHO must be neutralized from any formic acid due to aerial oxidation .