2- Weak acid - Strong base titration :- - PowerPoint PPT Presentation

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2- Weak acid - Strong base titration :-

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2- Weak acid - Strong base titration :-eg. CH3COOH (pKa = 4.74) and NaOH We have sample of 100 ml 0.1 N HAc and tirate against 0.1 N NaOH Before the titration pH is ... – PowerPoint PPT presentation

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Title: 2- Weak acid - Strong base titration :-


1
  • 2- Weak acid - Strong base titration -
  • eg. CH3COOH (pKa 4.74) and NaOH
  • We have sample of 100 ml 0.1 N HAc and tirate
    against 0.1 N NaOH
  • 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 H2O 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.
2
3- Strong acid - weak base titration - eg. NH4OH
(pKb 4.74)and HCl We have sample 100 ml 0.1 N
NH4OH and titrate against 0.1 NHCl a- Before
titration pH is due to the sample NH4OH pH is
calculated from eq. of weak base. pH pKw - 1/2
pKb -1/2 pCb 14 2.37 - 0.5
11.13 b-During titration eg. After adding 90 ml
HCl
NH4 OH HCl NH4Cl H2O NH4OH
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 NH4OH is neutralized and NH4Cl is
only present , therfore pH is calculated from
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.
3
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.
4
2- Displacement titration It is used for easily
hydrolysable salts - A- Salt of strong base and
weak acid eg. borax, Na2CO3 B- Salt of weak
base and strong acid eg. FeCl3, Al2(SO4)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 Na2B4O7
Borax hydrolyze in water to give
Na2B4O7 7 H2O
4H3BO3 2 NaOH
v.weak acid
tit. ?
2 HCl
using M.O ?titrate ? HCl using M.O
indicator.(let reading x)
eg.3. Na2CO3 sodium carbonate.
Na2CO3 HCl NaHCO3 HCl
NaCl H2O CO2 pH
8.3
3.8 Ph.Ph.
M.O.
5
Na2CO3 can be determined by titration against
HCl by 2 methods a- Using M.O. as indicator it
will give total CO3-2 b- Using Ph.Ph. as
indicator it will give 1/2 CO3-2 and considered
as half neutralization step. But care that the
1st step to NaHCO3 takes place on 2 separate
steps
Na2CO3 2HCl 2NaCl
CO2 H2O
Na2CO3 CO2 H2O 2 NaHCO3
2Na2CO3 2HCl 2NaHCO3
2 NaCl
i.e. For 1/2 neutralization, we must prevent the
escape of CO2 by 1- cooling
2- dilution 3-stirring
4- dipping the nozzle of burrete under the
surface of the soln.
6
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. NH3, formic acid. 2- When sample is
insoluble eg. ZnO, CaO, CaCO3, BaCO3 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
NH4Cl NaOH NaCl
NH3 H2O Add known excess NaOH , then boil to
remove NH3 and titrate excess NaOH using HCl and
M.R. as indicator.
Determination of nitrogen in organic compounds
(by Kjeldahl's method)
Organic cpd Conc.H2SO4 ? NH3
(NH4)2SO4 -Nitrogen of organic cpd
is reduced to NH3 by digestion with conc. H2SO4
in the presence of K2SO4 or Na2SO4 (to raise the
boiling point of acid) and CuSO4 or HgO as a
catalyst.
7
-The organic cpd. is oxidised to CO2 and the acid
is reduced to SO2 and nitrogen to NH3 which is
fixed with excess acid as (NH4)2SO4 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.
4NH4Cl 6HCHO
(CH2)6N4 4HCl 6H2O
4(NH4)2SO4 6HCHO
(CH2)6N4 2H2SO4 6H2O N.B. HCHO must be
neutralized from any formic acid due to aerial
oxidation .
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