analytical Chemistry just the basic one. consists of compleximetric titration and the principle titration

2. PRINCIPLES OF NEUTRALIZATION TITRATIONS
 Buffer solution
 Calculating pH in titrations of strong acids and
strong bases
 Calculating pH in weak acid (or base) titrations
COMPLEX-FORMATION TITRATIONS
 Indicators for EDTA titrations
 Some applications in human life

3.  Standard Solutions - strong acids or strong bases
because they will react completely.
› Acids: hydrochloric (HCl), perchloric (HClO 4), and
sulfuric (H2SO4) – also very hazardous.
› Bases: sodium hydroxide (NaOH), potassium
hydroxide (KOH)
 Variables: temperature, ionic strength of medium and
presence of organic solvents or colloidal particles

4.  Acid/Base Indicators: a weak organic acid or weak
Indicators
organic base whose undissociated form differs in color
from its conjugate form (In would be indicator).
HIn + H2O  In- + H3O+ or In + H2O  HIn+ + OH-
(acid color) (base color) (base color) (acid color)
Ka = [H3O+][In-]
[HIn]
[H3O+] = Ka[HIn]
[In-]

5. › HIn pure acid color: [HIn]/[In-] ≥ 10
› HIn pure base color: [HIn]/[In-] ≤ 0.1
~The ratios change from indicator to indicator~
› Substitute the ratios into the rearranged K a:
[H3O+] = 10 Ka (acid color)
[H3O+] = 0.1Ka (base color)
› To obtain the indicator pH range,
 acid color pH = -log (10 Ka) = pKa + 1
 base color pH = -log (0.1 Ka) = pKa – 1

7. Commonly Used Indicators
Indicator pH Range Acid Base
Thymol Blue 1.2-2.8 red yellow
Thymol blue 8.0-9.6 yellow blue
Methyl yellow 2.9-4.0 red yellow
Methyl orange 3.1-4.4 red orange
Bromcresol green 4.0-5.6 yellow blue
Methyl red 4.4-6.2 red yellow
Bromcresol purple 5.2-6.8 yellow purple
Bromothymol Blue 6.2-7.8 yellow blue
Phenol red 6.4-8.0 yellow red
Cresol purple 7.6-9.2 yellow purple
Phenolphthalein 8.0-10.0 colorless red
Thymolphthalein 9.4-10.6 colorless blue
Alizarin yellow GG 10.0-12.0 colorless yellow

8.  A titration curve is constructed by plotting pH of the solution
during titration as ordinates and the amount of acid or base
added.
 These curves are useful to indicate equivalence point
graphically.
graphically
 The nature of titration curve depends on the ionization
constants of acid and base employed in titration i.e., their
strength.
 The principles of acid–base equilibria are important for the
construction and interpretation of titration curves in
neutralization titrations.

9.  2 general types of titration
curves are encountered in
titrimetric methods.
 First type called a sigmoidal
curve: important observations
curve
are confined to a small region (±
0.1 to ± 0.5 mL) surrounding the
equivalence point.
 Second type called a linear-
segment curve, measurement
curve
are made on both side and
away from the equivalence
point.

11.  A buffer is a mixture of a weak acid and its conjugate base or a
weak base and its conjugate acid that resists change in pH.
 A solution containing a weak acid, HA and its conjugate base,
A-, may be acidic, base or neutral depending upon the position
of two competitive equilibria:
HA + H2O  H3O+ + A- @ A- + H2O  OH- + HA
Ka = [H3O+][A-] Kb = [OH-][HA]
[HA] [A-]
 If the first equilibrium lies father to the right than the second,
the solution is acidic. If the second equilibrium is more
favorable, the solution is basic.

12.  In principle, the calculations work but there are
uncertainties in numerical values of dissociation
constants & simplifications used in calculations.
 How to Prepare/Get:
› Making up a solution of approximately the desired pH and
then adjust by adding acid or conjugate base until the
required pH is indicated by a pH meter
› Empirically derived recipes are available in chemical
handbooks and reference works
› Biological supply houses offer a variety of such buffers

13. The effect of dilution
 The pH of a buffer solution is remains essentially
independent of dilution until the concentrations of the
species are decreased to the point so that we cannot
assume that the differences between the hydronium and
hydroxide ion concentrations is negligible when
calculating the concentration of the species.
The effect of Added Acids and Bases
 buffers are resistant to pH change after addition of small
amounts of strong acids or bases

14. There are 4 distinctly different types of calculations to derive a
titration curve for a weak acid @ weak base.
1.At the beginning: pH is calculated from the concentration of
beginning
that solute and its dissociation constant.
2.After various increments of titrant has been added: pH is
added
calculated by the analytical concentrations of the conjugate base
or acid and the residual concentrations of the weak acid or base
3.At the equivalence point: the pH is calculated from the
point
concentration of the conjugate of the weak acid or base ~ a salt
4.Beyond the equivalence point: pH is determined by the
point
concentration of the excess titrant

15. Determine the pH for the titration of 50 mL of 0.10 M acetic acid after
adding 0.00, 10.00, 50.00, and 50.10 mL of 0.10 M sodium hydroxide
HOAc + H2O  H3O+ + OAc-
OAc- + H2O  HOAc + OH-
Ka = 1.75 x 10 -5
Initial pH:
pH Ka = [H3O+]2 / cHOAc
[H3O+] = 1.32 x 10-3
pH = -log(1.32 x 10-3) = 2.88
pH after titrant has been added (5.00 mL NaOH):
NaOH
*the buffer solution now has NaOAc & HOAc*
cHOAc = mol original acid – mol base added
total volume

16. cHOAc = (50 x 0.10) – (10.00 x 0.10) = 0.067M
60
cNaOAc = mol base added
total volume
cNaOAc = (10.00 x 0.10) = 0.017M
60
*we can then substitute these concentrations into the dissociation-
constant expression for acetic acid*
[H3O+] = Ka x [HOAc]
[NaOAc]
[H3O+] = 1.75 x 10-5 x [0.067]
[0.017]
[H3O+] = 7.00 x 10-5 pH = -log(7.00 x 10-5) = 4.16

17.  Equivalence Point (50.00 mL NaOH):
*all the acetic acid has been converted to sodium acetate*
OAc- + H2O  HOAc + OH-
[HOAc] = [OH-]
In the present sample, the NaOAc concentration is 0.05M.
Thus : [OAc-]= 0.05 M
*we can substitute this in to the base-dissociation constant
(Kb) for OAc-*
Kb = [OH-][HOAc] = Kw
[OAc-] Ka
[OH-]2 = 1.00 x 10-14
0.05 1.75 x 10-5

18.  Beyond the Equivalence Point (50.10 mL NaOH):
* the excess base and acetate ion are sources of the hydroxide
ion, but the acetate ion concentration is so small it is
negligible*
[OH-] = cNaOH = mol base added – original mol acid
total volume
[OH-] = (50.10 x 0.100) – (50.00 x 0.100)
100.10
[OH-] = 1.00 x 10-4
pH = 14.00 – (-log(1.00 x 10-4)) = 10.00

19.  The Effect of Concentration: the change in pH in the
Concentration
equivalence-point region becomes smaller with lower
analyte and reagent concentrations (0.1 M NaOH versus
0.001 M NaOH)
 The Effect of Reaction Completeness: pH change in the
Completeness
equivalence-point region becomes smaller as the acid
become weaker (the reaction between the acid and the
base becomes less complete)
 Choosing an Indicator: the color change must occur in
Indicator
the equivalence-point region

22. What is the pH of a solution that is 0.4 M in formic acid and 1 M
in sodium formate?
HCOOH + H2O  H3O+ + HCOO- Ka = 1.80 x 10-4
HCOO- + H2O  HCOOH + OH- Kb = Kw/Ka = 5.56 x 10-11

[HCOO-] ≈ cHCOONa = 1 M
[HCOOH] ≈ cHCOOH = 0.4 M
[H3O+] = (1.80 x 10-4) x (0.400) = 7.20 x 10-5
(1.00)
pH = -log(7.20 x 10-5) = 4.14

23. Calculate the pH of a solution that is 0.2 M in NH3 and 0.3 M in
NH4Cl.
NH4+ + H2O  NH3 + H3O+ Ka = 5.70 x 10-10
NH3 + H2O  NH4+ + OH- Kb = Kw/Ka = 1.75 x 10-5

[NH4+] ≈ cNH4Cl = 0.3 M
[NH3] ≈ cNH3 = 0.2 M
[H3O+] = (5.70 x 10-10) x (0.3) = 8.55 x 10-10
(0.2)
pH = - log (8.55 x 10-10) = 9.07

24. o Buffer Capacity - the number of moles of strong acid or strong
base that causes one liter of the buffer to change pH by one
unit.
o Calculate the pH change that takes place when a 100 mL
portion of 0.05 M NaOH is added to a 400 mL buffer consisting
of 0.2 M NH3 and 0.3 M NH4Cl (see example for “Buffers Formed
from a Weak Base and its Conjugate Acid”)
An addition of a base converts NH4+ to NH3:
NH4+ + OH-  NH3 + H2O
The concentration of the NH3 and NH4Cl change:
cNH3 = original mol base + mol base added
total volume
cNH3 = (400 x 0.2) + (100 x 0.05) = 0.170 M
500

25. o Calculate the pH change that takes place when a 100 mL
portion of 0.05 M NaOH is added to a 400 mL buffer consisting
of 0.2 M NH3 and 0.3 M NH4Cl (see example for “Buffers
Formed from a Weak Base and its Conjugate Acid”)
cNH4Cl = original mol acid – mol base added
total volume
cNH4CL = (400 x 0.30) - (100 x 0.05) = 0.230 M
500
[H3O+] = (5.70 x 10-10) x (0.230) =7.71 x 10-10
(0.170)
pH = -log (7.71 x 10-10) = 9.11
∆ pH = 9.11 – 9.07 = 0.04

26. › Pre-equivalence: calculate the concentration of the acid
Pre-equivalence
from is starting concentration and the amount of base
that has been added, the concentration of the acid is
equal to the concentration of the hydroxide ion and you
can calculate pH from the concentration.
› Equivalence: the hydronium and hydroxide ions are
Equivalence
present in equal concentrations
› Post-equivalence: the concentration of the excess base
Post-equivalence
is calculated and the hydroxide ion concentration is
assumed to be equal to or a multiple of the analytical
concentration, the pH can be calculated from the pOH

27. Do the calculations needed to generate the hypothetical titration
curve for the titration of 50 mL of 0.05 M HCl with 0.10 M NaOH
› Initial Point: the solution is 0.05 M in H3O+, so
Point
pH = -log(0.05) = 1.30
› Pre-equivalence Point (after addition of 10 mL reagent)
cHCl = mmol remaining (original mmol HCl – mmol NaOH added)
total volume (mL)
= (50 mL x 0.05 M) – (10 mL x 0.10 M)
50.0 mL + 10.00 mL
= 2.5 x 10-2 M
pH = -log(2.5 x 10-2) = 1.602

28. › Equivalence Point – neither HCl nor NaOH is in excess. So
the concentration of OH- and H3O+ is equal.
[OH-] = [H3O+], pH = 7
› Post-equivalence Point (after addition of 25.10 mL reagent)
cHCl = mmol NaOH added – original mmol HCl
total volume solution
= (25.10 mL x 0.10 M) – (50 mL x 0.05 M)
50.0 mL + 25.10 mL
= 1.33 x 10-4 M
pOH = -log(1.33 x 10-4) = 3.88 pH = 14 – pOH = 10.12

29. › Pre-equivalence: calculate the concentration of the base
from is starting concentration and the amount of acid that
has been added, the concentration of the base is equal to
the concentration of the hydronium ion and you can
calculate pOH from the concentration, and then the pH
› Equivalence: the hydronium and hydroxide ions are present
Equivalence
in equal concentrations, so the pH is 7
› Post-equivalence: the concentration of the excess acid is
calculated and the hydronium ion concentration is the same
as the concentration of the acid, and the pH can be
calculated

31.  Metal ions are Lewis acids - accepting electrons pairs from
electron-donating ligands that are Lewis bases.
 Monodentate ligand: binds to a metal ion through only one
ligand
atom.
 Multidentate ligand: attaches to a metal ion through more
ligand
than one ligand atom, also known as chelating ligand.
ligand
• Chelate effect - the
ability of multidentate
ligands to form more
stable metal complexes
than those formed by
similar monodentate
ligands

32.  A titration based on complex formation is called a
complexometric titration.
titration
 Following are structures of analytically useful chelating
agents:

33.  EDTA is an abbreviation for ethylenediaminetetraacetic acid,
acid
a compound that most widely used complexometric titrant.
 EDTA has 6 potential sites for bonding a metal ion: the 4
carbonyl groups and the 2 amino groups, each of the latter
with an unshared pair of electron. Thus, EDTA is a
hexadentate ligand.

34. • EDTA form strong 1 : 1 complexes with many metal ions; the
coordination is through the 4 O atoms and 2 N atoms

35.  The most common technique to detect the end point in EDTA
titrations is to use a metal ion indicator.
 are compounds whose color changes when they bind to a metal
ion. Useful indicators must bind metal less strongly than EDTA
does.
 Masking agent:
 In a direct titration, analyte is titrated with standard EDTA. The
analyte is buffered to a pH at which the conditional formation
constant for the metal-EDTA complex is large and the color of
the free indicator is distinctly different from that of the metal-
indicator complex.

38. There are several methods for the determination
of cations with EDTA.
Direct titration
Back titration
Displacement titration

39.  The solution containing the metal ion to be determined is
buffered to the desired pH and titrated directly with the
standard EDTA solution.
 It may be necessary to prevent precipitation of the hydroxide
of the metal (or a basic salt) by the addition of some auxiliary
complexing agent, such as tartrate or citrate or
agent
triethanolamine.
 At the equivalence point the magnitude of the concentration
of the metal ion being determined decreases immediately. This
is generally determined by the change in colour of a metal
indicator or by amperometric, spectrophotometric, or
potentiometric methods.

40.  Back-titration procedures are used when no suitable indicator
is available, when the reaction between analyte and EDTA is
slow, or when the analyte forms a precipitate at the pH
required for its titration.
 In such cases, an excess of standard EDTA solution is added
until the reaction is judged complete. The excess of the EDTA is
back-titrated with a standard metal ion solution; a solution of
zinc chloride or sulphate or of magnesium chloride or sulphate
is often used for this purpose.
 The end point is detected with the aid of the metal indicator
which responds to the zinc or magnesium ions introduced in
the back-titration.

41.  Displacement titrations may be used for metal ions that do
not react (or react unsatisfactorily) with a metal indicator,
that are more stable than those of other metals such as
magnesium and zinc.
 An unmeasured excess of a solution containing the Mg @ Zn
complex of EDTA is introduced into analyte solution. If the
metal cation M2+ forms a more stable complex than Mg or Zn,
the following displacement reaction occurs :
 The amount of magnesium ion set free is equivalent to the
cation present and can be titrated with a standard solution
of EDTA and a suitable metal indicator.

42.  Complexometric is widely used in the medical
industry because of the microliter-size sample
involved. The method is efficient in research
related to the biological cell.
› Ability to titrate the amount of ions available in a
living cell.
› Ability to introduce ions into a cell in case of
deficiencies.

43.  Complexometric titration is an efficient method for
determining the level of hardness of water. Caused
by accumulation of mineral ions, pH of water is
increased.
 Softening of hard water is done by altering the pH of
the water reducing the concentration of the metal
ions present.