Transcript Chemical Analysis

Chemical Analysis
E. Bottari, Chemistry Department,
“La Sapienza” University of Rome
Malta, Summer School 2007, 20th August – 9th September
Analytical chemistry - Methods
Qualitative analysis
Quantitative analysis
Traditional analysis
Instrumental methods of analysis
Suitable reactions for analysis
Qualitative analysis
 Recognition of chemical species by means of colour, reaction
producing a colour, reaction producing a precipitate, reaction
involving a change of a physical parameter.
 Colored ions are: Cu2+ (blue), Cr3+ (green), CrO4=(yellow), Cr2O7=
(orange), MnO4-(viole), MnO4=(green), Ni2+(green), Co2+(pink, or
blu), Mn2+(pink), and generally ions of transition metals.
 Precipitates (slight soluble compounds): sulphurs of eavy metals
(like: As, Sb, Hg, Cu, Pb, Cd, Sn, Bi, Zn, Ni, Co, Mn), BaSO4,
Hg2Cl2, AgCl, PbCl2, Ag2CrO4, many hydroxides of eavy metals.
Traditional - Quantitative analysis
Volumetric analysis
A + B = C.
A solution of B, at known concentration is added to a known volume
of A, until the number of equivalent of B is equal to those of A.
The added volume of B is measured and the concentration of A can
be calculated.
Gravimetric analysis
A + B = C.
An excess of B, at unknown concentration, is added to A so that A is
completely transformed in C, that can be weighed.
Quantitative analysis – Volumetric 1
 Titration (traditional)
To perform a volumetric analysis (titration) is necessary to have a
solution at known concentration a glass vessel, a burette
(calibrated tube able to measure volume, equipped with a tap) and
an indicator (chemical species able to change colour when change
the composition of the solution.
 A titration is a chemical operation which allows to obtain the
unknown concentration of a reagent, A, by adding the reagent B
and by stopping the addition when the equivalent number of A
and B are exactly equal.
 The point corresponding to meq* A = meq B is called equivalent
point or point of equivalence.
*meq = milli equivalents = VACA = VB CB
Quantitative analysis – Volumetric 2
Reaction must be:
 Quick
 Stoichiometric with known coefficients
 Univocal
 Collateral reactions must be absent
 Complete
 In correspondence to the equivalent point, a sharp change of
the followed parameter must occur
 This change can be put into evidence by the presence of a
colorimetric indicator or by means of an instrumental method
Quantitative analysis – Volumetric 3
Standard
It is very hard to perform a volumetric analysis by comparing a
solution with another, without a reference point. It is
necessary to have solutions at known concentration, prepared
by dissolving a known weight in a measured volume.
Compounds having such property are called mother
substances or primary standards.
 A standard must be very pure, stable, not reacting with air or
water (the solvent). It must be possible to dry it at 110°C. It
must follow all the rules of volumetric analysis and have a
high equivalent weight.
Volumetric Analysis
Kind of reactions
Solutions of electrolytes
Acid – base
Precipitimetric (formation of a precipitate)
Chelometric (formation of a chelate)
Redox
Electrolytes
 Electrolytes are all the compounds dissociable in ions. For ex.
NaCl  Na+ + Cl- or Na2SO4  2 Na+ + SO4=
 Electrolytes can be strong or weak.
 Strong electrolytes are completely dissociated, weak ones are
only partially dissociated and generate an equilibrium.
 Salts are strong electrolytes, like NaCl. It exists as Na+ and Cl Acids and bases can be strong or weak according their
properties. Mineral acids are generally strong, organic acids are
weak. H3PO4 is a weak acid and many inorganic acids are weak.
NaOH and KOH are strong bases, while NH3 is a weak base.
 Dissociation of weak acids or bases is regulated by a constant.
What is an acid, what is a base?
•An acid is a compound able to yield protons
•A base is a compound to catch protons
•A compound able to yield or catch protons, depending
on the experimental conditions, is called ampholite or
amphiprotic substance.
•Free protons in a condensed matter do not exist, so that
in a solution the proton comming from an acid must be
taken by a base contemporanily present. Often the solvent
can carry out such task.
Example: HA1 + B2 = HA2 + B1.
There are two conjugated couples acid – base:
HA1 = H+ + B1 and B2 + H+ = HA2
What is Water?
Water is able to yield or to acquire a protons, according the
following relation:
H2O + H2O  H3O+ + OH(1)
Eq. (1) is the sum of the two following:
H2O  H+ + OH- (acid behaviour of water)
H2O + H+  H3O+ (basic behaviour of water)
As eq. (1) is an equilibrium reaction, it is regulated by a
constant k = [H3O+] [OH-][H2O]-2, but [H2O] is constant in
the solvent water and by involving it in the constant k, it can
be written: Kw = [H3O+] [OH-] = 10-14 at 20°C.
In a neutral solution [H3O+] = [OH-]= Kw½ = 10-7.
What is pH?
pH is an usual expression suitable to indicate the acidity of
a solution. Practically it is the negative decimal logarithms
of the free hydrogen ionic concentration, i.e. pH = -log [H3O+]
In the above slide, it was shown that in neutral solution,
[H3O+] = [OH-]= Kw½ = 10-7
It can be deduced that pH = -log [H3O+]= -log 10-7 = 7.
A neutral solution has pH =7, an acid one has pH<7 and a
basic solution has pH > 7.
Similarly, it can be defined pOH and pKw and of consequence
pKw = pH + pOH
Calculation of pH 1
 Solution of strong acids or bases: free concentration is equal to the
analytical one: CH = cH = [H3O+]. COH = cOH = [OH-].
Solution of HCl = 0.1 M has pH =1. Solution of NaOH = 0.1 M has
pH =13.
 For solution of weak acid, like acetic acid CH3COOH, 0.1 M, the
free concentration [H3O+] must be calculated on the basis of the
following equilibrium (0.1 M it the total concentration):
CH3COOH + H2O  CH3COO- + H3O+
(2)
Acid1
Base2
Base1 Acid2
Calculation of pH 2
 The constant of eq. (2) is ka = [CH3COO-][H3O+] [CH3COOH]-1,
where [CH3COO-]= [H3O+], then it follows:
ka = [H3O+]2 [CH3COOH]-1 = [H3O+]2 (0.1 - [H3O+])-1. If [H3O+] is
negligible with respect to 0.1, it can be written:
[H3O+] = (ka CHA)½, where CHA represents the generalization of acid.
 For a base solution, i.e. acetate ions, CH3COO- 0.1 M, [H3O+]
must be calculated by means of the equilibrium:
CH3COO- + H2O  CH3COOH + OH- (3), with a constant
Kb = [CH3COOH][OH-] [CH3COO-]-1. By combining (2) and (3), it
can be obtained ka Kb = Kw and pOH = (kb CA-)½ (approx. formula).
In this case Kb is also called hydrolysis constant, indicated by Kh.
Buffer solutions
Solutions are called buffer when their pH does not change
appreciably, by adding little amounts of strong acid or bases.
Buffer solutions can be those having pH < 3 and pH > 11.
They can be also formed by a weak acid in the presence of
Its conjugated base. In this case in eq. (2) it is not possible
To write [CH3COO-]= [H3O+], then it follows:
[H3O+] = ka CCH3COOH CCH3COO--1
They can be also formed by a weak base in the presence of
its conjugated acid. For example NH3 and NH4+.
[OH-] can be calculated as follows:
[OH-] = kb CNH3 CNH4+ -1
Procedure for acid – base titrations
If the concentration H0 of the solution of a reagent HA must be
determined, with accuracy of 0.1 %, a measured volume V0
of the HA solution is transferred in a titration vessel and two
drops of indicator are added. A solution of NaOH, at exactly
known concentration OHT, is gradually added by means of a
Burette, till the indicator changes colour for the addition of a
drop more. The final point of the titration is reached.
The added volume of NaOH is measured by the burette, VT.
In this point the number of equivalent of both reagents is equal.
The following realtion can be written: H0 V0 = OHT VT.
VT is measured, OHT and V0 are known, H0 can be calculated.
Determination of olive oil acidity
This analysis allows to classify oil in a particular category,
with different commercial value according the acidity content.
Olive oil can be classified in the first value category, if its
Content of acidity (expressed as oleic acid) is  0.80 %.
Analysis is performed similarly to that previously described.
V0 measured of oil is transferred in a vessel for the titration.
A solution of OHT standard is gradually added till the change
of the colour of a suitable indicator is reached.
At this point, the volume VT of added titrant OHT, is read on
the burette and the initial acidity of oil H0 = VT OHT V0-1 can
be calculated.
Reaction involving a slight soluble
compound (precipitate)
By mixing a solution of AgNO3 with one of NaCl, a precipitate
of AgCl takes place, according to the reaction: Ag++Cl- = AgCl,
or AgNO3 + NaCl = AgCl + NaNO3. The equilibrium in this case
is shifted to right because a slight soluble substance is formed.
According to the equilibrium rules, it can be written:
k = [Ag+][Cl- ] [AgCl]-1, but AgCl is solid in equilibrium with
The ions in solution and its free concentration can be assumed
as constant, which involved with k, gives: Ks = [Ag+][Cl- ].
Ks is called solubility product.
The reaction between Ag+ and Cl- frequently used to determine
The quantity of silver present in a sample, in a similar way seen
For the reaction acid – base.
Chelometric reactions – Hardness of water 1
It was previously described the “dative” bond, i.e. a covalent
bond, where a couple of electron coming from the same ion
or compound (for example :NH3) is put in common between
the transition ion (for example Cu2+) and the donor (:NH3) to
form a complex ion or a molecule.
Next to :NH3 which is able to give only a couple of electrons,
many compounds exist containing several atoms able to give
each a couple of electrons, so that a molecule of such compounds
is able to bind a metallic ion with several bonds, forming five
membered rings particularly stable.
The most used compound having this property is called EDTA.
The most important application of EDTA is the water hardness
determination, i.e. the calcium(II) + magnesium (II) dosage.
Hardness of water 2 - Chelometry
The knowledge of hardness of water is very important in many
practical cases: beer industry, metallic tubes for water, formation
of limestone in caldrons, water heaters, and in general, in washing
machines. Hardness is due to the formation of calcium or magnesium
carbonate, slight soluble compound forming limestone.
EDTA standard solution is able to titrate solutions containing
Calcium and magnesium ions to determine their concentration.
Analysis is carried out similarly to those previously described, but with
different indicator and at well defined and buffered pH.
It is possible to know separately the calcium and magnesium present in
the same solution. The same analysis is performed to know the calcium
and magnesium concentration in milk, or cheese.
Redox reactions 1
This kind of reaction takes place involving the electrons
transfer from the reducing to oxidant, as follows:
Ox1 + n1 e = Red1 and Red2 = Ox2 + n2 e, if n1 = n2, the
complete reaction can be written: Ox1 + Red2 = Red1 + Ox2
Oxidant and reducing compounds have different straight,
which can be experimental proved.
A solution of Cu2+ is blue. If you put a piece of Zn or of
Pb inside the copper solution, you can observe that after
some minutes solution becomes colourless, Zn or Pb are
dissolved and a red slight soluble compound is formed.
This means that Zn or Pb are oxided by Cu2+ reduced to Cu0
Red. The following reactions occur:
Cu2+ + Zn = Cu0 + Zn2+ or Cu2+ + Pb = Cu0 + Pb2+
Redox reactions 2
• If you repeat the same experiment by using Pb2+ and Zn, you
find that Zn is oxidized by Pb2+ to Zn2+ according to the
reaction: Pb2+ + Zn = Pb0 +Zn2+
•If you repeat the same experiment by using Ag+ and Cu, you
find that Cu is oxidized by Ag+ to Cu2+ according to the
reaction: Ag+ + Cu = Ag0 +Cu2+
From the above presented examples, you can deduce the
following sequence, as straight of oxidation:
Ag+ > Cu2+ > Pb2+ > Zn2+, vice versa the straight as reducing
The straight as OX of a couple (i.e. Ag+ + e = Ag0) is
represented by the redox potential. The table of standard
potentials collects all the couples with their values.
Redox reactions 3
Nernst equation shows the dependence of the redox potential
on the reagents (ox and red form) free concentration:
Nernst equation for a generic couple: Ox1 + n1 e = Red1:
E1 = E°1 + RT (n1 F)-1 ln {[Ox1] [Red1]-1}, which at 25°C is:
E1 = E°1 + 0.05916 (n1)-1 log {[Ox1] [Red1]-1}
The following Redox couple are frequently used in analysis :
•MnO4- + 5 e + 8 H+= Mn2+ + 4 H2O
•Cr2O7 + 6 e + 14 H+ = 2 Cr3+ + 7 H2O
•I2 + 2 e = 2 I-;
IO3- + e + 6 H+ = I- + 3 H2O
•2S2O3= = S4O6= + 2 e;
C2O4==CO2 +2 e;
S= = S + 2 e;
•Sn2+ = Sn4+ + 2 e;
Fe2+ = Fe3+ + e;
•NO3- + 3 e + 4 H+ = NO + 2 H2O; 2HgCl2 + 2 e =Hg2Cl2 +2Cl-
Instrumental Analysis
Electro analytical analysis
Optical analysis (spectroscopy)
Thermal analysis
Chromatography
Electro analytical analysis
 Electrolysis (electro gravimetric analysis).
 Electromotive Force Measurements (Galvanic
elements – Piles) (pH measurements). Direct and
Titrations.
 Conductometry.
 Coulometry (direct and indirect).
 Polarography.
Spectroscopy
Emission (flame, voltaic arc, sparkly).
Absoption (Molecular: UV violet, IR).
Absorption (Atomic: flame, furnace).
X Ray or more sophisticated methods.
Chromatography
On column
On paper
TLC (Thin layer chromatography)
GC (gas chromatography)
HPLC (High performance liquid chromat.)
Many applications: analysis of fat fraction of many substances,
Like: Milk, Different kinds of oil, Butter, Meet, Eggs, Vitamins
Cosmetic, Pesticides, Dioxins, Herbicides, etc…