Download Analytical Chemistry

Analytical Chemistry
Dr. Manal A. Tooma
First year
Chemistry is the study of matter, including its composition and structure, its
physical properties, and its reactivity. There are many ways to study chemistry, but,
traditionally divide it into five fields: Organic, inorganic, physical, analytical, and
biochemistry.
Analytical chemistry is a field of science consisting of a set of powerful ideas
and methods that are useful in all fields of science, analytical chemistry concemed
with determinins the composition of substances. It comprises of two branches:
Oualitative analvsis: deals with the determination of the identity of
chemical constituents in samples.
-
Ouantitative analvsis: deals with the determination of the amounts
(concentration or composition) of various substances in samples. There are
many methods of determining these amounts, all based on chemical or physical
properties of atoms, molecules and ions.
In general chemical analysis methods can be classification to:
.
Gravimetric methods determine the mass of the analyte or some
compound chemically related to it. This method is base on the measurement
of
the reaction product by precipitation.
.Volumetric method. the method is based on determining the volume of
a solution of known strength
thatrrtiredreacting with specific amount of the
,
sample such as acid-base titratioSy'
.
ElectroanalVtical methods involve the measurement of such electrical
properties as voltage, cunent, resistance, and quantity of electrical charge.
.
spectroscopic methods are based on measurement of the interaction
between electromagnetic radiation and analyte atoms or molecules or on the
production ofsuch radiation by analytes. Finally, there is a group
of
Analytical Chemistry
Dr. Manal A. Tooma
First year
miscellaneous methods that includes the measurement of such quantities as
mass-to-charge ratio of molecules by mass spectrometry, rate of radioactive
decay, heat ofreaction, and rate of reaction, sample thermal conductivity,
optical activity, and refractive index.
Fundamental Concepts
Atomic weieht of element: The mass of a single atom in grams is much
too small a number for convenience, and chemists therefore use a unit called an
atomic mass unit (amu) also known as a dalton (Da). One amu is defined as exactly
one-twelfth the mass of carbon isotope '2C and equal to 1 .66054 x l0-24 g.
Example:prove that carbon weighing 1.0 x10-3 g contains 5.01x
1.0 x10-3
e"
*#m=
G;#\
1Ole
carbon atom?
5'01 x 10re c atoms
Molecular weight (Mwtl: is the average mass of a substance's molecules.
Numerically, is equal to the sum of the atomic weights of all atoms in the molecule'
molecular weight = sum of atomic weight
M.wt of water (H2O):2(1.008) +l(16'00):18'02
M.wt of hydrogen (H2) : 2(1'008) :2'016
These weights are all relative to the mass of
r2c atom as 12.000 with (SI)
system of units' .Molecular weight is expressed by g/mole
\.2
H.IY: Calculate the M.wt' of
CzHr+Oz, CzHsNzSO+, NazCOr, PzO:
Mole (moD: its Avogadro number (6.022137 x
somethine consists of 6.022137
x
1023
1023)
of species, so one mole of
units of that substance'
Analvtical Chemistrv
Dr. Manal A. Tooma
First year
The mass
of 1 mole of an element is equal to its atomic
.
no.of mole 1
mot
mass of substance in g
------ M.*;--
36ss
ofHCI"
36.5 /mol
g
Examples: Calculate the number of moles in 200 mg of
-
in
,/ /
-Nr
P---e
mass in srams
D/
CaO? -s-'
10 g of P2O5?
in 5 mg of NaCl?
in 25 mg olH2Soa?
in 400 mg of K2CrO3?
L
Equivalent weieht: is the molecular weight divided by the number of reacting units
(no.of equivalent
- valency)
e
quiv arent
w
e
i
g
ht
-
mole cular
w eig
ht
no.of equivalent
For acids: the number of reacting units is the number of hydrogen ions that
will
fumish (acidic hydrogen).
For bases: the number of reacting units is the hydrogen ions that will react with it'
(Bases hydroxide)
For salt: the number of positive parts in the salt Multiplied by its oxidation number'
For redox reaction: the number ofreacting units is based on the number of electrons
that the oxidizing or reducing agent
will take on or supply' It
can be computed also as
the change in oxidation number (transfer electrons)'
H.ftCalculate
the Eq.wt. of HzSOr, HrPO+, HCI' CH3COOH, Na2CO3' NaHCO:,
Ca(OH)2, NaOH, KOH, Ba(OH)2, K2CrOa, FeCr2Oa, CaClz,FezO:, (Cr*3- Cr*7),
(KMnoa- Mnoz), (Mno4- -Mn*2),1K2Cr2or- 2Cr*3) AgNO3, HgCl2 '
-t
Analvtical Chemistrv
First year
Dr. Manal A. Tooma
C o ncentr atio
No.of mole =
ns express io ns
mass
of th substance in gm
wt
M.wt.
M.wt.
moles
of solute
Molarlty lMl =
volum of solution in Liters
wt
M=-+M.wt.
N
omaltty IN tt-
wt
1000
V
meq.of solute
volum of solution in liteTs
s
inml
.,
',v
wt
= Eq.wt.
-*-
1000
V i.n mI
volume mI
Density and SpeciJic Gravity
Density (p): the mass per unit volume at the specified temperature. It is usually
expressed in g/ml or g/Cm3 at 20
oC
oC
and O '99823 glml at 2Q'C
Density of water: is taken as 1 .0 g/ml at 4
Specific gravity (sp.g.): is defined as the ratio between the densities of solution
at2}octo that of water
at 4oC, hence it is a dimensionless quantity'
Numerical values of density and specific gravity are equal since
d*u1., at 4
oC
:1 .00
s.lml
The
wetght fractton -- r"#
Z*ff)=
Themole -n,
fraction
n
=-=
of solute
x 100
mass of sample
mass
1oo
moles of solute
total moles of samPle
= lmobo/o-
1.00
I
Analvtical Chemistrv
First year
Dr. Manal A. Tooma
The volume
fractton
volume of solute
o/o{ =
* 100
vt
total volume of sample
-
weight
. ---:Lwt
fracti"on -- o/o--voLume'
V
The
Parts per
of solute
* 100
total volume of sample
mass
million (ppm): Used to express trace concentration of solute. It can be
expressed either
PPm
/wt\
l-l
\wt/ =
of solute(g)
* 106
mass of sample(g)
mass
mass
of solute(mg)
--------"------
rJUtrL -
mass of sample(kg)
mass of solute(g)
PP /wt\
\v)=@*''u-
PPm
=
^A
pg
solute(mg)
S
-- or ---i
or
t*e s"rrrat"A
mass
""t
a
of
"f
There are a number of expression in which the concentration of solution can be
expressed
Molaritv: It is defined
Molarity [M] =
as the number of moles
moles of solute
vol.of solution in liters
of solute in each liter of solution
Molarity tMl
-#-;ffi
Mpls.lifr;the solution which contains I mole of solute per kilogram of solvent.
Lr
Molali*,
t*l
t'Irrr -
moles of solute
1000 gof solvent
Analvtical Chemistrv
First year
Dr. Manal A. Tooma
Normalitv: A one normal solution (1 N) is that contains one equivalent weight
of solute per liter of solution.
Normality [N]
-
no of equivalent of
vol.of solution in
solute
liters
The relation between Normality and
'
Normality [N] = - -!- * +00,
Eo.wt. V in ml
Molarity: N: M (valency)
H.W:
1) What
volume of 3 .25 M sulfuric acid is needed to prepare 0.500 L of
0. I 30
M
H2S04?
(
i' Zl How do we prepare 75.0 ml of 0.96 M sulfuric acid from 18 M acid?
^r,
3) What volume of 6.39 M sodium chloride contains 51.2mmol sodium chloride? ({.
4) Rubbing alcohol is an aqueous solution containing 70% isopropyl alcohol by
volume. How would you prepare 250 ml rubbing alcohol from pure isopropyl
alcohol?
5) How many grams ofglucose
and ofwater are in 500 gof a 5.3%oby-mass
^r/
glucose solution?
Calculate the molar concentration (Molarity) of a solution contains I ppm of
6)
lead (Pb),
(l
M.utt:207?
7) A sample with mass of 2.6gcontains 3.6 pg of zinc; calculate the
\?'
concentration of zinc in PPm?
8) 150mL of an aqueous sodium chloride solution contains 0'00459
ir'
NaCl. Calculate the concentration of NaCl in parts per million (ppm)'
9)
What mass in mg of potassium nitrate is present in 0.25kg of a 500ppm
KNO31uql?
.
10)
{
1I
A student is provided with 500m1 of 600ppm solution of sucrose. what
volume of this solution in millilitres contains 0.159 of sucrose?
) Convert 78 ppm
of Ca2* ions to
mol/L?
d"
(t
Analytical Chemistrv
First year
Dr. Manal A. Tooma
Concepl of chemical equilibrium
If two reactants A
and B are mixed together a chemical reaction can take place
to yield the product C&D
The chemical reaction can be presented as follows:
aA +
bB
------->
cC + 6P
a, b, c, d: stoichiometric coefficients
In many chemical reactions the reverse reaction (i.e.) reaction between
c with
D to form A&B the reaction is also possible hence the reaction is presented by:
aA+bB
<+------> cC+dD
when forward and reverse reaction becomes equal, the system has reacted to
the condition known as "chemical equilibrium" the reaction appear to stop when
equilibrium condition is actually reached since there is no change in the concentration
of all substances. The equilibrium constant the equilibrium condition is often
expressed by a factor or known as equilibrium constant
&:
&:
[B]b
ku [C]" [D]o
kr [A]"
kr [A]" [B]b
:
rate of the forward reaction
rate of the backward reaction
ku [C]"
[D]d when the system reach equilibrium
K:# = ffis
-
When R; decreases with time,
R6
increases with time and the two rate become
equal at equilibrium.
-
Equilibrium constants are useful for calculating the concentration of varies ions
in equilibrium
Analvtical Chemistrv
First year
Dr. Manal A. Tooma
Faclors affecting on chemical equilibrium
Chemical equilibrium is a very delicate system that represents a perfect balance
between forward and reverse reaction. A small disturb in the equilibrium may shift the
equilibrium position either to right forming more products or to left forming more
reactants. This reaction by the system is ofcourse temporary and eventually the
system
will
come back to equilibrium. This phenomenon can be expressed in the form
ofLe Chatelier's Principle (ifan external
the system
will tend to react in such way
stress is applied to a system at equilibrium
as to relieve the applied stress and tries to
reestablish the equilibrium). In chemical reaction terminology, the "stress" means
change in concentration, pressure, volume or temperature.
Change in Concentration Consider the following equilibrium reaction
Nz+3H2 +-> ZNH,
If we add either Nz or H2, we
increase the collisions between Nz and Hz thereby
forming more product NHl. This is how the equilibrium counteracts the applied
stress; we say the equilibrium shifts from left to right.
If we add more
NH1, we
increase the concentration of NH3' As a result, some NH3 decomposes and
forms more reactants. We say the equilibrium shifts from right to left'
If we remove either Nz or H2, now there is less concentration of Nz and H2' To
offset the applied stress, the equilibrium shifts from right to left. If we remove
some NH3, the equilibrium shifts from left to right to counteract the applied
StTESS.
lr t 2
'1\, ---d
Chanses in Pressure and Volume Pressure does not have any effect on
concentrations of species that are present in solid, liquid or solution form. On
the other hand, the change in pressure affects the concentrations ofgases.
According to ideal gas law, pressure and volume are inversely proportional to
each other; the greater the pressure, the smaller the volume, and vice versa'
Analvtical Chemistrv
First year
Dr. Manal A. Tooma
Thus, it is just enough to understand the effect of change in pressure on the
equilibrium system.
Nz+3H2 <+ 2NH3
lmol+3mol
2mol
vol + 3 vol
2vol
1
What
will
happen
if we increase the pressure (decrease the volume) on the
system at constant temperature? In order to understand the affect ofpressure,
it
is very important to understand the total volume of reactants and products.
According to Avogadro's law, the number of moles is directory proportional to
the volume of the gas. Think that one mole is one volume. Therefore, there are
4 volumes
(l
vol + 3 vol) on the reactant side and 2 volumes on the
product
side. The increase in pressure always affects the side that has more volume.
Hence, increase in pressure shifts the equilibrium from left to right. The
Dressure has no effect
if the total volume of reactants is equal to the total
volume of the products as in the following example.
Hzs +
Changes
in Temperature: A
Izs <+ ZHI,
change in concentration, pressure or volume alters
the position of the equilibrium but not the magnitude (value) of the equilibrium
constant. However, the change in temperature changes the value of the
equilibrium constant. To understand the effect of temperature, we must know
whether the reaction is endothermic (absorption of heat) or exothermic (release
ofheat). Let us consider the equilibrium reaction between dinitrogen tetroxide
and nitrogen dioxide:
NzO+
(g) ---_-=-
2NO2@)
This reaction is endothermic in the forward direction and exothermic in the
reverse direction
(g) ------------r> 2NO2 (g)
2NO2 @) ----------> NzOq (g)
NzO+
aH=
AH
=
58.0 kJ/mol
-58'0 kJ/mol
-
Analytical Chemistry
First year
Dr. Manal A. Tooma
To understand the effect oftemperature (heat), let us re-write the above
equations treating the heat as chemical reagent. Thus
heat+NrOo
2NO2
(g) ------------> 2NO2 (g)
G)
N2Oa (g) + heat
Therefore, increase in temperature favors the endothermic reaction (forward
-_-r>
reaction, i.e. left to right) while decrease in temperature favors the
exothermic -
reaction (reverse reaction, i.e., right to left). What does it mean? It means that
the value ofthe equilibrium constant increases when the heat is added (increase
in temperature) and decreases when the heat is removed (cooling the system)
that can be explained by the following equilibrium constant expression:
.
r<"
=
-
[Nor]z
fi,
or)
The Effect of a Catalvst The function of a catalyst is to speed up the reaction
by lowering the activation energy. The catalyst lowers the activation energy
of
the forward reaction and reverse reaction to the same extent. Due to this, there
is no shift in equilibrium or the change in the value ofthe equilibrium constant.
Therefore, we conclude that the catalyst has no effect on the equilibrium
system.
(When the initial concentration is
>
100k, we will neglect the small change in
concentration. IJ not we can solve it by general quadratic
formula) x
-
initial
-ot'lo'-+ac
2a
Type of equilibrium conslants
There are several types ofequilibrium constants according to the type
of
reaction as below:
-
Dissociation constant (IQ)
Formation constant (K1)
Acid & base dissociation constant (K" , Ku)
Ionization constant of water (K*)
10
Analytical Chemistrv
First year
Dr. Manal A. Tooma
The following examples give an idea for calculations using equilibrium
constants
t( l.
the chemical substances (A) and (B) are reacted to produce (C) and (D)
according to the reaction A +B
<--+
C +D
if
0.2 mol of (A) and 0.5mol
of
(B) are dissolved in 1 liter solution and the reaction takes place with
equilibrium constant
K:
0.3 calculate the concentrations of reactants and
products at equilibrium
2. Calculate the equilibrium concentration ofA
and B in a 0.10 M solution
weak electrolyte AB with an equilibrium constant of 3xl0'6
3.
Calculate the equilibrium constant for a reaction Aaq
+
AB
Baq
<------+
a
of
A+B?
3Coo
between 0.10 M A and 0.30 M B, given that the product, C has a concentration
of
0.20 M and that these values are those when the reaction is in a state
of
eq ui I i brium.
4.
(ans. 0.267 )
Calculate the equilibrium constant
(K)
for the following reaction:
HrG)+Iz(e)+2HI(e)
when the equilibrium concentrations at 25 "C were found to be:
lHrl :0.050s M, [Ir] :0.0498
M,
[HI] :0.389 M ?? ans:60.02
5. Calculate the equilibrium constant (K'o) for the following reaction:
Nz(g)+3Hdg)+2NHr(g)
when the equilibrium concentrations at 500
t3Ht :0.037
6.
M,
K
were found to be: [N2]
:0.03 M,
[NH3] = 0.016 M ?? ans:0.017M
Calculate the equilibrium constant (K"o) for the following reaction:
2SOz(g)+Oz(g)+2SO3@)
when the equilibrium concentrations at 800
M,
[SO2]
:0.003
M,
K
were found to be: [O2]
:
0.0035
[SO3] = 0.05 M ??
L1,
Analytical Chemistry
First year
Dr. Manal A. Tooma
Acids, Bases and buffers
Acids - base Theories
l-
Arrhenius Theory (1884)
Acid is any substance that ionized in water (partially or completely)
to give hydrogen
ions
Base: is any substance that ionized in water to give
hydroxyl ions
According to this definition, HCI is an acid NaoH is a
base as shown below:
HCI
rf + ct-
_+
NaOH ----+Na*
+ 6p
Acid chemically reacts with a base as follow:
HCI + NaOH
------f
Na* + 61- a g16
2- Brsnsted-Lowry Theory (1923)
Acid: is any substance that can donate
a proton.
Base: is any substance that can accept a proton.
3- Lewis Theory (1923)
Acids: is a substance which can accept an electron pair by combining with a second
substance with al unshared pair of electrons.
A base: Is a substance which shares it unshared electron - pair during chemical
reaction
BCl3 +:NH3
C6B:NH.,
--_'
Conjugate Pairs:
Acids: are base pairs in the ionization reaction are called pairs coniuqate base
The conjugate base ofthe acid and acid is the conjugate base
Acid = H'+ base
2H2O =
H3O'+
Off
12
Analytical Chemistry
First year
Dr. Manal A. Tooma
NH4*:11+ + NH:
H2O: conjugated acid for OH-
NH3: is conjugated base for (NI{4-)
or
Nlla":
is conjugated acid for (NH3) base
Acid- Base strength: when acid or base is dissolved in water it will dissociate or
ionize. The degree ofionization depends on the strength ofthe acid
A strong electrolyte is completely dissociated
A weak electrolyte is partially dissociated
weak acid has a relatively small dissociation constant (Ka) where's strong acid has a
large dissociation constant.
The strength ofan acid depends on its type and is not related to the concentration.
(e.g.): HCI strong acid regardless of whether its concentration. 1M or l0-a
M
Strong base: is a base with relatively large K5
Weak base: is a base with relatively small K5
Acid bases equilibria in water
Kw: tfl tOHl : 1.0 x 1O-ta(ionization
trfl toHl : 1.0 x lo-'a
Ffl = toHl:
Conc. of
1.0
x
content of water)
1o-7
[Ff] in a solution is often expressed
as the
pH of the solution which is the
negative of the logarithm of [H-]
pH:-log
[Ff]
pOH: - 1og [OFI]
pKw:pH+pOH : l4
H.l/. : Calculate the pH of 2xl0-3 M of hydrochloric
acid solution?
In order to determine the pH of weak (partially ionized), strong (completely ionized)
acids or bases and salts, different calculations must be followed as shown in the
followine:
Analytical Chemistry
First year
Dr. Manal A. Tooma
1- Strong acid / base
Since the strong electrolye is completely dissociated for that the concentration
of IFfl/ [OH-] is equal of the initial concentrarion.
Exp: What is the pH of a 0.010 M HCI solution?
Since HCI is a strong acid, the hydronium ion concentration
will
be equal to
the HCI concentration:
lH,o-l:0.010 M
The pH can be found by taking the negative log ofthe hydronium ion
concentration:
pH
2-
:
:
-log[H3O-]
lVeak acid
-log(0.01 0)
:
2.00
/ base
It's important to realize that a weak acid is not the
same thing as a dilute
solution ofa strong acid. Whereas a strong acid is 100% dissociated in aqueous
solution, a weak acid is only partially dissociated. It might therefore happen that the
concentration from complete dissociation of a dilute strong acid is the same as that
from partial dissociation of a more concentrated weak acid.
Equilibrium exists between the weak acid, water, H:O', and the anion of the weak
acid. The equilibrium lies to the left hand side ofthe equation, indicating that not
much H3O* is being produced. The fact that very little H3O* is being produced is the
definition of a weak acid.
The Ku for a weak acid is small, usually a number less than 1'
HA
(ad
+ H2O
0) €
t,^
nt,.
--
Aiuql + H3O*1uq;
lH3O+llAl
IHA]
L4
AnalWical Chemistrv
First year
Dr. Manal A. Tooma
Note that water has been omitted from the equilibrium equation because its
concentration in dilute solutions is essentially the same as that in pure water and pure
liquids are always omitted from equilibrium equations
Exp: The initial concentration of HNO2 is 0.45
HNO2luqy
Init
+ Hrofl) 5
NO2 ("q)
0.45
0.45
Ku
:
4.5 x
10-a
H3O*1aqy
0
Change -x
Equal.
*
M,
*x
+X
-x
Ka= [NO,-ILH3O-I
li{No,l
_ (x)(x) _,,,<,.,^.4
0.45
-
x
Simplifies to: since K" *100 < Ci
x:
[No2]
:
tH.l
: "f 4
c'
[]rNot :0.45-0.014
roHt=#:# -
:,l4sffix
0.436,lJ{
7.7x'J.0-73
H.IY: Codeine (CrsHzrNO:),
a
0.45 = 0.014
a drug used in painkillers and cough medicines, is
naturally occurring amine that has kb
:l
.6
x
10-6.
Calculate the pH and the
concentrations of all species present in a 0.001 2 M solution of codeine.
I)
Analytical Chemistrv
First year
Dr. Manal A. Tooma
We've seen
that the strength of an acid can be expressed bv its
valuc of K" and the strength of a base can be eipressed by its value of ks. For a lonjugate acid-base par, the hvo eq,rili!5ig6 constants are related in a simple way thar
makes it possible to calculate either one from the other. L.et's consider the conjirgate
acid-base pair NH.1' and NH.1, for example, $'here K., refers to proton transfer fiom
the acid NHl - to n'ater and K6 refers to proton transfer from u'ater to the base NH3
Thc sum of the two reactions is simply the dissociation of water:
N]+fldiil + H2O(/)
lnH:trrl) +
Net
H2O1/,1
2H2O1/;
-'.
H,"O+(rr(i)
+
N!*+*tnq)-t
NHttr4i
OH*(rr4)
::*:: Hro-(d(l) + OH-tnf)
Kn
*
[H3O*][NH3] _
5.6
x
10*ro
lNHr*l
[NH{,'][OH ] _
J{6 _
i.8 x 10 "5
INH:l
K," = [H3O+][OH*] = 1.0 >( 10-14
The eqrrilibrium constant fnr the net reacfion equals the product of the equilibrium
constants for the reactions added:
Kn;\Kb= [{3C)-]tr{r{51
lNHr*l
(5.5
x
10-10X1.8
k*:
{N}*al[oH ] :[H:O'][OH]:K".
{NHt
;< 10-5) = 1.0 x 10-14
ku. k5
ku: k*/ k6 or k6:
kw/ka
3- pH ofSalt
Salts such as NaCl that are derived from a strong base (NaOH) and a strong acid
(HCl) yield neutral solutions because neither the cation nor the anion reacts
appreciably with water to produce or ions. As the conjugate base of a strong acid, has
no tendency to make the solution basic by picking up a proton from water. As the
cation of a strong base, the hydrated ion has only a negligible tendency to make the
solution acidic by transferring a proton to a solvent water molecule.
The following ions do not react appreciably with water to produce either H3O- or OHions:
I
Analvtical Chemistrv
First year
Dr. Manal A. Tooma
. Cations from strong bases: Alkali metal cations of group 1A: Li*, Na*, K*
Alkaline earth cations of group 2A:
Mg*2 , Ca*2, Sr*2, Ba*2 except for Be*2
. Anions from strong monoprotic acids: Cl-, Br-, I-, NO3-, and ClO3-
Salts that contain only these ions give neutral solutions in pure water
(pH:
7).
_
Salts such as NlIaCl that are derived from a weak base (NH3) and a strong acid (HCl)
produce acidic solutions. In such a case, the anion is neither an acid nor a base, but the
cation is a weak acid
NH4
*tuq)
f
HzOtrt
<-----------+ H3O*1uq; -I- NH31uq;
Finally, salt such as (NHa)2CO3 in which both the cation and the anion can undergo
proton-transfer reactions. Because is a weak acid and is a weak base, the pH ofan
(N}I4)2CO3 solution depends on the relative acid strength of the cation and base
strength ofthe anion:
We can distinguish three possible cases:
. ku>
ku
If
ku for the cation is greater than k5 for the anion, the solution
will contain
an excess of H3O' ions (PH< 7 ).
. ku<
ku If ku for the cation
an excess of OH- ions (PH>
. ku=
is less than k5 for the anion, the solution
will contain
7 )'
kr If k" for the cation and k5 for the anion are comparable' the solution will
contain approximately equal concentrations of H3O* and OH- ions (pH:7 )'
-
Calculate the pH of a 0.10 M solution of AlCl3; k" for Al(H2O;u*3 , is 1'4 x 10r'
-
calculate the pH of a 0.10 M solution of NacN; ku forHCN is 4.9 x10-r0
-
Calculate the pH of 0.20 M NaNO2; ku for HNO2is 4'6x 1 0'-'
T7
Analvtical Chemistrv
First year
Dr. Manal A. Tooma
The human eye is not sensitive to color differences in solution containing a
mixture of HIn and In- , particularly when the ratio [HIn]/ [In-] is greater than
about 10 or smaller than 0.1
So we can write that the average indicator, HIn, exhibits its pure acid color
.
lHlnl 10
when
t;
and is base color when
ffi
luInl
HrO'
,
H,O'
-
= K^=
'ltn-l
= l0
lHtnl
: K"=
unj
:0.1 K"
K"
[Htnl 1
-
ffi =*
for full acid
for full base
: PK" -1
pH = - log (0.1 K") : PKu +1
pH:
- log (10
K")
Indicator PH range
:
PKa +1
pH (acid- base): 2
--1
-
4-Reaction of strong acids and strong bases
M HCI with 0' 1 M NaOH
Plot titration curve for the titration of 50 ml of 0'05
(9, 10, 2j, 24,25,26,30, 4p ml.
. 2- Calailarethe pH during the titration of 50 ml of 0'05M NaOH with 0'l
',.
volume of reagent
after the addition afteladding the following
M HCI
(24'5'25'25'5)
ml?
3-
when mixing 20 ml of 0'2 M HCI
Calculate the pH of the solution that result
with 25 ml of 0.132M NaOH, 0'323M NaOH?
19
Analytical Chemistry
First year
Dr. Manal A. Tooma
4- Plot titration curve for the titration of 50 ml of 0.1 M HCI with 0.2 M NaOH.(0.
r0,20, 24, 25, 26, 30,40)ml.
Titration
I
-
for
weak acids or bases
Plot curve for the titration of 50 ml of 0.1M acetic acid with
0. 1
M NaOH (0.
10,25,40,50, 60, 70) ml.
2-
Calculate the pH for the titration of 50 ml of 0.05 M NaCN with 0.1 M HCy(0,
10,25,26) ml.
3 -
Calculate the pH of solution prepared by :a) dissolving 43 g of lactic acid in
water and diluting to 500 ml
Titrationfor weak acids and strong bases (buffer)
If
ions into a
a strong base is added, the buffer converts the excess hydroxide
base by reacting
with the Brsnsted acid:
s"
o !Ho49!
llofore titration
tH1=JKaiH6'Ee1
With titration
pH:
At equivalent Point
loHl:.p too.-l [oAc1 > looH
After equivaience
pKa
*
foe
loAc-l
ffi
Henderson-Hasselbalch Equation
[OHl=[excesslitrant]
poiut excess tihant
q
for titration of 50m1 of
calculate the pH and plot the resulting titration curve
Ka =
0.1M HOAC by addition of 0, 10, 25' 50 and 60 ml of 0'lMNaOH'
1
.75x
10-s.
20
Analvtical Chemistrv
Dr. Manal A. Tooma
First year
Titrationfor weak
Ifa
base and strong acid (buJfer)
strong acid is added, the buffer converts the excess protons into an acid by
reacting with a weak conjugate base
Befbre titration
IOHI:
With titration
pH
At equivalent point
[H*] =
Alter cquivalcnce
: pKa + loe
INH"i
fi;fr
lNtr4l > loo
#
fH*J=[excesstitrant]
'''--i
e
ltNH-r
l([ q ':---"
100
rft[NH3]
,-
'
-\
i
.
'.
i" r r..+.."
Calculate the pH and plot the resulting curve for the titration of 20 ml of
0'1lM
ammonia by the addition of 0,5, 11,15,20 ,22,25,3},35,and 40ml of 0'1M HCI
Kb: 1.79x
l0-5?
'ii\
./
consider the titration of 100.0 ml of 0.016 M HOCI with 0.0400 M NaOH'
How many milliliters of 0.0400 M NaOH are required to reach the equivalence
M NaOH
point? Calculate the pH: (a) After the addition of 10'0 ml of 0'0400
(b) Halfway to the equivalence point (c) At the equivalence point
-
acid titration, the
Shows the pH titration curve for a typical weak base-strong
l0-5
40.0 ml of 0.100 M NH3 with 0'100 M HCI' Kb 1'8 x
:
titration of
H.t/:
or [H3O-]
1- For each of the following solutions, calculate [OH] from [H:O-],
from [OH-] Classifi each solution as acidic' basic' or neutral'
[H:ot]
:
toHl
:1.x10-r0M, 0.01M, 5.6x10-eM,
3.4 xl0-e M , 8.6 xl0-5
M,
M, 2'5 xl0-4 M, 2'0M
1'5xl0'M, l'0xI0-7M
1.0 x10-7
2I
Analvtical Chemistrv
First year
Dr. Manal A. Tooma
2- Water superheated under
What is [H3O
*]and
pressure to 200 oC and 750 atm has
Kw:
I .5
Xl0-1'
[OH-] at 200'C? Is the water acidic, basic, or neutral?
3- Water at 500 "C and 250 atm is a supercritical fluid. Under
these conditions,
Kw is approximately I .7 Xl0-re Estimate [HrO *] and [OH-] at 500 oC. Is the
water acidic, basic. or neutral?
4- Calculate the pH to the correct number of significant figures for solutions of
Question
I?
5- Calculate the concentration to the correct number of
[H3O
"], [OH] significant
figures for solutions with the following pH values: 4.1 , 10.82 ,0.00 , 14.25
,
5.238, -1.0,9,14.25 -0.3, 10.75
,/6-
A solution of NaOH has a pH of 10.50. How many grams of CaO should be
dissolved in sufficient water to make 1.00 L of a solution having the same pH?
7- A solution of KOH
has a pH
of
10.00. How many grams of SrO should be
dissolved in sufficient water to make 2.00 L of a solution having the same pH?
8- Calculate the pH ofsolutions prepared by:
(a) Dissolving 4.8 g of lithium hydroxide in waterto give250 mL of solution
(b) Dissolving 0.93 g of hydrogen chloride in water to give 0.40 L of solution
(c) Diluting 50.0 mL of 0. I 0 M HCI ro a volume of 1 .00 L
(d) Mixing 100.0 mL of HCI and 400.0 mL of HCIOa (Assume that volumes are
additive.)
9- Calculate the pH ofsolutions prepared by:
22
Analvtical Chemistrv
First year
Dr. Manal A. Tooma
(a) Dissolving 0.20 g of sodium oxide in water to give 100.0 mL of solution
(b) Dissolving 1.26 g of pure nitric acid in water to give 0.500 L of solution
(c) Diluting 40.0 mL of 0.075 M Ba(OH)2 to a volume of 300.0 mL
(d) Mixing equal volumes of 0.20 M HCI and 0.50 M HNO3 (Assume that volumes
are additive.)
l0-Look up the values of k for C6H5OH, HNO3, CH3CO2H, and HOCI, and
aftange these acids in order ofi a) Increasing acid strength
Increasing conjugated acid strength
1
l- Look up the values of k for HCO2H, HCN,
HCIO4, and HOBr, and arrange
these acids in order of: a) Increasing acid strength,
b) Increasing conjugated acid strength
12- The pH of 0.040 M hypobromous acid (HOBr) is 5.05. Set up the equilibrium
equation for the dissociation of HOBr, and calculate the value of the aciddissociation constant.
13-Lactic acid (C3II6O3), which occurs in sour milk and foods such as sauerkraut,
is a weak monoprotic acid. The pH of a 0.10
M solution of lactic acid is 2.43.
What are the values of ka and oka for lactic acid?
14-Acrylic acid (HC3H3O2) is used in the manufacture of paints and plastics. The
pka of acrylic acidis 4.25.
(a) Calculate the pH and the concentrations of all species tH3Ol, C:H:O:-,
HC3H3O2, and
OH-in 0.150 M acrylic acid.
(b) Calculate the percent dissociation in 0.0500 M acrylic acid.
\\,,
st year -
Enviroffnental Chemistry
Dr. Manal A. Tooma
Photochemistry
Photophysics and photochemistry both deal with the impact of energy in the form
),
otphotons on materials. Photochemistry focuses on the chemistry involved as a material
idimpacted by photons while photophysics deals with physical changes that result from
theimpact
of
photons.
we will
focus on some
tophotophysics and photochemistry followed
by
principleswill be related to
of the basic principles related
general examples. Finally these
ways there is a great similarity
between amaterial's
the material is small
ormacromolecular.
to transferthe
effects ofradiati
Iex.
Ph
magnetic
involves
ofPlanck's
thewavel
icles;
ejection of el
particle
behavior while the
properties.
The Grotthus-Draper law statE
when
a
bal reactions onlyoccur
.t --
photon of light is absorbed.
for the First Law ofphoto_
chemistry, that is, only light that is
absorbed can have aphotophysical/photochemical
effect.
'.tirTffiasis
We can write this as follows.
M + lig61-214*
O.r,
,.*
- Environmental Chemistry
Dr. Manal A. Tooma
where M* is M after it has taken on some light energy that it has acquired energy duringa
photochemical reaction. The asterisk is used to show that M is now in an excited state.
Optical transmittance, T, is a measure of how much light that enters a sample is absorbed.
T:
I/IO
If no light is absorbed then I = Io. Low transmittance values indicate that lots ofthe lisht
has been absorbed.
Most spectrophotometers give their results in optical absorbency, A, or opticaldensify
(same) which is defined
as:
A=
= log
Transmittance T =
Percent Transmittance %T =
(1/T): -log T
I
Io
L
ylggoTo
Io
Absorbance A = -logT
,I
= - loq'Io
,In
= loq-I---:Beer's la
;
is often
bH
absorbaqce of chromophoies, in
rhores where k
Beer's law
€.%
i_s
a constant.
A:kc
rdlationship between
I iJ
:
and
-,
known
absorbance
The optical path,
ln
looking at the color iri
at the deep end
because the optical path is
Iaw wherek' is another
empirical constant.
A
=k'l
To the eye some colors appear similar but may differ
in intensity when c and I arethe
same. These solutions have a larger molar
absorption coefficient, e, meaning theyadsorb
more. The larger the adsorption coefficient the
more the material adsorbs.
The Beer-Lambert law combines the two laws giving
-2-
First year - Environmental Chemistry
Dr. Manal A. Tooma
A=elc
The proportionality constant in the Lambert's law is e.
The extinction coefficients of chromophores vary widely from less than 100 l/mol.cm,for a
so-called forbidden transition, to greater than 105 l/mol.cm for fully allowedtransitions.
We can redefine the elements of the Beer-Lambert law where I is the samplethickness and
c is the molar concentration of chromophores. This can be rearranged todetermine the
penetration depth of light into a material. Here I is defined as thepath length where 90Yo of
the light of a particular wavelength is
l (in Prm):
es
I giving.
104 e c
',ipsudaru"i4a*p,i
In absorption, the
light and the
transmitted light
angle,
generally 90o,
the
resulting po
energy and is sai
ii
it
of cold body radiation. Otder TV
principle
tne.emlssron ot light occurs when
i
Iorescence and fluorescence! In a
accel
gunisitting behindJhe screen. In
the el
the
cool.
sets,
isht
when hit
off light.
The same
of the screen
is coated with tho
green,
and blue. The kinetic
r andre-emitted as
visible light to be seen by us.
Jluorescence involves the morecurar
uuroiiffiffi
photon that triggers theemission of a
photon oflonger wavelength (less energy). The
energy differenceends up asrotational,
vibrational, or heat energy loses.
Here excitation is described as
So
Where
So is
*
hv"* a Sland emission asSl a hv". + go
the ground state, S1 is the first excited state.
Firrt ye* - Enuironmental Chemistry
Dr. Manal A. Tooma
The excited state molecule can relax by a number of different, generallycompeting
pathways. One of these pathways is conversion to a triplet state that cansubsequently relax
through phosphorescence or some secondary non-radiative step.
Relaxation ofthe excited state can also occur tlrough fluorescence quenching.
Molecular oxygen is a particularly effrcient quenching molecule because of its
unusualtriplet ground state.
Watch hands that can be "seen in the dark" allow us to read the time in the dark.
These watch hands typically are
paintedyi
paint. Like fluorescence,
phosphorescence is the emisgi
hit by electromagnetic
radiation.Unlike
afterglow for some
time after the
liCht
emission is 10:
The energy
Jab
1935; a
given in
tn
figure
represents the electrohic ground
J
qr
!
n
u
!
fl
H
M
4-
h o r e's c en c e
Inlerttal conversion
Itt2s
ll;?st
-
r,.
,rrr,
,.*
- Environmental Chemistry
Dr. Manal A. Tooma
Classiftcafion of Analytical Methods
Qualitative instrumental analysis is that measured property indicates
pres ence
of
analyte
in matrix
Quantitative instrumental analysis is that magnitude of measured property is proportional
to concentration of analyte in matrix
Species of interest:
All constituents including
analyte and Matrix-analyte (concomitants)
often need pretreatment - chemical extraction, distillation, separation, precipitation
(A) Classical:
Qualitative - identifi cation
Quantitative - mass
(B)
Qualitative
physical
property (
qualitati
Instrument
to the
concentration o
than
indicator to locate
Spectro chemical tech niq ues
Is one of the optical methods widery used for
anarysis, i,
ment of the wavelength frequency or energy
of electromagnetic
ij*:g
the measure-
radiation (EMR) that
either absorbed or emitted by the sample-the
EMR spectrometry is divided into severar
energy regions as shown :_
-5-
Dr. Manal A. Tooma
c
IArnt
I
I
10t"
-: r,
lo
"r3
!-
1-0j rloa
4
Absorption Spectrum:
D'-.-
lo't
-.
a
x_0'
vlSbte Llght
obs!€Es*F
I
lor
lO'
__ W'avclangth (m eters)
velength) l,
called absorption
e can only
In atoms
Visible is
Wand
In M'
with
many
AE:
Fs:
AE el
For each
Absorption spectra affected
(l)
-"t*
a,' i ii.r
Number of atoms in moleculeEEier
(2) Solvent molecules blurred features
-6-
F-_.-
E
l-.,1
r
ij R 1 :tt'
First year - Environmental Chemistry
U
V U ltravio I et
Dr. Manal A.
Tooma
.
(',r
so e ctro metm
I
{"
when two atoms share to make new bond; there are electron in both atom
will
participate to form this bond the electrons occupied new orbital called molecular bonding
orbital with low energy and antibonding orbital with high energy and the electron how
don't share called nonbonding.
when the molecular absorb energy in
o-'rl*, rl-lt'
uv
range; electronic transfer
will happened (o-o*,
,n- o', n- ,{, fi-o'1(200-380nm)
-F
- Itis
the
length
).
Wav
Vf/here:c is the speed
-
of
Energy is related to wavelength and frequency by the
fo'owing formulas:
E
- nt,- Il
),"
Were: h : Planck's constant, 6.6 x l0-3foules_sec
-
Note that energy is directly proportional to frequency
and inversely proportional to
wavelength.
First year - Environmental Chemistry
'
Dr. Manal A. Tooma
IR is a device that measures the intensity of the electromagnetic spectrum which is
between the visible and Microwave (400-4000 cm-';and expressed either
transmittance or
A:
absorption:
log (1/ T)
An increase in wavenumber corresponds to an increase in energy, when the molecule
absorb the energy
will
start to vibrate; stretching (changing the length ofthe bond or
bending changing the angle ofthe bond)
.
we can calculate the
L
t:4
'IK
K:5*1
l05dyne/cm
for
C:
=3
x
l01o
i.1r...,
&
*ra
-ts
w.*
asymrlrtricsl rtrE{c-lrhg
2925
cm-l
r!'mmcticsl rarcaihilg
2850 (Jn-l
rcisioridg, or bcndiog
in-tl.'lc
l465gtrl
+
q,P
,I
/
".,.
ourof-phna
crn-l
twirrioS, o. bcndhg
l35Gl150
io-plaoe
?20erri
rocting, or-bcDding
wrggi||& or bending out-of_ptr
Stretching ard bending vibntional modes for a CH2 group.
-8-
l35gllJ6gtrl
r.-f=
.]/
t:"t .#
;',#
First year - Envirorunental Chemistry
Dr. Manal A. Tooma
con
sample.
element.
flame into
absorbed by
ement
interest is
makes
the method relati
ofenergy
at the characteristic
the concentration
of the element in the samold
ost atomic absorption
instruments are also equipped for
Atomic absorption spectrophotometer consisting of a light source emitting the line
spectrum of an element, a device for vaporising the sample, a means of isolating an
absorption line and a photoelectric detector with its associated electronic amplifuing and
measuring equipment
.
-9-
of
1.. ,.,
-.
First year - Environmental Chemistry
Dr. Manal A. Tooma
deute riurtr
liBht sou'te
msn*trrqmgtor
trESOrUrnte
d€lcctor
hnlf rilvcrod mirrtrr
cpilrry shsmbcr
ofitrliftcr
fucl
ncbulizcr
oxidant
dfflxrrar
S.'
odo
on
them
the
relative
strongly
mobile
the
morQ
the
stationary
Tvne ofchrometosraph
Gas
Jiquid
moblle phose
g3.s
stadonarv ohase
liquid
Gas -solid
gas
solid
Liquid-liquid
liquid
liquid
Gel-permeation
-10-
liquid
Gel
First year - Environmental Chemistry
Dr. Manal A. Tooma
W(tp):isthetimerequiredforthemobilephasetoremovethecomponent
from the stationary phase.
series ofpeaks as shown
Tx: can be computed from such plot as the time between sample injection and recording
the peak. The time corresponds to the apffofsuch peak represents the tx ofthat component
Concentration: they are area under the
onal to theconcentration, hence the
amount of sample can be
Ifthe peak is
Efftciency
The
(n) in the
Theoreti
the
The
by the
Higher effrciency
d be as short
as possible to avoid a
Two equations are used to
the half peak width
n= 76(3
w
Or
N = 5.55
peak width (w) and
f-L)'
\wrlz/
(p) must be accurately determined .
(w): distance along the base line between extrapolated
tangents.
-
rt-
First year - Environmental Chemistry
Dr. Manal A. Tooma
wlthe width of the peak at half of the peak height
pH-meter
A pH Meter is a device used for potentiometrically (voltmeter which is calibrated to
convert voltage to pH units) measuring the pH, which is either the concentration
or
the activity ofhydrogen ions, ofan aqueous solution. It usually has glass
a
electrode plus
a calomel reference electrode, or a combination electrode. pH
meters are usually used to
measure the pH of liquids, though speci
ggggglimes used to measure the pH of
semi-solid substances.
The electrode
has a silver-
inside
a
glass electrode,
contained
thin
(typically
electrode
oride.
The
,soit
contains
're
testing
electrode
the di
potassium
When
hydrogen ions
the metal ions insidd
the unknown solution.
to how a glass electrode works.
the
some
of
electrode into
, and it's the key
on the inside surface ofthe
glass electrode from the potassium
chloride solution. The two solutions on either
side of
the glass have different acidity, so a different
amount of ion-swapping takes place on
the
two sides ofthe glass. This creates a different
degree ofhydrogen_ion activity on the
two
surfaces of the glass, which means a different
amount of electricar charge builds up
on
them' This charge difference means a tiny
voltage (sometimes cared a potential
difference, typically a few tens or hundreds
of millivolts) appears between the two sides
-t2-
of
First year - Environrnental Chemistry
Dr. Manal A. Tooma
the glass, which produces a difference in voltage between the silver electrode (5) and
thereference electrode (8) that shows up as a measurement on the meter.
(1)Solution tested;
metal salts,
inside which
silver/silver
glass. (7) H
glass. (8)
this
5srmed
in.
thc test
sllutiorffi
of the
potassium chloride qolution inter
i,
rence in volhge betlveen the two
rss and
reading. (9) Rdference iilechode acts as
or refe:
of the
3
[,-_
$:z
Glass pH
relation:
Constant k
l) pH meter is adj
2) pH ofunknown
--t5-
iffinrs!