GENERAL CHEMISTRY For First Years of Faculties of Science, Medicine and Pharmacy Part 1
GENERAL CHEMISTRY For First Years of Faculties of Science, Medicine and Pharmacy Part 1
GENERAL CHEMISTRY For First Years of Faculties of Science, Medicine and Pharmacy Part 1
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Prepared by:
Prof. Dr. Talaat I. El-Emary
Prof. of Organic Chemistry
Chemistry Deaprtment
Faculty of Science
Assiut University
Assiut, Egypt
٢
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٣
CHAPTER 1
Matter is anything has mass and occupies space. Mass is a measure of the
quantity of matter in a sample of any material.
٤
The relationship between matter and energy is given by Albert Einstein's
equation:
E = mc2
Were, m is the mass of matter, and c is the velocity of light.
States of matter
Matter is conveniently classified into three states, solid state, liquid state and
gaseous state.
In the solid state, the substances are rigid and have definite shapes, and the
volumes do not vary much with changes in temperature and pressures.
In the liquid state: the substances are not rigid and have no definite shapes, it
takes the shape of the container in which it is placed in, and the volumes vary
with changes in temperature and slightly compressed with pressures.
In the gaseous state : the substances are not rigid and have no definite shapes
and volume, it fills and takes the shape of the container in which it is placed in,
and the volumes vary much with changes in temperature and compressed with
pressures.
Both chemical and physical changes are always accompanied by either the
absorption or liberation of energy. Energy is required to melt ice, and energy is
required to boil water. Conversely, the condensation of steam to form liquid water
always liberates energy, as does the freezing of liquid water to form ice.
٥
M.P. = 0 OC B.P = 10 0 OC
MEASURMENTS IN CHEMISTRY
The International System of Units (SI) is based on the seven fundamental units
listed in table 1-1 & 1-2 & 1-3 and all other units of measurements are derived
from these.
Table 1-1
٦
The distances and masses in metric system illustrate the use of some common
prefixes and the relationships among them are listed in table 1-2 &1-3.
Table 1-3
٧
Table 1-4
SI UNITS OF MASS
Table 1-5 illustrate conversion factors relating length, volume and mass units
Table 1-5
Volume
1 mL = 1 cm-3 = 1 gal = 4 qt = 8 pt 1 liter = 1.057 qt
10-3liter 1 ft3 = 28.32 liter
1 m3 = 106 cm 1 qt = 57.75 in3
= 103 liter
Mass
1 kg = 103 g 1b = 16 oz 1b = 453.6 g
1 mg = 10-3 g 1 short = 2000 1b 1g = 0.03527 oz
1 metric tone = 103 kg ton 1 = 1.102 short
metric ton
tone
Dimensional analysis
Unit factor method
Unit factor may be constructed from any two terms that describe the same amount
of whatever we my consider. For example, 1 foot is equal to exactly 12 inches, by
definition, and we may write an equation to describe this equality:
٨
1 ft = 12 in
1 ft 1 2in 12
1
1 ft 1 ft or 1 ft
1 yd 1 yd 1 mile 4 qt 2000 1b
, , , ,
3 ft 36 in 5280 ft 1 gal 1 ton
Conversion from decimal to exponential
In scientific (exponential) notation, we place one nonzero digit to the left of the
decimal as in the following examples
4 ,3 00 ,00 0 . = 4.3 x 10 6
6 is places to the left, therefore, exponent of 10 is 6
0 .00 0 34 8 = 3.4 8 x 10 -4
4 is places to the right, therefore, exponent of 10 is -4
Solution:
14 37 m m = 1 4 3 7 x 10 -3 m
٩
14 37 m m = 1 4 3 7 x 10 -3 m
x 10 -3
1 cm = 10-2 m
1dm = 10-1 m
1 cm 10-2 m
-2
and
10 m 1 cm
If we multiply our given 172 cm by the second factor, the units cm will cancel
and we will have converted centimeters to meters.
10-2 m
172 cm x 1.72 m
1 cm
Similarly, we use the second equation to convert meters to
decimeters.
1 dm
1.72 m x -1 17.2 dm
10 m
We can also " string together" conversion factors and obtain the same net
result.
١٠
10 -2 m 1 dm
172 cm x x -1 17.2 dm
1 cm 1 0 m
to m eter
to decimeter
Example 1.3: Calculate the number of cubic centimeter in 0.225
dm3.
Solution:
0.225 dm3 = ? cm3
1 dm = 10-1 m
1 cm = 10-2 m
(1 dm)3 = (10-1 m)3
1 dm3 = 10-3 m3
(1 cm)3 = (10-2 m)3
1 cm3 = 10-6 m3
3 10-3 m 3 1 cm 3
0.225 dm x 3
x -6 3 255 cm 3
1 dm 1 0 m
Example 1.4: How many millimeter are in 1.39 x 104 meters?
Solution:
1000 mm
? mm 1.39 x 104 m x 1.39 x 107 mm
1m
Example 1.5: A sample of gold weigh 0.234 mg. what is its weight
in g? in cg?
١١
Solution:
1g
? g 0.234 mg x 2.34 x 10 4 g
1000 mg
1 cg
? cg 0.234 mg x 0.0234 cg or 2.34 x 10 - 2 cg
10 mg
Solution:
1 dm 2
? dm 2 215 cm 2 x( ) 2.15 dm 2
10 cm
Example 1.7: How many cubic centimeters are there in 8.34 x 105
cubic decimeters?
Solution:
10 cm 2
? 1 cm 2 8.34 x 105 dm 2 x ( ) 8.34 x 108 cm 2
1 dm
Example 1.8: A common unit of energy is the erg. Convert 3.74 x
10-2 erg to the SI units of energy, joules and kilojoules. One erg is
exactly 1 x 10-7 joule.
Solution:
-2 1 x 10-7 J
3.74 x 10 erg x 3.74 x 10-9 J
1 erg
-9 1 x 10-3 kj
3.74 x 10 J x 3.74 x 10-12 kj
J
Example 1.9: Express 1.0 mL in gallons.
١٢
1L 1.06 qt 1 gal
? gal 1.0 mL x x x 2.7 x 10-4 gal
1000 mL 1L 4 qt
mass M
density or D
volume V
Densities may be used to distinguish between two substances or to assist in
identifying a particular substance. They are usually expressed as g/cm3 or g/mL
for liquids and solids, and as g / L for gases. These units can also be expressed as
g cm-3, g mL-1, respectively.
Density is constant for a given substance at a given temperature. It is also
an intensive property, that is, a property that does not depend on the amount of
material examined. Both mass and volume are extensive properties, in that they
depend on the amount of material in the sample. Obviously, the volume occupied
by two kilogram of lead is twice that of one kilogram of lead, but both samples
have the same density. Table 1- 5 illustrate the densities of some common
substances.
١٣
Example 1-7: A 47.3 mL sample of liquid weighs 53.74 g. What is
its density?
Solution:
M 53.74 g
D 1.14 g/mL
V 47.3 mL
Example 1-8: If 100 g of the liquid described in example 1-7 is
needed for a chemical reaction, what volume of liquid would you
use?
Solution:
M M 100 g
D , so V 87.7 mL
V D 1.14 g/mL
The specific gravity of a substance is the ratio of its density to the density of
water.
DSubstance
Sp. Gr.
D Water
The density of water is 1.000 g/ml at 3.980C, the temperature at which the density
of water is greatest. However, variation in the density of water with changes in
temperature are small enough that we may use 1.00 g/mL up to 250C without
introducing significant errors into our calculations.
Specific gravities are dimensionless numbers, as the following example
demonstrates.
Example 1-9: The density of table salt is 2.16 g/mL at 200C. What
is its specific gravity?
DSalt 2.16 g/ mL
Sp. Gr. 2.16
D Water 1.00 g/ mL
١٤
Example 1-10: Battery acid is 40.0% sulfuric acid H2SO4, and
60% water by weight. Its specific gravity is 1.31. Calculate the
weight of pure sulfuric acid, H2SO4, in 100 ml of battery acid.
Solution:
We may write
Density = 1.31 g /mL
The solution is 40.0% H2SO4 and 60% H2O by weight. From this information we
may construct the desired unit factor.
40.0 g H 2SO 4
100 g soln.
We first used the density as a unit factor to convert the given volume of solution
to mass of solution, and then used the percentage by weight to convert the mass of
solution to mass of acid.
١٥
1.0 K 1.0 0 C
K ( x 0 C 273.150 C) OR 0
C (xK - 273.15 K)
1.00 C 1.0 K
W ate r bolis
(at norm al 2 12 0 F 1 00 0 C 3 73 0K
atm o spheric
pr essu re )
1 80 0
1 00 0
1 00 0
W at er
f r e ez e s 3 20 F 00 C 2 73 .15 0 K
K elvin
F a hr enhe it C e lsius
(c ent ig ra de)
1.80 F 1.00 C
and
1.0 0 C 1.80 F
0 1.80 F 0
? F( x C X 0
) 32 0 F
1.0 C
01.0 0 C
? C 0
( x 0 F - 320 F )
1.8 F
Example : When the temperature reaches "100 0F) in the shade," it's hot. What is
this temperature on the Celsius scale?
١٦
1.00 C
? 0C 0
( 1000 F - 320 F )
1.8 F
0 1.0 0 C
? C 0
( 1000 F - 32 0 F ) 380 C
1.8 F
Example : When the absolute temperature is " 400 K). What is Fahrenheit?
0 1.00 C
? C (400 K - 273 K) 1270 C
1.0 K
1.80 F
? C 127 C X 0 320 F 2610 F
0 0
1.0 C
Another expressions for the relationships among the three scales
of temperature.
5
t ( 0 C) [t ( o F) 32]
9
t (k) t ( o C)
273.15
0
F 32 ( 0 C) (1.8)
SIGNIFICANT FIGURES, EXACT, PRECISE AND
ACCURATE
Significant figures are digits believed to be correct by the person who makes a
measurements. We assume that the person is competent to use the measuring
device. Suppose one measures a distance with a meter stick and reports the
distance as 343.5 mm. what does this number mean?. In this person's judgment,
the distance is greater than 343.3 mm but less than 343.7 mm, and the best
estimate is 343.5 mm. The 343.5 mm contains four significant figures-the last
digit, 5, represents a best estimate and therefore doubtful, but it is considered to
be a significant figure. In reporting numbers obtained from measurements, we
report one estimated digit, and no more. Since the person making the
measurement is not certain that the 5 is correct, clearly it would be silly,
meaningless, and wrong to report the distance as 343.53 mm.
١٧
Strictly speaking, the distance should be reported as 343.5 ± 0.2 mm
because it is difficult to estimate distances less than 0.2 mm on an ordinary meter
stick.
The term exact is used for quantities that have an infinite number of significant
figures. For example, by definition there are exactly 1,000 cubic centimeters in 1
liter. We could write that there are 1,000.00000000….. cubic centimeters in
1.00000000…. liter, with no limit to the number of zeros. For example, the inch is
defined as exactly 254 x 10-2 meter, or 254 centimeters. Exact quantities either are
defined arbitrarily or result from counting objects one by one; no measured
quantities, other than by simple counting, are exact.
Exact = دﻗﯿﻖ، ﻣﻀﺒﻮط
The term precise refers to how reproducible measurements of the same
quantity are. Suppose that two calibrated 250 cm3 beakers are each tested three
times to determine their volumes. Beaker A is found to have volumes of 248, 249,
and 247 cm3. The set of values for beaker A is more precise, varying only ± 1
from the average value of 248 cm3.
ﻣﺪﻗﻖ،ﺻﺤﯿﺢ
The error is a positive quantity whether the measured value is higher or lower
than the true (or most probable) value. The percent error is
error
% error x 100
true (or most probable) value
2 cm 3
% error x 100 0.8 %
250 cm 3
١٨
the measurement is. Now, notice what happens to these zeros when we rewrite the
number in scientific notation.
We see that the zeros disappear; their function was just to locate the decimal point
and they are really not measured digits. A rule that we can apply, therefore, is that
zeros that lie to the left of the first nonzero digit are not counted as significant
figures.
How about the zero in the measured value 12.30 m? This zero is counted
as a significant figures because it would not have been written unless the digit had
been estimated to be a zero. Also note that this zero does not disappear when the
number is written in scientific notation.
This leads to another general statement: in a measured quantity, a zero that lies to
the right of the decimal point and also to the right of the first nonzero digit always
counts as a significant figures. If we put all this together, in the measurement
below only the nonzero digit plus the zeros above the line is count as significant
figures
0.0054070 m
And the measurement therefore contain five significant figures. Below we see
three different ways of expressing a length of 1200 m, each expressing a different
number of significant figures.
In almost all cases, the numbers that we obtain from measurement are used to
calculate other quantities, and we must exercise care to report the calculated result
in a way that neither overstates nor understates the amount of confidence we have
in it. This mean that we must be careful to report the computed value with the
proper number of significant figures.
To see problems can arise, suppose we wished to calculate the area of a
rectangular carpet whose sides have been measured using two different rulers, to
be 6.2 m and 7.00 m long. We know the area is the product of this two numbers.
notice that this error in the measured length would cause a change in the second
digit of the answer (it changes from a 3 to a 4). An uncertainty in the second digit
of the length causes an uncertainty in the second digit in the answer.
The length of the other dimension, 7.00 m, has an implied uncertainty of
about ± 0.01 m. if the 7.00 m were in error by this amount, instead of the other
measurement, how much would this influence the answer? As before, let's
recalculate the area, this time using 7.01 m.
An error in the third digit of 7.00 m causes a change in the third digit of the
answer, so if only the 7.00 m were in error, the answer would have its uncertainty
in the third digit.
For multiplication or division, the product or quotient should not have more
significant figures than are present in the least precise factor in the calculation.
As an illustration, consider the calculation below, in which we will assume that all
the numbers are measured values. A calculator gives the answer shown.
21.95
1.347452425
3.62 x 4.5
The least precise factor (i.e., the one with the fewest number of significant
figures) is the value 4.5, which contain two significant figures. The answer,
therefore, should contain no more than two significant figures and should be
rounded* to 1.3.
For addition and subtraction, the procedure used to determine the number
of significant figures in an answer is different. Here, the number we write as the
result of a calculation is determined by the figures with the largest amount of
uncertainty. For example, consider the following sum:
4.371 m
+ 302.5 m
306.871 m (before rounding)
٢٠
The first quantity, 4.371 m, has an uncertainty of ± 0.001 m; the second an
uncertainty of ± 0.1 m. When we add these quantity, we expect the answer to be
uncertainty by at least ± 0.1 m, so we must round the answer in this case to the
nearest tenth; it should be reported as 306.9 m. The rule for addition and
subtraction, therefore, can be stated as follows.
when wish to round off a number at a certain point, we simply drop the
digit that follows if the first of them is less than 5. Thus, 6.2317 rounds to
6.23, if we wish only two decimal places. If the first digit after the point of
round off is larger than 5, or if it is a 5 followed by other nonzero digits,
then we add 1 to the preceding digit. Thus, 6.236 and 6.2351 both round to
6.24. Finally , when the digit after the point of round off is a 5 and no
other digits follow the 5, then we drop the 5 if the preceding digit is even
and we add 1 if it is odd. Thus, 8.165 rounds to 8.16, but 8.175 rounds to
8.18.
A little practice will enable you to determine with confidence the number
of significant figures in different quantities. The following examples illustrate
some typical cases:
1.062 4
751 3
0.006110 centimeters 4
1.2 x 108 stars 2
$ 683,462.02 8
7,685,000 4
7.6850 x 106 people 5
٢١
physical means. For example, the mixture of salt and water can be separated by
evaporating the water and leaving the solid salt behind. A mixture of sand and salt
can be separated by dissolving the salt in water, filtering out the sand, and then
evaporating the water to obtain the solid salt. Very fine iron powder can be mixed
with powdered sulfur to give a mixture. The iron may be removed by magnet, or
the sulfur may be dissolved in carbon disulfide, which does not dissolve iron.
M A T TE R
M ix tu res Pu re s ub st a n ces
P y sical ch a n g es
(va ria b le co m p o sitio n ) (co n sistan t co m po s iti o n )
H o m o g en eo u s H etrero g en eo u s C om poun ds
m ix tu res ch em ica l ch a n g es
m ix tu res El em en ts
O il
W ater
O il an d water f orm
a h eteogen eou s
m ixtu re
٢٢
The periodic table of the elements
Compounds are pure substances consisting of two or more different
elements in a fixed ratio. All compounds can be decomposed or resolved into their
constituent element by chemical changes. The physical and chemical properties of
a compound are different from the properties of its constituent elemental. A
chemical reaction between iron and sulfur gives iron sulfide.
Water can be decomposed into hydrogen and oxygen by electrical energy
Heating
A chemical reaction
٢٣
Dil. H2SO4
Valve
Oxygen
gas Hydrogen
gas
Anode
(oxidation)
Platinum
electrode
Battery Battery
Electrolysis of water
٢٤
CHAPTER 2
H H O O F F I
I
H2 O2 F2
H y d ro ge n I
O x y ge n Flou rine Io dine
m o le cu le m olec ule m olecu le m olec ule
٢٥
P
P P
P ho sp h ro us
m o le cu le
S S
S
SS S
S
S S
S
Su lf u r
m o lec u le
H
c H
O
H C O H
H O
H 2O C O2 C H4
( W at er ) (C arb on d ioxid e) (M eth ane )
H H
H
H C C O
H H
C2H6O H
(E th yl a lc oh o l)
٢٦
Each of the elements has been assigned a chemical symbol that we can think
of as a shorthand way of representing the element. The symbol consist of one or
two letters that usually bear a resemblance to the English name of the element.
For instance, carbon = C, Chromium = Cr, Chlorine = Cl, Calcium = Ca, and Zinc
= Zinc. Notice that the first letter is capitalized, but if there is a second letter, it is
not. Some of the elements have symbols have Latin names. Some examples are
Potassium (L Kaluim) = K, Sodium ( Natrium) = Na, Silver (Argentium) = Ag,
mercury (hydrargyrum) = Hg, and copper (cuprum) = Cu.(cf. the periodic table of
elements).
A chemical compound is represented symbolically by its chemical formula.
For example, water is represented by H2O, carbon dioxide by CO2, methane
(natural gas) by CH4, and aspirin by C9H8O4. The formula H2O, for instance,
describes a substance containing two hydrogen atoms for every oxygen atom.
Similarly, the compound CH4 contain one atom of carbon for every four atoms of
hydrogen.
Often, two or more atoms are able to join tightly together so that they behave
as a single particle called a molecule. If the atoms are of different elements, as in
water (H2O) or methane (CH4), it is a molecule of a compound. If the atoms of
the same element, it is a molecule of an element. Some common and important
elements that occur in nature as molecules composed of two atoms are, hydrogen,
H2; oxygen, O2, nitrogen, N2; fluorine, F2; chlorine, Cl2; bromine, Br2; and iodine,
I2 .
A chemical equation is written to show the chemical changes that occur
during a chemical reaction. In a sense, it's a " before and after" description of the
reaction. For example, the equation
Zn + S Zn S
reactan ts p rod u ct
describes the reaction between Zinc (Zn) and Sulfur (S) to produce zinc
sulfide (ZnS), a substance used on the inner surface of TV screens. The
substances on the left of the arrow are called reactants and are the chemical
present before the reaction takes place. Those on the right of the arrow are called
the products and are the substance present after the reaction is over. (In the
reaction above, there is only one product). The arrow is read as "react to yield" or
simply "yield". Thus, the equation above can be read as "zinc plus sulfur react to
yield zinc sulfide" or "zinc plus sulfur yield zinc sulfide", or "zinc recta with
sulfur to yield zinc sulfide."
Sometimes it is necessary to indicate whether the reactants and the products
are solids, liquids, gases, or dissolved in solvent such as water. This is done by
placing the letters s = solid, l = liquid, g = gas, or aq = aqueous (water) solution in
parentheses following the formula of substances in the equation. For instance, the
equation
٢٧
Many of the equations that we write contain numbers, called coefficients,
preceding the chemical formulas. An example is the reaction of hydrogen (H2)
with oxygen (O2) to form water.
2H 2 + O 2 2 H 2O
Atomic masses
The law of definite proportions is easy to explain. Let's imagine that two
elements, say A and B , form a compound in which each molecule of the
substance is composed of one atom of A and one atom of B. Let's also suppose
that an atom of A is twice as heavy as atom of B, so if an atom of B were arbitrary
assigned a mass of 1 unit, then the mass of an atom A would be 2 units. Below we
see how the masses of A and B vary for various numbers of molecules.
Now, notice that no matter how may molecules we have, each with the same 1- to
1- ratio by atoms, the ratio by mass is the same. This is exactly what the law of
٢٨
definite proportions says: In any sample of compound, regardless of sise, the
elements are always present in the same proportion (ratio) by mass.
Atomic masses
Because atoms are too small to be seen and too small to have their masses
measured individually on laboratory balance in a unit of grams, an atomic mass
scale was devised in which mass is measured in atomic mass units (the SI
symbol is u).
The atomic mass unit is defined as 1/12th of the mass of atom of carbon-
12 isotope. The atomic masses of elements are given along with the symbols of
the elements in the periodic table. The numbers in these table are relative average
atomic masses, expressed in atomic mass units.
The law of multiple proportions can be expressed as follows: " Suppose we have
samples of two compounds formed by the same two elements. If the masses of
one elements are the same in the two samples, then the masses of the other
elements are in a ratio of small whole numbers. This can be understand by the
following example. We know that carbon forms two different compounds with
oxygen called carbon monoxide and carbon dioxide. In 2.33 g of carbon
monoxide, we find 1.33 g of oxygen combined with 1.00 g of carbon. Notice that
the masses of oxygen that are combined with the same mass of carbon (1.00 g) are
in a ratio of 2 to 1 (a ratio of small whole numbers).
2.66 g 2
1.33 g 1
The result is consistent with the atomic theory if a molecule of carbon monoxide
(CO) contain 1 atom of C and 1 atom of O, and a molecule of carbon dioxide
(CO2) contain 1 atom of C and 2 atoms of O, so the ratio of the oxygen masses
has to be 2 to 1.
We really can't work with individual atoms because they are so tiny., so
we must increase the sizes of the samples to the point where we can see them and
manipulate them, but we do this in a way that maintains the proper ratio of atoms.
One way we could enlarge amounts in a chemical reaction would be to
work with dozens of atoms instead of individual atoms.
٢٩
1 at om C + 1 at om O 1 m olecu le C O
1 d ozen 1 d ozen O at om s 1 d ozen C O m olecu le
( 12 atom s C ) (12 at om s O ) ( 12 m olecu les CO )
1 dozen = 12 objects
1 mol = 6.022 x 1023 objects
The atomic weight of carbon to four significant figures is 12.01 atomic mass unit
(amu) or (u). How many atoms are in 12.01 g of carbon? Modern experimental
methods show that this number of atoms is 6.022 x 1023. This huge number is
called Avogadro's number, named in honor of Amadeo Avogadro, the brilliant
contemporary of Dalton. The weight of 6.022 x 1023 atoms of oxygen is 16.00 g;
the weight of 6.022 x 1023 molecule of carbon monoxide (CO), is 28.01 g; and the
weight of 6.022 x 1023 molecule of carbon dioxide (CO2), is 44.01 g. a mole or
(mol, for abbreviation) of a substance is the amount that contains 6.022 x 1023
units of substance.
6.02 x 10 23 atom Ca
1 m ol C a
1 atom Ca x x 40.1 g C a = 6.66 x 10 -2 3 g C a
6.02 x 10 2 3 atom s Ca 1 m ol Ca
Example: How many moles of calcium (Ca) are required to react with 2.5 mole
of chlorine (Cl) to produce the compound CaCl2 (calcium chloride)?
٣٠
2.5 m ol C l (?) m ol C a
We know from the formula that 1 atom of Ca combines with 2 atoms of Cl. Thus,
1 atom C a 2 atom s C l
Because moles combine in the same ratio as atoms
1 m ol C a 2 m ol C l
We can obtain the answer as follows:
1 m ol C a
2.5 m ol C l X 1.25 m ol C a
2 m ol C l
Example: How many grams of Ca must react with 41.5 g of Cl to produce CaCl2?
Solution:
1 mol C a 2 mol Cl
1 m ol C a = 41.1 g C a
Now we have all the information needed to solve the problem, but before we
actually go to the solution, let's diagram how it will work.
gra m s C l m oles C l m o le s C a
(la b u n it) (ch em ical u n it) g ra m s C a
(ch em ica l u n it) (lab u n it)
we follow the outline above, being careful to set up the conversion factors so that
units cancel correctly.
1 m ol C l
41.5 g C l X = 1.17 m ol C l
35.5 g C l
1 m ol C a
1.17 m ol C l X 0.585 m ol C a
2 m ol C l
٣١
40.1 g Ca
0.585 m ol Ca X = 23.5 g C a
1 m ol Ca
1 m ol Cl 1 m ol C a 40.1 g Ca
41.5 g C l X X X
35.5 g C l 23.5 g C a
2 m ol C l 1 m ol C a
to m oles of C l
to m oles of C a
to gram s of Ca
1 mol Si = 28.1 g Si
1 mol Si
28.1 g Si
Thus,
1 m ol S i
30.5 g S i X = 1.09 m ol S i
28.1 g S i
Example: How many moles of carbon atoms are needed to combine with 4.87
mol Cl to form the substance C2H6?
٣٢
4.87 m ol C l ? m ol C
We have conversion factors to choose from
2 m ol C 6 m ol C l
6 m ol C l , 2 m ol C
1 m ol C 3 m ol C l
3 m ol C l , 1 m ol C
To get rid of the units " mol Cl" we must use the one on the left.
2 m ol C
4.87 m ol C l X 1.62 m ol C
6 m ol C l
We need 1.62 mol C.
2.65 m ol C 2 C l 6 ? m ol C
And then use it to construct the necessary conversion factor.
2 m ol C
2.65 m ol C 2 C l 6 X 5.30 m ol C
1 m ol C 2 C l 6
The simplest way of obtaining the weight of one mol of a substance is merely to
add up the atomic masses of all the elements present in the compound. If the
substance is composed of molecules (for example, CO, H2O, or NH3), the sum of
the atomic masses is called the molecular mass (or molecular weight-the terms
are used interchangeably). Thus, the molecular mass of CO2 is obtained as
C 1 X 12.0 u = 12.0 u
2O 2 X 16.0 u = 32.0 u
CO2 total 44.0 u
٣٣
Similarly , the molecular mass of H2O = 18 u and that of NH3 = 17.0 u.
The weight of one mol of a substance is obtained simply by writing its molecular
mass followed by the units, grams. Thus,
One mole samples of several different compounds, contains the same number of
the formula units, although the number of atoms is different form sample to
sample. They differ because the number of atoms per formula unit is different for
each compound. Atoms or groups of atoms that have acquired an electrical
charge are called ions. Since solid NaCl is composed of Na+ and Cl+ ions, it is
said to be an ionic compound.
For ionic compounds, the sum of the atomic masses of the elements
present in one formula unit is known as the formula mass or formula weight.
For NaCl, this is 22.99 + 23.45 = 58.44.
Na 22.99 u
Cl 35.45 u
NaCl 58.44 u
One mole of NaCl ( 6.022 X 1023 formula units of NaCl) would contain 58.44 g
NaCl. Use of terms formula mass, of course, is not restricted to ionic compounds.
It can be applied to molecular substances, in which case the terms formula mass
and molecular mass mean the same thing.
Solution: we calculate the formula mass of Na2CO3 from the atomic masses of its
elements
2 Na 2 X 23.0 = 46.0u
1C 1 X 12.0 = 12.0 u
3O 3 X 16.0 = 48.0 u
Total 106.0 u
This can be used to make conversion factors relating grams and moles of Na2CO3,
which we need to answer the questions
٣٤
(a) to covert 0.250 mol Na2CO3 to grams, we set up the units to cancel.
106.0 g Na 2 C O 3
0.250 m ol N a 2 CO 3 X = 26.5 g N a 2 C O 3
1 m ol N a 2 CO 3
1 m ol Na 2 C O 3
132 g N a 2 C O 3 X = 1.25 m ol Na 2 C O 3
106.0 g N a 2 C O 3
Solution: the total mass of 1 mol of CHCl3 is obtained from the molecular mass
In general,
weight of part
% by weight X 100
weight of whole
Then,
weight carbon
%C X 100
weight CHCl3
12.01 g C
%C X 100 10.06 % C
119.37 g CHCl3
٣٥
1.008 g H
% H X 100 0.844 % H
119.37 g CHCl 3
106.35 g Cl
% Cl X 100 89.09 % Cl
119.37 g CHCl 3
Total % = 100.00
Solution:
Wt. of 2 Na = 2 X 23 = 46
Wt. of S = 32
Wt. of 4 O = 4 X 16 = 64
Molecular mass of Na2SO4 = 46 + 32 + 64 = 142
46
Percentage of Sodium (Na) X 100 32.39 %
142
32
Percentage of Sulfur (S) X 100 22.54 %
142
64
Percentage of Oxygen (O) X 100 45.08 %
142
Example: calculate the mass of iron in a 10.0 g sample of iron oxide, Fe2O3,
commonly refereed to as rust
Solution: In one mole of this compound, there are 2 mol Fe and 3 mol O.
Therefore,
٣٦
2 F e: 2 X 5 5.8 5 g = 11 1.7 g F e thi s m a ss o f F e
3 O : 3 X 16.00 g = 48.0 g O i s in
1 m o l F e 2 O 3 = 15 9.7 g F e 2 O 3 th is m ass of Fe 2 O 3
We can write
1 1 1 .7 g F e 1 5 9 .7 g F e 2 O 3
111.7 g F e
10.0 g F e 2 O 3 X 6.99 g F e
159.7 g F e 2 O 3
Chemical formulas
There are different kinds of chemical formulas, and each conveys certain kinds of
information. This include the elemental composition, the relative numbers of each
kind of atom present, the actual members of atoms of each kind in a molecule of
the substance, or structure of a molecule of the substance.
A formula that uses the smallest set of whole-number-subscripts to specify
the relative number of atoms of each element present in a formula unit is called
simplest formula. It is also called an empirical formula because it is normally
derived from the results of some experimental analysis. The formulas NaCl, H2O,
and CH2 are empirical formulas.
A formula that states the actual number of each kind of atom found in a
molecule is called a molecular formula. H2O is a molecular formula (as well as
an empirical formula) since a molecule of water contains 2 atoms of H and 1 atom
of O. The formula C2H4 is a molecular formula for a substance (ethylene)
containing 2 atoms of carbon and 4 atoms of hydrogen. The simplest formula
CH2 is not unique to C2H4, however. A substance whose empirical formula is
CH2 could have as a molecular formula CH2, C2H4, C3H8, and so on.
A third type of formula is a structural formula, for example,
H O
H C C O H
٣٧
In a structure formula the dashes between different atomic symbols represent the
"chemical bonds" that bind the atoms to each other in the molecule. A structural
formula gives us information about the way in which the atoms in a molecule are
linked together and allow us to write the molecular and empirical formulas. Thus,
for acetic shown above, we can write its molecular formula (C2H4O2) and its
empirical formula (CH2O).
Therefore,
1 m ol N
2.34 g N X = 0.167 m ol N
14.0 g N
1 m ol O
5.34 g O X = 0.334 m ol O
16.0 g O
We might write our formula N0.167O0.334. However, since the formula should have
meaning on a molecular level where whole numbers of atoms are combined, the
subscripts must be integers. If we divide each subscript by the smallest one we
obtain
1 m ol O
5.34 g O X = 0.334 m ol O
16.0 g O
N 0.167 O 0.334 NO 2
0.167 0.167
٣٨
Problem: what is the empirical formula of a compound composed of 43.7% P and
56.3% O by weight?
1 mol P
43.7 g P X 1.41 mol P
31.0 g P
1 mol O
56.3 g O X 3.52 mol O
16.0 g O
The formula is
Whole numbers can be obtained by doubling each of the subscripts. Thus, the
empirical formula is P2O5.
Problem: 1.025-g sample of a compound that contains carbon and hydrogen was
burned in oxygen to give carbon dioxide and water vapor as products. These
products were trapped separately and weighed. It was found that 3.007 g of CO2
and 1.845 g H2O were formed in this reaction. What is the empirical formula of
the compound?
12.01 g C
3.007 g CO 2 X 0.8206 g C
44.01 g CO 2
Similarly, in 1 mol of H2O (18.02 g), there are 2 mol (2.016 g) of H, so the mass
of H in the 1.845 g of H2O is
٣٩
2.016 g H
1.845 g H 2 O X 0.2064 g H
18.02 g H 2 O
Now we have the masses of C and H in the sample, so we can calculate the
number of moles of each,
and
1 mol C
0.8206 g C X 0.06833 mol C
12.01 g C
1 mol H
0.2064 g H X 0.2048 mol H
1.008 g H
C 0.06833 H 0.2048
0.06833 0.06833
Solution: First we calculate the mass of carbon and hydrogen in the CO2 and
H2O.
12.01 g C
0.1910 g CO 2 X 0.05212 g C
44.01 g CO 2
٤٠
2.016 g H
0.1172 g H 2 O X 0.0131 g H
18.02 g H 2 O
When we add the masses of C and H, we get a total of 0.06523 g, so the mass of
oxygen in the sample must have been
Now we convert the masses of C, H, and O to moles of each of the element. Thus,
for carbon we have
1 mol C
0.05212 g C X 0.04340 mol C
12.01 g C
Similar calculation for other two elements give 0.0130 mol H and 0.00218 mol O.
This empirical formula is therefore
Molecular formulas
It is possible for more than one compound to have the same empirical formula.
For example, the molecules C2H4, C3H6, and C5H10 all have a 1-to-2 ratio of
carbon to hydrogen atoms and the empirical formula CH2
٤١
C4H8 56.0 = 4 X 14.0
CnH2n n X 14.0
Ionic compounds don't have molecular formulas because they don't contain
molecules.
Solution:
Molecular mass (Formula mass (FM.))
n
Empirical mass
The empirical mass of NO2 is 46.0. the number of times the empirical
formula, NO2, occurs in the compound is
92.0
2
46.0
The molecular formula is the (NO2)2 = N2O4 (dinitrogen tetroxide). N2O4 is the
preferred answer because (NO2)2 implies a knowledge of the structure of the
molecule (i.e., that two NO2 units are somehow joined together).
٤٢
CHAPTER 3
CHEMICAL REACTIONS
AND THE MOLE CONCEPT
reactan ts p rod u ct
describes the reaction between Zinc (Zn) and Sulfur (S) to produce zinc sulfide
(ZnS), a substance used on the inner surface of TV screens. The substances on the
left of the arrow are called reactants and are the chemical present before the
reaction takes place. Those on the right of the arrow are called the products and
are the substance present after the reaction is over.
In order to write an equation, we must be able to write the formulas of the
reactants.
If an experiment has just been carried out, the equation might be meant to
describe what has just occurred in a chemical reaction. In this case, the reactants
are known because the chemist knows what chemicals were placed in the vessel.
The products, however, must be collected and identified (by a chemical analysis,
for example) before a valid equation can be written.
Often we write chemical equations to help us plan experiments. One way
that chemical equations are particularly useful in planning experiments is that
they allow us to determine the quantitative relationships that exist among the
amounts of reactants and products. To be helpful in this way, however, chemical
equations must be balanced, which means that they must obey the law of
conservation of mass by having the same number of atoms of each kind on both
sides of the arrow.
٤٣
Step 1. First write an unbalanced equation, being careful to write the correct
formula for each substance involved.
Step 2. Balance the equation by adjusting the coefficients that precede the formula
of the reactants and products so that there are the same number of atoms
of each kind on both sides of the arrow.
It is very important to remember that in carrying out step 2, you must not alter the
formulas of the reactants or products
For example, consider the reaction between hydrochloric acid (HCl) and
sodium carbonate (Na2CO3). The product of the reaction are sodium chloride
(NaCl), gaseous carbon dioxide (CO2) and water.
To obtain the balanced equation for the reaction, we proceed as follows.
Step 1. Write the unbalanced equation, being sure to give the correct formulas for
the reactants and products.
Na 2 CO 3 HCl NaCl H 2 O CO 2
Step 2. Place coefficients in front of chemical formulas to balance the equation.
Although there are no set rules to tell you where to start, it is often best
to seek out the most complex formula in the equation and begin there by
given it coefficient 1. in this case, we start with the Na2CO3. there are
two atoms of Na in this formula, so to balance the Na we need to place a
coefficient 2 in the front of NaCl on the right. This gives
Na 2 CO 3 HCl 2 NaCl H 2 O CO 2
Although this balances the Na, it cause the Cl to become out of balance, but we
can correct this by placing a 2 in the front of HCl on the left.
Na 2 CO 3 2HCl 2 NaCl H 2O CO 2
Notice that this also brings hydrogen into balance, and a quick counts for
each elements reveals that the equation is now balanced.
Problem: Balance the following equation for the combustion of octane, C8H18,
which is a compound of gasoline:
C8 H18 O 2 CO 2 H 2 O
Solution: First, we assign C8H18 (the most complex formula) a coefficient of 1.
Next, we see that we need to have 8CO2 on the right to balance the carbon and 9
H2O on the right to balance the hydrogen (9 H2O contain 18 H atoms because
each H2O contains 2 H atoms). This gives
٤٤
C8 H18 O 2 8 CO 2 9 H 2 O
Now we can work on the oxygen. On the right there are 25 O atoms (2 X 8 + 9 =
25). On the left the oxygen (O) atoms come in pairs. This means that we must
have 12 1 pairs (O2 molecules)to have 25 O atoms on the left. This gives us
2
1
C8 H18 12 O 2 8 CO 2 9 H 2 O
2
Finally, we can eliminate the fractional coefficient by doubling each coefficient.
2 C8 H18 25 O 2 16 CO 2 18 H 2 O
C 2 H 5OH 3 O 2 2 CO 2 3 H 2O
On a molecular, submicroscopic level, we can view this as a reaction between
individual molecules. But we can just as easily scale this up to lab-sized amounts
by applying the same mole ratio.
The coefficient in a chemical equation provide the ratios by which moles of one
substance react with or form moles of another.
For example, the equation for combustion of C2H5OH gives six chemical
equivalents that we can use to form conversion factors for calculations.
3 mol O 2 2 mol CO 2
3 mol O 2 3 mol H 2O
٤٥
2 mol CO 2 3 mol H 2 O
Let's look at some examples that illustrate how we use this information in a
chemical calculation.
Problem: How many moles of oxygen are needed to burn 1.80 mol C2H5OH
according to the balanced equation
C 2 H 5OH 3 O 2 2 CO 2 3 H 2O
Solution: the coefficient of the equation gives us the relationship
3 mol O 2
1.8 mol C 2 H 5 OH X 5.40 mol O 2
1 mol C H
2 5 OH
We need 5.40 mol O2.
Problem: How many moles of CO2 will be formed when 0.274 mol C2H5OH
burns?
Solution: Now we look at the coefficient for C2H5OH and CO2, which give us
2 mol CO 2
0.274 mol C 2 H 5OH X 5.48 mol CO 2
1 mol C 2 H 5 OH
٤٦
Problem: How many moles of water will form when 3.66 mol CO2 are produced
during the combustion of C2H5OH?
2 mol CO 2 3 mol H 2 O
Therefore,
3 mol H 2 O
3.66 mol CO 2 X 5.49 mol H 2 O
2 mol CO 2
General way to tackle stoichiometry problems that involve a chemical
reaction.
U se ratio of
U se f orm ula coef icien ts U se f orm ula
m as s of A to con vert m ass of B
to convert m oles of A
to m ass to convert
to m oles of t o gram s
B
4 Al 3 O 2 2 Al 2O3
How many grams of O2 are required to react with 0.300 mol Al?
٤٧
0.300 mol Al ? g O 2
The balanced equation provide a path between moles of Al and moles of O2.
4 mol Al 3 mol O 2
We know that
1 mol O 32.0 g O 2
3 mol O 2 32.0 g O 2
0.300 mol Al X X 7.20 g O 2
4 mol Al 1 mol O 2
4 Al 3 O 2 2 Al 2O3
Calculate the number of grams of Al2O3 that could be formed if 12.5 g O2 react
completely with aluminum.
12.5 g O2 ? g Al2 O 3
The central route we have to follow
3 mol O 2 2 mol Al 2 O3
The three relationships needed to solve the problem are
1 mol O 2 32.0 g O 2
3 mol O 2 2 mol Al 2 O3
1 mol Al 2 O3 102 g Al 2O 3
٤٨
We use them to construct conversion factors and arrange them so that the units
cancel correctly.
1 mol O 2 2 mol Al 2 O 3 102 g Al 2 O 3
12.5 g O 2 X X X 26.6 g Al 2 O 3
32.0 g O 2 3 mol O 2 1mol Al O
2 3
The following diagram summarizes the steps involved.
X 2 m ol A l 2 O 3
3 m ol O 2
0 .3 91 m ol O 2 0.260 m ol A l 2 O 3
X 1 m ol O 2 X 2 m o l A l2O 3
32.0 g O 2 3 m ol O2
N o d ir ect
12 .5 g O 2 calcu la tion 26.6 g A l 2 O 3
The theoretical yield of a given product is the maximum yield that could be
obtained if the reactants gave only that product with no side reactions.
The percentage yield is a measure of the efficiency of the reaction and is defined
as
actual yield
percentage yield X 100%
theoretical yield
Example: 3.48 g of CO2 was obtained when 1.93 g a mixture of ethylene (C2H4)
and 5.92 g O2 is ignited according to the following equation. Calculate the
percentage yield of CO2
C 2 H 4 3 O 2 2 CO 2 2 H 2 O
Solution:
٤٩
3.48 g CO 2
percentage yield of CO 2 X 100%
5.43 g CO 2
64.1 %
moles of solute
molarity ( M )
liters of solution
M = mol / L or = mmol / L
Solution:
10
molarity ( M ) 5 M
2
Example: What is the molarity of a solution that has 18.23 g HCl in 2.0 L?
Solution:
Solution:
٥٠
The formula mass of NaOH is 40.0 g/ mol, so
1 mol NaOH
2.00 g NaOH X 0.0500 mol NaOH
40 g NaOH
When expressed in liters, 200 mL becomes o.200 L. The molarity is therefore
1000 mL soln
0.0200 mol NaOH X 80.0 mL soln
0.250 mol NaOH
Thus, if we measure out 80.0 mL of this solution, it will contain the desired
0.0200 mol of NaOH
0.400 mol
0.400 M means
1000 mL soln
٥١
Now we apply the ratio as a conversion factor to cancel " mL soln"
40.0 g NaOH
0.0200 mol NaOH X 0.800 g NaOH
1 mol NaOH
Thus, 50.0 ml of 0.400 M NaOH contains 0.800 g NaOH.
Example: What is the molality of solution that contains 330 g of CaCl2 per
kilogram of solvent?
Solution:
Example: How many grams of NaOH are needed to make 2 molal solution?
Solution :
m = number of moles of solute/ Kg of solvent
Number of moles = wt/ mol.wt
Mol.wt of NaOH = 23 + 16 + 1 = 40
So,
wt
2 40
1
wt 80 g
٥٢
N = number of equivalents of solute/ liter of solution
N = equivalents / L
mol.wt
equivalent weight (eq. wt)
valnce (n)
In case of acids: e.g H2SO4 eq.wt = 98/2 = 49
In case of bases: e.g. Ba(OH)2 eq. wt = 171/2 = 85.5
In case of salts: e.g. NaCl eq.wt = 58.5/1 = 58.5
NaCO3 eq.wt = 106/2 = 53
? g /40
0.5
2L
٥٣
? g 40 g
Thus, the number of grams of NaOH that required to prepare 2.00 L of 0.5 N
solution is 40 g.
Mole fractions
Is the ratio of number of moles of one component to the total number of moles of
all components in the solution
nA
XA
n A n B ..... n n
XA = mole fraction of A
nA = number of moles of component A
nA + nB + ……..+ nn = total number of moles
Example: What is the mole fraction of ethyl alcohol (C2H5OH) and H2O if 0.5
mol of ethyl alcohol and 1.5 mol of H2O is mixed?
Solution:
1.5
XH 2O 0.75
0.5 1.5
0.5
X C 2 H 5 OH 0.25
0.5 1.5
Dilution of solutions
٥٤
The process involves mixing a concentrated solution with additional solvent to
give larger final volume is known as dilution. Throughout this process, the
number of moles of the solute remains constant, and only the volume increases.
If we multiply a solution's molarity M by its volume V, we obtain the
number of moles of the solute
mol
M XV X L mol
L
Since the number of moles of solute stays the same during a dilution, the product
of the final molarity and volume (M1XV1) must be equal to the product of the final
molarity and volume (M2XV2). This give the useful equation
M1 X V1 = M2 X
V2
Example: What is the concentration of solution produced by diluting 100 mL of
1.5 M NaOH to 2 L?
Solution:
M1 X V1 = M2 X
V2
(1 = Initial, 2 = Final)
M1 = 1.5 M
M2 = ?
V1 = 100 mL
V2 = 2000 mL
Solution:
M1 X V1 = M2 X V2
M1 = 18.0 M M2 = 3.00 M
V1 = ? V2 = 750 ml
M 2 X V1
V1
M1
٥٥
(3.00 M )(750 mL)
V1
18.0 M
V1 125 mL
To prepare the solution, we dilute 125 mL of the concentrated H2SO4 to a total
final volume of 750 mL.
Example: How much water must be added to 25.0 mL of 0.500 M KOH solution
to produce a solution whose concentration is 0.350 M?
Solution:
M1 X V1 = M2 X
V2
M1 = 0.500 M M2 = 0.350 M
V1 = 25.0 mL V2 = ? ml
V2 37.5 mL
Since the initial volume was 35 mL, we must add 10.7 mL water to get final
volume 37.5 mL of 0.350 M concentration.
٥٦
CHAPTER 4
Metals
More than 70% of the elements are metals, some examples are iron (Fe),
aluminum (Al), copper (Cu). Chrome (Cr) gold (Au), lead (Pb), sodium (Na)---
etc. They have distinctive appearance, shiny with a luster. They have abilities to
deform without breaking when hit with hammer and to stretch when pulled. The
ability to deform when hammered is called malleability. The ability of metal to
stretch when pulled from the opposite directions is called ductility.
Metals are good conductors of electricity. They are also good conductors
of heat. Some metals are very reactive like sodium, while other like gold are very
unreactive. Some metal are very hard, and other are very soft. Chromium (Cr) and
iron (Fe) are example of hard metals; gold and lead are example of soft ones.
Sodium is also a soft metal. The extremes of melting point are even more
impressive. Tungsten has the highest melting point of any element, 3400 0C.
Mercury has the lowest point of any metal, -38 0C. Mercury is fluid at room
temperature, commonly used in thermometers.
Nonmetals
Metalloids
Metalloids (also called semimetals) are elements that have properties between
those of metals and those of nonmetals. The best-known example is the element
silicon. Two others are arsenic (As) and antimony (Sb).
Metalloids are typically semiconductors-they conduct electricity, but not
nearly as well as metals.
٥٧
4.2 The first periodic table
Mass Charge
Grams Atomic Mass Coulombs Electronic
units (u) charge
Proton 1.67 X 10-24 1.007276 + 1.602 X 10- 1+
19
Isotopes
As mentioned before, not all atoms of the same element have identical masses as
first suggested by Dalton. It is referred to these different kinds of atoms as
isotopes. The existence of isotope is a common phenomenon, and most of the
elements occur naturally as mixtures of isotopes.
A particular isotope of an element is identified by specifying its atomic
number given the symbol Z, and its mass number A.
We represent an isotope symbolically by writing its mass number as a
superscript and its atomic number as subscript, both preceding the atomic symbol.
A
Z X
For example, a carbon atom (Z = 6) that has 6 neutrons would be given the
symbol 126 C . This is the carbon-12 isotope that is serves as the basis of the atomic-
mass scale.
The mass number (Z) of normal hydrogen is 1, for deuterium 2, and for
tritium 3, and the atomic number for each is 1, so, we can write their symbols as
follows
1
1 H, 21H, 31H
Atomic numbers and the modern periodic table
When the elements are arranged in the periodic table in order of their atomic
numbers, all the inconsistencies that had occurred in Medeleev's table disappear.
The periodic table that is used today is shown in the next separate sheet.
The numbers printed above the chemical symbols are atomic numbers, and those
below are atomic amass. Like Medeleev's table, it consists of a number of rows
called periods, which are identified by Arabic numerals. The vertical columns are
called groups, each containing a family of elements. The groups are identified by
Roman numeral and letter, either A or B.
The groups labeled with the letter A (Groups IA through VIIA) and Group
0 are referred to collectively as the representative elements. Those labeled with
the letter B (Groups IB through VIIB) plus Group VIII are called transition
element
Finally, there are two long rows of elements placed just below the main
part of the table. These elements, known as inner transition elements, actually
belong in the body of the table.
٥٩
The reaction between sodium (Na) and chlorine (Cl) atoms to form sodium
chloride (NaCl) is a clear example for the reactions of metals with nonmetals to
form ionic compounds.
qq
F 2
r
Table 4.2 lists the ions formed by most of representative elements. Notice
first that the atoms in any particular group form ions with the same charge.
Group number
IA IIA IIIA IVA VA VIA VIIA
Li+ Be 2+
C 4-
N3-
O 2-
F1-
Na+ Mg2+ Al3+ Si4- P3- S2- Cl-
K+ Ca 2+
Se 2-
Br-
Rb+ Sr2+ Te2- I-
Cs+ Ba2+
We can use the periodic table to help us remember the ions formed by the
nonmetals. The number of negative charges is just the number of step to the right
that you have to go in the table to get the noble gas column. Consider oxygen, for
example. It is in Group VIA, and to get the noble gas column requires moving
two steps to the right.
٦٠
1 2
O F Ne
Oxygen forms an ion with a negative two charge, O2-. Similarly, to move from
fluorine to the noble gas column is just one step, and fluorine forms the ion F-.
The ions formed by the transition metals is different from that formed by
representative elements, because the transition elements to form more than one
ion. Some of these ions are listed in Table 4.3.
The reason for the third rule is that there are no molecules in an ionic compound.
Each positive ion is surrounded by and attracted equally to some number of
negative ions, and vice versa, so we can't say that any particular positive ions
belongs to any particular negative ion. All we can do is specify the ratio in which
the ions occur in the compound.
Let's look now at a few simple examples of how the rules are applied.
1. The compound formed from calcium and oxygen. Looking at the periodic table
we see that calcium is in Group IIA, soothe calcium ion is Ca2+. Oxygen is a
nonmetal in group VIA, so its ion is O2-.To make an electrically neutral
compound, we take the ions in a 1-to-1 ratio, and the formula is CaO.
2. The compound formed from zinc and chlorine. Zinc is a transition metal and it
forms the ion Zn2+. Chlorine is Group VIIA, so its ion is Cl-. To give an
٦١
electrically neutral compound, we must have two Cl- for each Zn2+, so the formula
is ZnCl2.
3. The compound formed from of aluminum with oxygen. Aluminum forms Al3+
and oxygen forms O2-. In the compound, we need a ratio of ions that makes the
total positive and negative charges the same. This requires that we take two Al3+
ions for every three O2- ions, because
There is a very simple shortcut that can use to obtain the formula of an ionic
compound. It involves exchanging superscripts for subscripts.
3 + 2 -
Al O
A l2O 3
Ions that contain more than one atom
Cations
NH4+
H3O+
Anions (alternative names in parentheses)
CO32- Carbonate ClO2- Chlorite
-
HCO3 Hydrogen ClO- (or OCl-) hypochlorite
carbonate
(bicarbonate)
C2H42- Oxalate PO43- Phosphate
-
CN Cyanide HPO42- Hydrogen
phosphate
NO3- Nitrate H2PO4- Dihydrogen
phosphate
NO2- Nitrite CrO42- Chromate
OH- Hydroxide Cr2O72- Dichromate
SO42- Sulfate MnO4- Permanganate
HSO4- Hydrogen sulfate C2H3O2- Acetate
(bisulfate)
SO32- Sulfite
HSO3- Hydrogen sulfite
(bisulfite)
ClO4- Perchlorate
٦٢
ClO3-
+ 3 -
2 PO4
Ca
C a 3 (PO 4 ) 2
4.5. Reactions of nonmetals with each other; the formation of molecular
compounds
The nonmetals react not only with metals but also with each other. However,
when two nonmetals combine to form a compound, electrically neutral molecules
are formed instead of ions. An example of such a reaction occurs between oxygen
and hydrogen to form water.
2 H 2 ( g ) O 2 ( g ) 2 H 2O( l )
The number of compounds formed among the nonmetals is enormous. There are
millions of organic compounds, for example, whose molecules are composed
chiefly of carbon and hydrogen, but that often contain other nonmetals as well
(principally oxygen, nitrogen, sulfur, and the halogens).
٦٣
Table 4.6. Empirical formulas of some oxides of nonmetals
Some of the properties associated with many simple ionic and covalent
compounds in the extreme cases are summarized Table 4.7 shown below.
Table 4.7 Some of the properties associated with many simple ionic
and covalent compounds
٦٤
4Fe(s) 3O2 (g) 2 Fe 2O3(s)
Iron begins as electrically neutral atoms that lose electrons to become Fe3+ ions.
When the oxide is reduced to give metallic iron, the reverse process must happen,
so Fe3+ ions must gain electrons to give Fe atoms. It is this loss and gain of
electrons, which occurs in many similar reactions, that is now associated with the
terms oxidation and reduction.
Mg Mg 2 2e- (oxidation)
O 2 4e- 2O 2- (reduction)
Two terms that we often use in discussing redox reactions are oxidizing agent
and reducing agent.
The oxidizing agent is the substance that takes electrons from the substance that
is oxidized, thereby causing oxidation to take place. That's what O2 does in the
reaction between Mg and O2, it takes electrons from Mg and causes Mg to be
oxidized, so O2 is oxidizing agent. Notice that the oxidizing agent (O2) becomes
reduced in the reaction.
The reducing agent is the substance that gives electrons from the substance that
is reduced, thereby causing reduction to take place. That's what Mg does when it
reacts with O2; it gives electrons to O2 and causes O2 to be reduced. Notice that
the reducing agent (Mg) is oxidized.
Oxidation numbers
٦٥
not the compound is ionic or molecular, which means that for molecular
substances the oxidation numbers are really fictitious charges.
Problem: Assign oxidation numbers of each of the atoms in the following: (a)
FeCl3, (b) KNO3, (c) H2O2, (d) Fe2(SO4)3, (e) Cr2O72-, (f) ClO3-, and (g)NaS4O8.
Solution: the oxidation number of the Fe can then be obtained by the summation
rule (rule 3).
(a) FeCl3
Cl 3 X (-1) = -3
Fe 1 X (x) = x
Sum = 0
For the sum to be zero, the oxidation number of the iron must be +3
(b) KNO3
potassium (Group IA) forms ions with a charge of 1+ (K+), so the
oxidation number of K must be +1 (rule 2). Oxygen is assigned an oxidation
number of -2 (rule 6). We get the oxidation number of N by applying rule 3.
K 1 X (+1) = +1 (rule 2)
O 3 X (-2) = -6 (rule 6)
N 1 X (x) = x
Sum = 0 (rule 3)
٦٦
(c) H2O2 is hydrogen peroxide compound formed between nonmetals, so we
expect it to be molecular. Since there is no ions, we can't use rule 2. Rules 5 and 6
refer to H and O, so we can use these, but there is a conflict. If we take H to be +1
according to rule 5, then O must be assigned an oxidation number of -1 in order
for the sum of oxidation numbers to be zero. On the other hand, if we take O to be
-2 according to rule 6, then H must be +2 for the sum to be zero. As mentioned
earlier, however, when two rules conflict, we apply the one higher up on the list.
Therefore, we take H to be +1 and the oxidation number of O in the compound is
-1.
(d) Fe2(SO4)3
Fe 2 X (+3) = +6 (rule 2)
S 3 X (x) = 3x
O 12 X (-2) = 24 (rule 6)
Sum = 0
3x + (+6) + (-24) = 0
3x = +18
x = +6
(e)For the Cr2O72- ion, the sum of oxidation numbers must equal the charge on the
ion. Therefore,
Cr 2 X (x) = 2x
O 7 X (-2) = -14 (rule 6)
Sum = -2 (rule 3)
2x + (-14) = -2
x= +6
In this ion, the oxidation number of Cr is +6. However, you should keep in
mind that the atoms in a polyatomic ion are held together by the same kinds of
attractions that hold together atoms in molecules. There are really no Cr6+ ions
within the Cr2O72- ion. As in molecular substances, the oxidation numbers of
Cr and O here are not their actual charges.
(f) ClO3- ion is composed of two nonmetals, which means that the ion is held
together by the same kind of attractions that exist within molecules. This means
that we can't just assume that the Cl exist as Cl- ion within the ClO3- ion, so we
can't apply rule 2 in this case. However, rule 6 tells us that oxygen has an
oxidation number of -2, so we can calculate what Cl must be in order for the sum
of oxidation numbers to be equal to the charge on the ion.
Cl 1 X (x) = x
O 3 X (-2) = -6 (rule 6)
Sum = -1 (rule 3)
It's not hard to see that chlorine must have an oxidation number of + 5.
(g)NaS4O8,
٦٧
sodium is an alkali metal and must be Na+, so its oxidation number is +1.
oxygen is -2 according to rule 6. therefore,
Na 2 X (+1) = +2 (rule 2)
S 4 X (x) = 4x
O 6 X (-2) = -12 (rule 6)
Sum = 0
(+2) + (4x) + (-12) = 0
4x = 0
x = +5/2
The oxidation number of sulfur is +5/2. notice that oxidation numbers don't have
to be whole numbers (although they usually are).
In redox reaction, there is a change in the oxidation number or oxidation state (we
use the terms interchangeably) of two or more elements. Consider, for example
the reaction between magnesium and oxygen
2Mg O 2 2MgO
0 0 2 -2
S O 2 SO 2
0 0 4 -2
The oxidation number of sulfur (S) increase from 0 to +4, so the sulfur is
oxidized. The oxidation number of O decreases from 0 to -2, so O2 is reduced.
This means that O2 is oxidizing agent and S is reducing agent.
٦٨
Example 4.2. identifying oxidation and reduction in reaction
Problem: In the following reaction, which substance is oxidized and which is
reduced? Which substance is the oxidizing agent and which is the reducing
agent?
Now we look for changes in oxidation numbers. We see that the oxidation
number of Cl changes from -1 to 0. this is an increase in oxidation number, so
the substance oxidized is HCl. We also see that Cr changes from +6 to +3. this
is a decrease in oxidation number, so the K2Cr2O7 is reduced.
The oxidation agent is the substance that's reduced; it is K2Cr2O7. The
reducing agent is the substance that's oxidized is HCl.
٦٩
CHAPTER 5
CHEMICAL REACTIONS
IN AQUEOUS SOLUTION
When the reactants are dissolved in a solvent, we said that we have a
solution. One of the most important solvents for chemical reactions is water.
It is a common substance, and it is also a good solvent for many different
kinds of chemicals, both ionic and molecular.
٧٠
5.2 Electrolytes
Water is generally a good solvent for ionic compounds, and water solutions
(aqueous solutions) that contain these substances have some unusual
properties, one of which is that they conduct electricity. A compound such as
NaCl that gives an electrically conducting solution is said to be an electrolyte.
When an ionic compound dissociate in water, its ions are not really
entirely free. Instead, they become surrounded by water molecules and said to
be hydrated. We indicate this by writing (aq) after formulas of ions. For
example, the dissociation of sodium chloride (NaCl) that occurs when the
solid is dissolved in water can be represented by the equation
The two examples of electrolytes discussed above, NaCl and HCl, are
essentially completely dissociated in aqueous solution. The electrolytes that
are completely dissociated in solution are called strong electrolytes.
There are also many molecular substances that have no tendency at all to
undergo ionization when dissolved in water. Sugar and ethyl alcohol, for
example, do not produce ions in solution, the solution doesn't conduct
electricity and these solutes are therefore called nonelectrolytes.
Between the extremes of strong electrolytes and nonelectrolytes exists a
large collection of compounds called weak electrolytes. These compounds
produce aqueous solution that conduct electricity, but only very weakly. An
example is acetic acid HC2H3O2.
In a solution of acetic acid, only a small fraction of all the acetic acid
molecules are present as ions, produced by the reaction
٧١
-
HC2 H3O2(aq) H 2O H 3O (aq) C2 H 3O 2 (aq)
For example, in a 1.0 M solution of HC2H3O2, only about 0.42% of the solute
has undergone this reaction. The rest of acetic acid exists as uncharged
molecules. In this solution there are two reactions occurring simultaneously,
-
HC2 H 3O 2 H 2O H3O C 2 H 3O 2 (1)
H 3O C 2 H 3O 2 H 3O H C 2 H 3O 2 H 2 O (2)
H 2O + H 2O H 3 O + ( aq) + O H - ( aq)
Table 5.1 Some weak electrolytes
٧٢
5.3 Reactions between ions
Ag (aq) NO3 (aq) Na (aq) Cl (aq) AgCl(s) Na (aq) NO3 (aq)
This is called the ionic equation and is obtained by writing the formulas of
any soluble strong electrolytes in "dissociated form" and the formulas of any
insoluble substances in molecular form."
The net ionic equation is obtained by removing the spectator ions Na+ and
NO3- as follows,
Ag (aq) NO3 (aq) Na (aq) Cl (aq) AgCl(s) Na (aq) NO3 (aq)
٧٣
(vitamin C), sulfuric acid (H2SO4), hydrochloric acid (HCl), nitric acid
(HNO3).
Among important bases are ammonia (NH3), sodium hydroxide (NaOH),
potassium hydroxide (KOH).
Acids and bases have certain properties that help us identify them, for
example, a solution of an acid has a sour taste. On the other hand, bases have
a bitter taste. (Caution: you should never test for acids and bases by tasting a
chemical in the laboratory. It might not be healthy). Another property of acids
and bases is their effect on indicators (chemicals whose colors depend on the
acidity or basicity of their solutions). A typical example is the dye called
litmus. Litmus is a chemical that has a blue color in a basic solution and a
pink color in an acidic solution.
Stated in modern terms, the Arrhenius definitions of acids and bases are as
follows: An acid is a substance that increases the concentration of hydronium
ion, H3O+, in an aqueous solution, and a base is a substance that increases
the concentration of hydroxide ion, OH-.
Acids
In general, acids are molecular substances that produce hydronium ion by
reaction with water. For example, hydrogen chloride (HCl) is an acid, because
when it is dissolved in water, it reacts with the solvent to produce H3O.
Or more simply
H C 2 H 3 O 2 ( aq) H + ( a q) + C 2 H 3 O 2 - ( a q)
This is an equilibrium, and in a solution of HC2H3O2 only a small fraction of
the solute is dissociated into ions. This means that the concentration of H3O+
in the solution is low. As a result, acetic acid and other acids that are weak
electrolytes are called weak acids.
٧٤
Both HCl and HC2H3O2 are able to furnish only one hydrogen ion (one
proton) per molecule of the acid. Such acids are said to be monoprotic acids.
There are also many acids that are able to furnish more than one proton per
molecule of the acid. As a class, they are referred to as polyprotic acids. Two
examples are sulfuric acid, H2SO4, and phosphoric H3PO4.
Sulfuric acid is also called a diprotic acid because each molecule of it is
able to give up two protons. This happens in two distinct steps
-
H 2SO4(aq) H (aq) HSO4 (aq)
H 3 PO 4 ( a q) H + ( aq) + H 2 PO 4 - ( aq)
H 2 PO 4 - ( aq) H + ( aq) + H P O 4 2 - ( aq)
H P O 4 2 - ( aq) H + ( a q) + P O 4 3 - ( aq)
Notice that the second step in dissociation of H2SO4 is an equilibrium
(only about 10% of HSO4- is actually dissociated). Despite this, sulfuric acid
is considered a strong acid because the first dissociation step is complete.
Phosphoric acid, is a weak acid because all three of its dissociation steps are
equilibria that do not proceed very far toward completion.
There are substances that do not contain hydrogen, yet still produce acidic
solutions when dissolved in water. A common example is carbon dioxide
(CO2). When dissolved in water, it reacts as follows:
C O (g) + H 2 O H 2 CO 3 ( aq)
The compound H2CO3 is called carbonic acid; it is a weak diprotic acid that is
dissociated in two steps, the first of which is
Bases
There are two principal kinds of bases: ionic hydroxides and molecular
substances that react with water to produce OH-. Sodium hydroxide (NaOH),
٧٥
and calcium hydroxide (Ca(OH)2) are typical ionic hydroxides. In the solid
state they consist of a metal ion and hydroxide ion, and when they dissolve in
water, they dissociate.
N H 3 ( a q) + H 2 O N H 4 + ( aq) + O H - ( aq)
In this case, a proton is transferred from a water molecule to an ammonia
molecule. After the H2O loses an H+, the particle left behind is a hydroxide
ion OH-.
The reaction of NH3 with water is an equilibrium, and only a small
fraction of the NH3 placed into the solution is present as NH4+ and OH- ions.
Ammonia is a weak electrolyte, and because its solution have relatively few
OH- ions, it is also said to be a weak base. In general, molecular bases are
weak bases.
Acids (Those followed by an asterisk(*) are strong electrolytes and are completely
ionized in aqueous solution.)
Monoprotic acids HF Hydrofluoric HClO3 Chloric acid*
acid
HX H X
HCl Hydrochloric HClO4 Perchloric
acid* acid*
HBr Hydrobromic HIO4 Periodic acid*
acid*
HI Hydroiodic HNO3 Nitric acid*
acid*
HOCl Hydrochlorous HNO2 Nitrous acid
acid
HClO2 Chlorous acid HC2H3O2 Acetic acid
Diprotic acids H2SO4 Sulfuric acid* H2S Hydrosulfuric
acid
H 2 X H HX
HX H X 2
H2SO3 Sulfurous acid H3PO3 Phosphorous
acid (only two
٧٦
hydrogen can
be removed as
protons)
H2CO3 Carbonic acid
H2C2O4 Oxalic acid
Triprotic acids H3PO4 Phosphoric acid
(orthophosphoric
H3X H H 2 X acid)
H 2 X H HX 2
HX 2- H X 3
Typical acidic oxides SO2
(nonmetal oxides)
SO2 H 2O H 2SO3
SO3
SO3 H 2O H 2SO4
N2O3
N 2O3 H 2O 2HNO2
N2O5
N 2O5 H 2O 2HNO3
P4O6
P4O6 6H 2O 4H3PO3
P4O10
P4O10 6H 2O 4H3SO4
Bases
Molecular bases NH3 Ammonia
(weak bases) (N H 3 + H 2 O N H 4+ + O H -)
N2H4 Hydrazine
( N2H 4 + H 2O N 2H 5 + + O H -)
NH2OH Hydroxyl amine
( NH 2 + H 2 O N H 2 O H + + O H -)
Ionic bases Metal
(strong bases) hydroxides M(OH)n M n nOH-
NaOH
Ca(OH)2
Typical basic oxides Na 2 O
(metal oxides) M 2 O H 2 O 2MOH
K 2O
CaO
SrO MO H 2 O M (OH)2
BaO
Neutralization
The most important reaction that acids and bases undergo is their reaction
with each other, a reaction called neutralization. In aqueous solutions the
٧٧
neutralization reaction between a strong acid and strong base takes the form of
the net ionic equation
HA H+ + A-
acid p roton con ju gate b ase
Water itself is both BrØnsted acid and base.(Has amphoteric behavior, act as
acid and base). Careful measurements have shown that pure water ionizes ever
so slightly to produce equal numbers of hydrated hydrogen ions and
hydroxide ions:
H 2O + H 2O H 3O + + H O -
Or in simplified notation,
H 2O H + ( aq ) + H O - ( aq)
٧٨
reaction (right-to-left) shows that H3O+ (an acid) donates a proton to OH- (a
base) to form two H2O molecules. in the notation under the equation:
H 2 O + H 2O H 3O + + H O -
acid 1 b ase 2 acid 2 b ase 1
The subscripts refer to conjugate acid-base pairs. Note that each acid differs
from its conjugate base by a proton.
H 2O + H 2O H 3O + + HO -
acid b ase con ju gate acid con ju gate b ase
The following equations provide specific examples of this relationship.
The stronger the acid the weaker the conjugate base and the stronger the base
the weaker the conjugate acid. Acid and bases are ranked in order of their
comparative strengths as shown in Table 5.3.
٧٩
LEWIS ACIDS AND BASES; COMPLEX ION OF METALS
The American chemist Gilbert N. Lewis provide a further extension for the
acid-base concept. Thus, in the Lewis definition of acid and base: A base is
defined as a substance that can donate a pair of electrons to the formation of
covalent bond. An acid is a substance that can accept a pair of electrons to
form the bond.
A simple example of a Lewis acid-base reaction is the reaction of a
hydrogen ion with a hydroxide ion
.. - H H
H+ + H
..
O
..
O
The hydroxide ion is the Lewis base because it furnishes the pair of electrons
that become shared with the hydrogen. The hydrogen ion, on the other hand, is
the Lewis acid because it accept a share of pair of electrons when the O-H
bond is created.
Another example is the reaction between BF3 and ammonia,
H F H F
H N + B F H N B F
H F H F
In this case, the NH3 functions as the base and BF3 serves as the acid.
Compounds elements with incomplete valence shells, such as BF3 or AlCl3,
tend to be Lewis acids, while compounds or ions that have lone pairs of
electrons can behave as Lewis bases. When the Lewis acid-base reaction
occurs, a coordinate covalent bond is formed.
Some common Lewis bases are amines and alcohols.
٨٠
The Ionization of Water
As we've known in earlier chapter that water itself is a very weak electrolyte
because of the reaction
H 2O + H 2 O H 3 O + ( aq) + H O - ( aq)
This kind of reaction, in which two molecules of the solvent react with each
other to form ions, is called autoionization. Since autoionization is an
equilibrium, we can write an equilibrium expression for it as follows
Kc
H O OH
3
H 2OH 2O
The molar concentration of water, which The density of H2O is 1.00 g/ml,
appears in the denominator of this so 1.00 L has a mass of 1.00X
expression, is very nearly constant (=55.6 M) 103g. therefore,
in both pure and dilute aqueous solutions. 1.00X103 g H 2 O
Therefore, [H2O]2 can be included with the [H 2 O] X
1.00 L
equilibrium constant, Kc, on the left side of
1mol H 2 O
the equation. This gives 55.6 mol H 2 O/L
18.0g H 2 O
K c X[H2O]2 [H3O ][OH ]
The left side of this expression is the
product of two constants which, must also a
constant. This combined constant is written
as
K w Kc [H2O]2
The equilibrium law therefore becomes
Kw varies with temperature. At
K w K c [H3O ][OH- ] 370C(body temperature)
+ -
Because [H3O ][OH ] is the ion Kw = 2.42 X 10-14
concentration, Kw is called ion product
constant for water, or simply the ionization
constant or dissociation constant of water.
At 250C, Kw= 1.0 X 10-14, and is one
equilibrium constant that you should be
memorize.
H 2O H+ + HO -
٨١
When we use this simplified equation, the expression for the ionization
constant for water is
K w [H ][OH- ]
This equation can be used to calculate the molar concentration of both H+ and
OH- ions in pure water. From the stoichiometry of dissociation, we see that for
each 1 mol of H+ formed 1 mol of OH- is also produced. This means that at
equilibrium, [H+] = [OH-]. If we let x equal the hydrogen ion concentration,
then
x [H ] [OH ]
Therefore,
K w x.x x 2
Or, because Kw = 1.0 X 10-14,
x 2 1.0 X 1014
Taking the square root yields
x 1.0 X 107
Which means that the concentration of hydrogen ion and hydroxide ion in
pure water are
[H+] = [OH-] = 1.0 X 10-7 M (at 250C)
The pH concept
Hydrogen ion and hydroxide ion enter into many equilibria in addition to the
dissociation of water, so it is frequently necessary to specify their
concentrations in aqueous solutions. The concentrations may range from
relatively high values to very small ones (for example, 10 M to 10-14), and a
logarithmic notation has been devised to simplify the expression of these
quantities. In general, for some quantity X, the pX is defined as
1
pX log logX
X
For example, if we wish to specify the hydrogen ion concentration in a
solution, we speak of pH. This is defined as
1
pH log
log[H ]
[H ]
In a solution where the hydrogen ion concentration is 10-3 M, we therefore
have
٨٢
Notice that logs to the base 10
1
pH log -3 log[10-3 ] -(-3) (common logs) are used here,
[10 ] not natural logs.
pH 3
Similarly, if the hydrogen ion concentration is 10-8 M, the pH of the solution is 8
Following the same approach for the hydroxide ion concentration, we can
define the pOH of a solution as
1
pOH log
log[OH- ]
[OH ]
Just as the H+ and OH- ion concentration in as solution are related to each other,
so also are the pH and pOH. From the equilibrium expression for the dissociation
of water, we obtain the following by taking the logarithm of both sides.
log K w log[H ] log[OH- ]
Multiplying through by -1 gives
(-log K w ) ( log[H ]) (-log[OH- ])
By definition, -logKw = pKw, therefore,
pK w pH pOH
Since Kw = 1.0X 10-14, pkw = 14.00. this gives the useful relationship
٨٣
Measuring pH; acid-base indicators and pH meters
One of the earliest ways of judging the acidity or basicity of solution was the use
of substances called acid-base indicators. These are organic compounds whose
color depends on the pH of the solution in which they're dissolved. A typical
example is litmus, which is pink in an acidic solution and blue in a basic solution.
It is quite common today to find instruments called pH meters in labs of
all kinds. These are electronic devices that enable the pH of a solution to be
measured with a high degree of precision and accuracy.
Solution:
pH log[H ]
pH - log(2.0 X 10-3 )
pH 2.70
Example: calculating the pH of a solution of strong base
Problem: What is the ph of a 5.0 X 10-4 M solution of NaOH at 250C?
1.0 X 10-14
[H ]
5.0 X 10-4
2.0 X 10-11 M
From this we calculate the pH
pH - log(2.0 X 10-11 )
pH 10.70
Another solution: By definition
pOH log[OH- ]
[OH- ] 5.0 X 10-4 M
٨٤
pOH log(5.0 X 10 - 4)
3.30
Therefore,
pH 14 - 3.30
10.70
Example: Calculating [H+] and [OH-] from pH
Problem: A sample of orange juice was found to have a pH of 3.80. what were the
H+ and OH- concentrations in the juice?
Solution:
pH log[H ]
The antilogarithm of this gives the following relationship:
[H ] 10-pH
For the sample of orange juice, therefore,
[H ] 10-3.80
1.6X104
Once we have obtained [H+], we can calculate [OH-] from Kw.
1.0 X 10-14
[OH ]
1.6 X 10-4
6.3 X 10-11 M
Alternatively, we could first calculate pOH from pH and the take the
antilogarithm.
pOH 14 - pH
14.00 - 3.8 10.20
[OH- ] 10-pH
10-10.20
6.3 X 10-11 M
٨٥
Conjugate acid-base systems in aqueous solutions
The most important equilibrium in aqueous solutions is the ionization of water. If
we look at this equilibrium in BrØnsted-Lowry terms and consider one water
molecule the acid and the other the base, we have
H 2O + H 2 O H 3O + + H O -
acid b ase acid b ase
-
[H3O ][C2 H3O2 ]
Kc
[H2O][HC2 H3O2 ]
The concentration of water in diluted aqueous solution is constant, and it can
incorporated into the equilibrium constant to give a new constant that we call Ka.
This constant is called ionization constant or dissociation constant for the acid.
-
[H3O ][C2 H3O2 ]
K c X [H2O] K a
[HC2 H3O2 ]
We can also simplify this somewhat by using H+ in place of H3O+ and writing the
simplified equation as
H C 2H 3O 2 H + + C 2H 3O 2-
-
[H ][C2 H3O2 ]
Ka
[HC2 H3O2 ]
٨٦
Some anions, such as the bisulfate ion, HSO4-, are also acids. This anion is formed
in the first step in the ionization of H2SO4, reacts with water as follows.
H S O 4- + H 2O H 3 O + + SO 4 2-
Acids can even be cations, such as the ammonium ion, NH4+, which undergo
reaction.
N H 4+ + H 2O H 3O + + N H 3
We can write similar equilibrium laws for acids such as HSO4- and NH4+
2-
[H3O ][SO4 ]
Ka -
[HSO4 ]
The simplified equation for the reaction is
H S O 4- H + + SO 4 2-
[H ][NH3 ]
Ka
[NH4 ]
Usually, it is easier to write everything in the abbreviate simpler form
HA H+ + A-
[H ][ A- ]
Ka
[HA]
٨٧
Where Ka is called ionization constant or dissociation constant for the acid, A- is
the conjugate base
As with acids, bases also undergo the same kinds of reactions with water. An
example is ammonia (the conjugate base of NH4+).
N H 3 + H 2O N H 4+ + O H -
[NH4 ][OH- ]
Kc
[H2O][NH3 ]
Once again, the concentration of water appearing in the denominator is essentially
a constant that we incorporate into the equilibrium constant kc to give a new
constant that we will call kb.
[NH4 ][OH- ]
K c X[H2O] Kb
[NH3 ]
Many anions are bases and produce hydroxide ion in water just as ammonia does.
For example, acetate ion, C2H3O2-, which is conjugate base of acetic acid, reacts
with water as follows.
C 2H 3O 2- + H 2O H C 2H 2O 2 + O H -
B - + H 2O H B + O H-
٨٨
[HB ][OH- ]
Kb
[B]
You can apply this equation to solve equilibrium problems that involve weak
bases in water.
Ka X Kb = Kw
Problem: A student prepared a 0.1 M acetic acid solution and measured its pH to
be 2.88. calculate Ka for acetic acid.
H C 2H 3 O 2 H + + C 2H 3O 2-
-
[H ][C2H3O2 ]
Ka
[HC2 H3O2 ]
Equilibrium concentration
+
H 1.3 X 10-3M
C2H3O2- 1.3 X 10-3M
HC2H3O2 1.0 X 10-1 – 0.013 X 10-1 = 1.0 X 10-1 M
Note that when we compute the acetic acid concentration rounded to the proper
number of significant figures, the amount that has dissociated is negligible
compared tom the amount initially present. Thus,
0.10 M – 0.0013 M = (0.0987 M) = 0.1 M (rounded)
٨٩
Substituting the equilibrium concentrations into the expression for ka, we have
percent dissociation
mol/L of acid or base dissociated X100
mol/L of acid or base available
For the acetic acid in the preceding example, the mol/L dissociate equals the
mol/L of H+ formed, which is 1.3 x 10-3 M. the mol/L available is the original
concentration, 0.01 M. Therefore,
1.3 x 103 M
percent dissociation X 100 1.3%
0.10M
Example: Using the percent dissociation to calculate an
equilibrium constant
Problem: A 0.010 M NH3 solution was prepared and it was determined that the
NH3 had undergone 4.2% ionization. Calculate the ka for NH3.
N H 3 + H 2O NH 4 + + O H -
[NH4 ][OH- ]
Kb
[NH3 ]
At equilibrium
[NH4+] = [OH-]
Since the 0.010 M solution undergoes 4.2 % ionization, the number of moles per
liter of these ions at equilibrium is
٩٠
[NH4+] = [OH-] = 0.042 X 0.010 M = 4.2 X 10-4 M
Equilibrium concentration
NH4+ 4.2 X 10-4 M
OH- 4.2 X 10-4 M
NH3 1.0 X 10-2 M
Problem: What are the concentrations of all the species present in a 0.50 M
solution of acetic acid. For HC2H3O2, ka = 1.8 x 10-5.
H C 2H 3 O 2 H + + C 2H 3O 2-
-
[H ][C2 H3O2 ]
Ka 1.8 X 105
[HC2 H3O2 ]
We assign the initial [C2H3O2-] a value of zero. We also expect the initial
concentration of H+ to be zero.
In reaching equilibrium, some of HC2H3O2 will dissociate, so its
concentration will decrease. We don't know how much, so let's say that the
change in HC2H3O2 concentration is x. Since all the coefficient in the equation are
equal to 1, the concentrations of H+ and C2H3O2- will increases by x. this gives us
the quantities in the center column of the table. The quantities corresponding to
the equilibrium concentrations are obtained by adding the "change" to the initial
concentration algebraically
٩١
C2H3O2- 0.0 +x x
HC2H3O2 0.50 -x 0.50 - x
.
Substituting these values into the equilibrium expression gives us
0.50 - x = 0.50
Our equation the becomes
x 2 1.8 X 105
0.50
x 3.0 X 103
If we look back on our assumption, we see that x is in fact small compared to 0.50
and that, when rounded to the proper number of significant figures,
Kw
[OH- ]
[H ]
1.0 X 1014
3.0 X 10-3
3.3 X 10-12 M
Example: Calculating the pH of a solution that contains a weak
acid and strong acid
Problem: What is the pH of a solution that contains 0.10 M HCl and 0.10 M
HC2H3O2? For acetic acid ka = 1.8 x 10-5.
٩٢
Solution: The 0.10 M solution of HCl contains 0.10 mol/L of H+, because HCl is a
strong acid. This is the concentration of H+ before the HC2H3O2 is added. Then
we place 0.10 M HC2H3O2 into it, but we imagine that none of it dissociates. This
gives us the initial concentrations of H+ and HC2H3O2.
[H+]initial = 0.10 M
[HC2H3O2]initial = 0.10 M
Now, we have to consider the ionization of acetic acid. The chemical equilibrium
and equilibrium law are the same as in preceding problem.
H C 2H 3 O 2 H + + C 2H 3O 2-
-
[H ][C2 H3O2 ]
Ka 1.8 X 105
[HC2 H3O2 ]
Since there is no C2H3O2- present initially, some of it must form by the
dissociation of the HC2H3O2. Therefore, we let the concentration of HC2H3O2
decrease by x and the concentrations of H+ and C2H3O2- increases by x and then
add the initial concentrations and the changes to get the equilibrium quantities.
1.8 X 10-5
0.10 x
0.10
(0.10)(x)
1.8 X 10-5
(0.10)
x 1.8 X 10-5
We see that x is indeed small compared to 0.10, so the solution [H+] = 0.10 M.
This gives a pH of 1.00.
٩٣
Example: Calculating the pH of a solution of a weak base
Problem: What is the pH of a 0.010 M solution of the weak base diethylamine
(C2H5)2NH, for which kb = 9.6 X 10-4?
( C 2H 5) 2N H + H 2O (C 2 H 5 ) 2 N H 2 + + O H -
Therefore,
[(C2 H5 )NH2 ][OH- ]
Kb 9.6 X 104
[(C2 H5 ) 2 NH]
We take the (C2H5)2NH concentration specified in the problem as its
initial concentration and the concentration of (C2H5)2NH2+ to be zero. As usual,
we assume the solvent makes no contribution to [OH-]. Therefore, we take the
initial OH- concentration to be zero, too. This gives us the entries for the initial
concentration column in the concentration table. Next, we realize that some of the
diethylamine must react, so we let its concentration decrease by x, and we
increase the concentrations of (C2H5)2NH2+ and OH- by x. adding the "changes" to
the initial concentrations gives the equilibrium quantities.
If we aren't careful, we might accept this as the answer. However, if we check our
assumption, we find that x is not negligible compared to 0.010. In decimal form, x
= 0.0031, so 0.010-0.0031 = 0.007(rounded to the correct number of significant
figures). Certainly 0.007 is not the same as 0.010. in this problem, it was not safe
٩٤
to assume that the initial concentration of (C2H5)2NH is the same as its
equilibrium value.
0031 M is about 30% of the initial value of 0.010 M, so it is certainly not
negligible.
The next question, of course, is what do we do now? There are actually two
choices. One is to expand the equation into a quadratic equation and apply the
quadratic formula to it.
x = 2.7 X 10-3
x = -3.6 X 10-3
we can discard the second value because the concentration of (C2H5)2NH2+ and
OH- can't be negative and also because 0.010- (- 3.6 X 10-3) = 0.014 (rounded),
which is impossible. Therefore, we are left with x = [OH-] = 2.7 X 10-3 M. This
gives a pOH of 2.57 and a pH of 11.43.
-
H3O C2 H3O2 HC2H3O2 H 2O
Or more simply,
-
H C2H3O2 HC2 H3O2
٩٥
Notice that this reaction removes H+ and changes the base into its conjugate acid.
A similar reaction occurs if a strong base is added. The OH- supplied by the
strong base reacts with the acetic acid and converts it to its conjugate base.
HC2 H3O2 OH- C2 H3O 2 H 2O
Buffers with a ph higher than 7 can be prepared by using a base that is
stronger than its conjugate acid. A common basic buffer is formed by mixing
ammonia with an ammonium salt such as NH4Cl and contains the conjugate acid-
base pair NH4+ and NH3. If a strong acid is added to the buffer, it reacts as
follows,
H3O NH3 NH4 H 2O
H NH3 NH4 (simplified)
And if a strong base is added, the reaction is
OH- NH4 NH3 H 2O
Buffer calculations
Solution: There is only one equilibrium here that we must concerned with
H C 2H 3O 2 + H 2O H + + C 2H 3O 2-
[H ][C2H3O2 ]
1.8 X 10 5
[HC2 H3O2 ]
When NaC2H3O2 dissolves, it is completely dissociated. It is important to
remember that virtually all salts are 100% dissociated in solution. Therefore, 0.10
mol/ L of NaC2H3O2 gives 0.10 mol/ L of Na+ and 0.10 mol/ L of C2H3O2-. We
are interesting only in the C2H3O2-; the Na+ is simply a spectator ion and we
ignore it. The initial concentration of the C2H3O2- is therefore 0.10 M. As usual,
we take the initial concentration of the weak acid to be the value specified in the
problem; in this case [HC2H3O2]initial = 0.20 M. We also ignore the contribution
that solvent makes to [H+], so we set this value to zero. These are the values that
go into the first column of concentration table.
Next, we enter the changes in the center column. Since no H+ is present,
some HC2H3O2 must ionize; so let's allow x to equal the number of moles per liter
٩٦
of HC2H3O2 that dissociates to give H+ and C2H3O-. This will increase [H+] and
[C2H3O2-] by x and decrease [HC2H3O2] by x. The equilibrium concentration are
then found in the last column for our table.
As before, we look at ka and see that x will probably be small. We will therefore
assume that 0.10 + x ≈ 0.10 and 0.20 – x ≈ 0.20. substituting into the expression
for ka gives
x 0.10 1.8 X 10 5
0.20
x 3.6 X 10 5
Now let's check our assumption. If 3.6 X 10-5 is added to 0.10 and the result
rounded to the correct number of significant figures, the sum is 0.10. If 3.6 X 10-5
is subtracted from 0.20, the difference is 0.20 when rounded correctly. This means
that we were correct in our assumption that x was small compared to 0.10 and
0.20. the equilibrium concentrations are therefore
Solution:
HC2 H3O2 H
[C 2 H 3 O 2 ] Ka
The H+ concentration when the pH is 5.70 is
٩٧
ACID-BASE TITRATIONS
0.10 mol -3
X 25 mL 2.5 X 10 mol of H
1000 mL
When 10 ml of the 0.10 M NaOH is added, we in fact have added
0.10 mol -3
X 25 mL 2.5 X 10 mol of OH
1000 mL
The neutralization reaction
H OH- H 2O
Table: Titration of 25 ml of 0.1 M HCl with a 0.1 M NaOH solution
Volume volume Total Moles Moles of OH- Molarity of Ions in pH
of HCl of volume of H+ Excess
(mL) NaOH (mL)
(mL)
25.00 0.00 25.00 2.5 X 10-3 0 0.10 (H+) 1.00
25.00 10.00 35.00 2.5 X 10-3 1.0 X 10-3 4.3 X 10-2 (H+) 1.37
25.00 24.99 49.99 2.5 X 10-3 2.499 X 10-3 2.0 X 10-5 (H+) 4.70
25.00 25.00 50.00 2.5 X 10-3 2.50 X 10-3 0 7.00
25.00 25.01 50.01 2.5 X 10-3 2.501 X 10-3 2.0 X10-5 (OH-) 9.30
25.00 26.00 51.00 2.5 X 10-3 2.60 X 10-3 2.0 X10-3 (OH-) 11.30
25.00 50.00 75.00 2.5 X 10-3 5.0 X 10-3 3.3 X10-2 (OH-) 12.52
٩٨
The molar concentration of H+ is now
Solution:
N of KOH = n H = 1 X 0.06 = 0.06 N
N of H2SO4 = NH = 2 X 0.025 = 0.05 N
Na X Va = Nb X Vb
0.05 X Va = 0.06 X 525
Va = 630 mL
Therefore, the pH of neutralized solution = 7
Acid-Base Indicators
Indicators are usually weak organic acids or bases that change color over a
range of ph values. Not all indicators change color at the same ph, however.
If we denote an indicator by the general formula HIn, we have the
dissociation reaction
٩٩
H In H + + I n-
C O 2 + 2H 2 O H 2 CO 3
CO 2 H 3O + + H C O 3-
١٠٠
Metabolic acidosis
Can be caused by uncontrolled diabetes mellitus, diarrhea, aspirin overdose and
after heavy exercise.
Metabolic alkalosis
Caused by prolonged vomiting, excessive use of bicarbonate for treating an upset
stomach.
١٠١