Chapter 1:



Chemistry: Is the science that describes the chemical and physical properties of matter.

Matter: Any thing that has mass and occupies space.


Properties of matter

Physical properties: Are properties exhibited by matter in the absence of any change in chemical composition.[color, density, melting]

Physical changes: Are the changes in the physical properties.

Physical states of matter: Gas, Liquid, and Solid


Chemical properties: Are properties exhibited by matter as it undergoes change in

Chemical composition.

Chemical changes: Are the changes in the chemical properties.


Properties of matter can be classified according to whether they depend on the amount to:

1)      Intensive properties: they do not depend on the amount of matter.[Temp. Density, melting,..]

2)      Extensive properties: they depend on the amount of matter.[mass, volume,..]


Classifications of Matter


Matter can be classified to: Pure substances (elements and compounds) and mixtures (homogeneous and heterogeneous).(*)


Law of constant composition: The elemental composition of a pure compound is always the same.


Separation of mixtures: filtration, crystallization, distillation (simple and fractional)(*), and chromatography.


Units of measurements: International System (SI), Metric, And English.



SI units

Metric units


Kilogram (kg)

Gram (g)


Meter (m)

Meter (m)





Kelvin (K)



Cubic meter (m3)

Liters (L)

Larger and smaller units (prefixes)(*)

A) Basic Units:

1-Mass: kg, g

2-Length: m

3-Time: s

4-Temperature:Celsius (centigrade)(C), Kelvin(K), Fahrenheit(F)(*)

K=C+273, C=K-273, F=1.8C + 32, C=[F-32]/1.8


B) Derived unites:

1-Volume: (length)(width)(height). Solid (cm3), Liquids (mL), Gas (L)

2-Density: is mass per unit volume. D=m/v. Solid (g/cm3), Liquids (g/mL), Gas (g/L)

Example1: What is the density of 3.5 g solid that has a volume of 2.6 cm3.


Example2: What is the volume of 4.75g of a liquid that has a density of 0.800 g/mL.



Uncertainty in measurements


Exact numbers: are umber obtained from things that can be counted, and by definition.


Inexact numbers: obtained from measurements.


Precision: is how closely individual measurements agree with each other.(*)



Accuracy: is how closely a measured value agrees with the correct value.(*)



Significant figures: all figures that have a meaning and value in the number.

        Determining the number of significant figures in a number.


        Exponential (Scientific) Notation: is used to remove the ambiguity in some numbers.



Significant figures in calculations

1-     In multiplication and division the result must be reported with the same number of s.f as the measurement with the fewest s.f.


2- In addition and subtraction the result cannot have more digits to the right of the decimal point than any of the original numbers.


Dimensional analysis


Conversion factor: is a fraction whose numerator and denominator are the same quantity expressed in different units. Given[desired/given]


1)      Express 3.72 yd in mm, cm, m.


2)      Calculate the number of meters in 0.200 mi


3)      Convert 45 cm3 to in3


4) Suppose your automobile gas tank holds 18.0 gallons and the price of gasoline is $0.293 per liter. How much would it cost to fill your gas tank?


5) If the density of a piece of wood is 0.58 g/cm3. What is the density in lb/in3



Atoms, Molecules, and Ions


Atom: Is the smallest particle of an element that has the chemical properties of that element (Dalton' theory).


Element: is the substance that consists of atoms having the same chemical properties.


Names and symbols of elements:(table in the cover page)


The modern view of atomic structure: Atoms consist of three fundamental particles, electrons, protons, and neutrons.


Atomic number [Z]: is the number of protons in the nucleus.

Mass number [A]: is the sum of the number of protons and neutrons.

Isotopes: are different masses of the same element that have different number of neutrons.


Nuclide: is an atom of specific isotope.

Nuclide symbol: AZE

Some of the isotopes of carbon

Nuclide symbol

# of protons

# of neutrons

Mass number

# of electrons


















The periodic table

Is made of groups (vertical) and periods (horizontal) of elements according to their properties.

        Alkali metals.

        Alkaline earth metals.

        Transition metals.


        Noble gases (inert).

Elements in the periodic table also can be classifies to: Metals, Nonmetals, and Metalloids.

Molecules and molecular compounds


Molecule: is made of two or more atoms tightly bound together.(diatomic, polyatomic)(2.16)


Chemical formula: show the kinds of elements and their ratio (H2O)


Molecular formula: show the actual numbers and types of atoms in the molecule.


Empirical formula: shows the relative number of atoms of each type in a molecule.

Structural formula: shows which atom is attached to which within the molecule.(2.17)


Ions and ionic compounds

Ions: are atoms or group of atoms that gained or lost electrons.(cations and anoins)


A) Cations: Na+, Ca2+, NH4+

B) Anions: NO2-, X-, CN-, OH-,


Ions can be classified to: Single ions (Na+, Cl-,..) and poly-atomic ions (SO4-2, NO2-,...).


The charge on single ions: (figure 2.22, page 53)


The charges on poly-atomic ions:

X- (Halogen), CN- (Cyanide), OH- (Hydroxide), CH3COO- (Acetate), NO3 (Nitrate), NO2- (Nitrite), S2- (Sulfide), SO3-2 (Sulfite), SO4-2 (Sulfate), HSO4- (Hydrogen sulfate or bisulfate), H3O+ (Hydronium), ClO3- (Chlorate), CO3-2 (Carbonate), HCO3- (Hydrogen carbonate or bicarbonate), PO4-3 (Phosphate), HPO4-2 (Hydrogen phosphate or biphosphate), H2PO4- (dihydrogen phosphate), NH4+ (Ammonium), CrO4-2 (Chromate), Cr2O7-2 (Dichromate), MnO4- (Permanganate), Ag+ (Sliver), Zn+2 (Zinc)


Chemical bonds: the electrostatic attractions between atoms in a molecule.


1-     Ionic bond: (results from the reaction between metals and nonmetals)

Ionic compounds: Made of metals and nonmetals with ionic bond.


2-     Covalent bond: (results from the reaction between from nonmetals and nonmetals)

Molecular compounds: Made of nonmetals and nonmetals with covalent bond.


Electronegativity: is a measure of the ability of an atom to attract electrons


Oxidation state (number): The number of electrons gained or lost by an atom when it forms the compound.




Chemical compounds

Binary Ternary


Ionic binary Covalent binary Ternary acids Ternary salts


Salts Pseudo acids



1. Naming binary compounds: Are made of two elements.

A)     Binary ionic:

1- SALTS: contain metal and a nonmetal.

        Name cation first.

        Add "ide" suffix to the stem of the anion.

        Stems: B(bor), C(carb), N(nitr), O(ox), S(sulf), H(hydr), Cl(chlor).



Examples: KBr potasium bromide, NaH sodium hydride

CaCl2 calcium chloride, Li2O lithium oxide.

        If the oxidation number of the metal is variable we indicate that by the Roman numerals in parentheses.(Stock system)


Examples: Cu2O copper(I) oxide (cuperous). CuO copper(II) oxide (cuperic)

FeO iron(II) oxide (ferrous). Fe2O3 iron(III) oxide (ferric).


2-     PSEUDO:

They contain more than two elements, which one ion or more consist of more than one element but behave as simple ion.[OH-, CN-, NH4+]


3-     ACIDS: They contain hydrogen and a nonmetal.

        Pure acids (gas phase) are named as binary salts.

        Aqueous solutions are named by adding the "hydro" prefix to nonmetal name and the suffix "ic" followed by the word "acid".


Examples: HF (hydrofluoric acid), HCl, HBr, HI, HCN.


B)     Binary covalent: are made of two nonmetals.

        Name the less electronegative first.

        Add "ide" suffix to the stem of the more electronegative.

        The elemental proportions are indicated by using a prefix for both elements: mono, di, tri, tetra, penta, hexa, hepta, octa, nona, deca.


Examples: SO2 sulfur dioxide. CS2 carbon disulfide. Cl2O7 dichlorine heptoxide


2. Naming Ternary compounds:

A)    Ternary Acids (oxacids): contain hydrogen, oxygen, and a nonmetal.

        Give the stem from the nonmetal name.

        Add"ous" for lower oxidation, "ic" for higher oxidation followed by the word "acid".


Examples: H2CO3 carbonic acid. H2SO4 sulfuric acid. HNO3 nitric acid. H3PO4 phosphoric acid, HClO3 chloric acid


B)     Ternary salts: contain cation, oxygen, nonmetal.

        The cation is named first.

        The name of the anion is based on the name of the ternary acid from which it is derived. "ic" "ate". "ous" "ite".


Examples: NaHCO3 (sodium hydrogen carbonate), CaCO3, CuNO3, FePO3











Chemical reaction: is changing one or more substances into one or more different substances.(regrouping of atoms).


Chemical equation:(3.3) CH4 + 2O2 CO2 + 2H2O


Balancing equations:

2Al + 6HCl 2Al Cl3 + 3H2

4Al + 3O2 2Al2O3

3Mn + 2O2 Mn3O4

2Al + 3H2SO4 Al2(SO4)3 + 3H2

3Mg + 2H3PO4 mg3(PO4)2 + 3H2


Patterns of chemical reactivity:

1-     Reaction of metals with water (Displacement reactions):

2K + 2H2O 2KOH + H2


2-     Combustion reactions(oxidation-reduction reactions):

CH3CH3 + 3O2 2CO2 + 2H2O


3-     Combination reactions:(3.6) A + B C

2Mg + O2 2MgO


4-     Decomposition reactions: C A + B

CaCO3 CaO + CO2

Atomic and molecular weights

Atomic weight (AW): is the average atomic mass of all isotopes of the element expressed in (amu).

Molecular weight (MW): is the sum of the masses of all atoms in the molecule.

Formula weight (FW): is the same a the MW, but usually used for ionic compounds.



Percent composition from formula:

% Element = [(AW of element).n]/(MW of molecule) x100


Example: calculate the percent composition of HNO3 by mass?

%H = (1.0 g of H)/(63.0 g HNO3) x100 = 1.6 %

%N = (14.0 g of N)/(63.0 g HNO3) x100 = 22.2%

%O = (48.0 g of O)/(63.0 g HNO3) x100 = 76.2



The Mole

Mole: is the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12.0 g of carbon-12.


Avogadro's number: (NA) is the number of atoms in 12.0 g sample of C-12. 6.022X1023.


Molar mass (MM): is the mass of one mole of atoms in grams.


Mole=mass(g)/MW. Mole=MW, Mole=AW, mole=6.022x1023, MW=6.022x1023


Inter-converting Masses, Moles, and number of particles:(3.11)



Mass(g) Moles Number of molecules or atoms



1-How many moles of atoms does 245.2 g of iron metal contain?

2-How many atoms are contained in 4.390 moles of iron atoms?

3-How many moles of NaHCO3 are present in 508 g of this substance?

4-Calculate the mass, in grams, of 0.433 mol of calcium nitrate?



Empirical formulas from analysis


1. A compound contains 50.1% sulfur and 49.9% oxygen by mass, What is the

Empirical formula?



2. A sample contains 6.072 g Na, 8.474 g S, and 6.336 g of O. What is the empirical formula?


Molecular formula from empirical formula


Molecular formula = empirical formula x n

Where (n) = MW of molecular formula/MW of empirical formula.


3. Ethylene glycol (antifreeze), is composed of 38.7% C, 9.7% H, and 51.6% O. What is its molecular formula if its MW = 62.1 g/mol.


Combustion analysis


Exapmle:A 0.1014g sample of purified glucose(C,O,H) produced 0.1486g CO2 and 0.0609g of water. Determine the simplest and molecular formulas if the MW-180g/mol.


Quantitative information from balanced equations


1. From the equation: CH4 + 2O2 CO2 + 2H2O A. How many moles of water could be produced by the reaction of 3.5 moles of methane with excess oxygen?


2. What mass of oxygen is required to react completely with 24.0g of CH4?



3. What mass of CH4 produces 3.0x1023 H2O molecules?


4. How many grams of O2 are required to react with 0.300 moles of Al?

Al + O2 Al2O3


Limiting reactant: Is the reactant that used up first in chemical reaction.


Recipe of making a cheese sandwich:


2Bd + Ch Bd2Ch Where, Bd = slice of bread, and Ch = slice of cheese.


1. What is the maximum mass of Ni(OH)2 that could be prepared by mixing two solutions that contain 26.0 g of NiCl2 and 10.0 g of NaOH?

NiCl2 (129.7) + NaOH (40) Ni(OH)2 (92.7) + NaCl



Percent yield = [actual yield/theoretical yield] x 100


1.A 15.6g sample of C6H6 (78) is mixed with excess HNO3 we isolate 18.0g of C6H5NO2(123). What is the percent yield?




2. If 150g of carbon reacts with 250 g of chlorine and reaction goes in 87% yield. How many grams of CCl4 (154) are produced?





















Chapter 4

Aqueous reactions and Solution Stoichiometry


Solution: is a homogeneous mixture of two or more substances.(solute and solvent)

Aqueous solution: solution that has water as the solvent.


Properties of solutes in aqueous solutions

Ionization: The separation of the ionic molecules to their ions in aqueous solutions.


*The degree of ionization of a substance can be measured by the ability of its aqueous solution to conduct electricity:


Strong Electrolyte: ionizes completely in dilute aqueous solution.(soluble salts, strong acids and bases)

Weak electrolyte: ionizes slightly in dilute aqueous solution. (weak acids and bases)

Non-electrolyte: Does not ionize in dilute aqueous solution. (covalent or molecular compounds)


Precipitation reactions: reactions that yield solids.

Solubility guidelines: (Table 4.1)


Metathesis (exchange) reaction: single and double displacement.


Ionic equations and net ionic equations:


Acid-base reactions:


Bronsted-lowery definition for acids and bases:


Strong acids and bases: they ionize completely

Weak acids and bases: they ionize slightly


Neutralization reaction: is the reaction between an acid and a metal hydroxide that produces water and salt.

Acid-base reactions with gas formation: Na2S + HCl H2S(g) + NaCl


Oxidation-Reduction reactions: Oxidation is loss of electrons and Reduction is gain of electrons.

Oxidation numbers (states): The rules are in page 121.

1- The oxidation number of any free, uncombined element is zero; Fe, H2, O2,..

2-     The oxidation number of ions is the charge on the ion.

3-     The sum of the oxidation numbers of all atoms in the neutral compound is zero, and for ions is the charge on the ion.


Oxidation of metals by acids and salts


Activity series: is a list of metals arranged such that a metal can reduce the cations formed by any of the metals below it in the list.(Table 4.5)





Solution: is a homogeneous mixture of two or more substances. It consists of solute (solid, gas, liquid) and solvent (aqueous or non-aqueous).


Concentration: is the amount of solute present in a given mass or volume of solution.


Molarity: is the number of mole of solute per liter of solution.



1. Calculate the molarity of a solution that contains 3.65g of HCl in 2.00L of solution.



Preparing solutions:

A) From solid:


1. Calculate the mass of Ba(OH)2 required to prepare 2.50L of a 0.06000M solution of Ba(OH)2.



B) From Liquids: Dilution


1. Calculate the volume of 18.0M H2SO4 required to prepare 1.00L of a 0.900M solution of H2SO4.



Calculating the amount of solute in a solution:


1. Calculate the number of moles of H2SO4 and its mass in 500ml of 0.324M H2SO4 solution.



Solution Stoichiometry and chemical analysis:




1. Calculate the volume of .0324M solution of H2SO4 required to react with 2.792g of Na2CO3.



Titration: is the process in which a solution of one reactant (titrant) is carefully added to a solution of another reactant.

Standard solution: solution with an accurately known concentration.

Indicator: a substance that gives different color when pH changes.

Equivalence point: the point at which stoichiometrically equivalent amounts of acid and base have reacted.


1. Find the volume of 0.505M NaOH solution required to react with 40.0ml of 0.505M H2SO4.


2. What is the molarity of 40.9ml of Ba(OH)2 that is require to react with 38.3ml of 0.108M HCl.

Ba(OH)2 + 2HCl BaCl2 + 2H2O




































Chapter 5:



Is the study of the energy change that accompanies physical and chemical processes.


System: is part of the universe.

Surroundings: Is the rest of the universe.

State functions: P, V, T, E,..


The first law:

Energy is neither created nor destroyed in ordinary chemical reactions.

E = Ei Ef


Units of energy: Joule, Calorie, 1cal= 4.184 j

Calorimeters: Are devises used to measure the heat of the reaction.

(q)rxn = (specific heat)(mass)(DT) where DT = (Ti Tf)


Example: How much heat is required to raise the temperature of 250g of water from 22C to 98C. Heat capacity of water is 4.18j/g.K.


Enthalpy (H): is the heat (q) absorbed or released by the reaction under constant pressure.


Change in enthalpy (DH): (Hi Hf)


Calculations of DH

1) Using Hesss Law: DH for a multistep reaction will be equal to the sum of the enthalpy change for the individual steps.

Example 1: Calculate DH for the reaction: C + 2H2 CH4


Where: C + O2 CO2 DH= -393.5 kJ/mol

H2 + 1O2 H2O DH= -285.8 kJ/mol

CO2 + 2H2O CH4 + 2O2 DH= 890.3 kJ/mol


Example 2: Calculate DH for the reaction: 2C + H2 C2H2


Where: C2H2 + 5/2O2 2CO2 + H2O DH= -1299.6kJ/mol

C + O2 CO2 DH= -393.5kJ/mol

H2 + 1/2O2 H2O DH= -285.9kJ/mol


Example 3: Calculate DH for the reaction: NO + O NO2

Where: NO + O3 NO2 + O2 DH= -198.9kJ/mol

O3 3/2O2 DH= -142.3kJ/mol

O2 2O DH= 495.0 kJ/mol

Standard Enthalpy of formation: DHf

The amount of heat absorbed or released in a reaction in which one mole of the substance is formed under standard conditions.


H2 + Br2 2HBr DH= -72.8

1/2H2 + 1/2Br2 HBr DH= -36.4


2) Enthalpy of the reaction from Enthalpy of formation:


DH(reaction) = n DHf (products) + n DHf (reactants)


1. Calculate the DH(reaction) for the following equation if DH for CO2 is -393.51; DH for H2O is -285.83; DH for C3H8 is 103: and DH for O2 is 0.

C3H8 + 5O2 3CO2 + 4H2O (2220 kj/mol)


Foods and Fuels:

Dietary Calorie = 1000 cal = kcal

Carbohydrates 17 kj/g

Protein 17 kj/g

Fat 38 kj/g


Example: 5.11

How much energy is produced by burning 5.0 g of C3H8
























Chapter 6:

Electronic Structure of Atoms


Electromagnetic radiation: is the visible and invisible light which can be described in terms of waves:

The wave length(l ): is the distance between any two adjacent identical points of the wave and is measured in m, nm.

Frequency of the wave(n): is the number of waves passing a given point per unit time, and the units are 1/s, Hz.

C = l.n where C is the speed of light = 3.00x108 m/s

Example: the frequency of violet light is 7.31x1014 1/s, and that of red light is 4.57x1014 1/s. Calculate the wavelength of each. Violet is 410 nm, and for red is 656 nm.

The Energy of the light (E) is given by Planks equation: E=h.n, where (h) is Planks constant = 6.626x10-34 j.s.

Example: Calculate the energy of the red light (656 nm), and the violet light(410 nm). Red is 3.03x10-19 j, and for violet is 4.84x10-19 j.



Bohrs Model of the hydrogen atom:

Spectrum: is the wavelengths of a radiation. Which can be classified to:

A)    Continuous spectrum: when the spectrum contains all wavelengths (rainbow).

B)     Line spectrum: when the spectrum contains only specific wavelength.(H2, Hg, Na,..). Which can be classified to:

1)      Emission spectrum: the bright line spectra.

2)      Absorption spectrum: the dark line spectra.


*Each element displays its own characteristic set of lines in its absorption spectrum (fingerprints).

Bohrs Theory:

1)      electrons can only be in certain discrete orbits.

2)      Electrons can absorb or emit energy in discrete amounts as they move from one orbit to another.

3)      Each line in the absorption spectrum represents the difference in energies between tow energy levels for the electron.

Rydberg equation: is used to calculate the energy of the light emitted or absorbed when an electron moves from one energy level to another.

E = h.n = RH (1/ni2-1/nf2). Where RH is Rydberg constant = 2.18 x 10-18 j.

Example: Calculate the wavelength of light that correspond to the transition of the electron from n=4 to n=2. Is the light absorbed or emitted?


Quantum Numbers

There are four quantum numbers that describe the electronic configuration in the atom and its energy distribution.

1)      The principal quantum number(n):

It describes the main energy levels (shells), it may be any positive integer. N=1,2,3,..

2)      The azmuthal (subsidiary) quantum number (l):

It describes the shape of the region in space an electron occupies (sublevel, or subshell).

*Within each main energy level(n), l may take integral values from zero upto (n-1).

If n=1, then l=0, and when n=2, then l= 0,1


3)      The Magnetic quantum number( ml)

Designates the spatial orientation of an atomic orbital.

*Within each sublevel, Ml may take any integral values from l through zero to +l.

If l=0, then ml=0, and when l=1, then ml=-1,0,+1


4)      The spin quantum number(ms)

Refers to the spin and orientation of the magnetic field of an electron. Ms can take the values of 1/2 or +1/2.


Electron Configuration

Is the way in which electrons are distributed among various orbitals of an atom.

1)      The Aufbau principle: Orbitals are filled so that those of lowest energy are filled first.


2)      The Pauli exclusion principle: A maximum of two electrons may be placed in each orbital, but only when the spins of the electrons are opposite to each other (paired).


3)      Hunds rule: For degenerate orbitals (p,d,f). The lowest energy is attained when the number of electrons with the same spin is maximized


Valence shell: Is the outer shell.

Valence electrons: Are the electrons in the valence shell.


The elements can be classified according to their electronic configuration to:


1)      Noble gases: They are inert, they have 8 electrons in their highest occupied energy level (s&p).

2)      Representative elements: (Group A) They have partially occupied highest energy levels (s&p). They show regular variation in their properties with change in atomic number.

3)      D-transition elements: (Group B) They are filling the (d) orbitals after they filled the highest energy levels. They have variable properties.

4)      Inner-transition elements: (Lanthanides and Actinides) They are filling the (f) orbital.










Chapter 7.

Periodic properties of the Elements


The properties of the elements are periodic functions of their atomic numbers.


1)      Atomic Radii: (size of atoms and ions)

Isoelectronic series: Ions possess the same number of electrons.(The radius of the ions decreases with increasing the atomic number).


2)      Ionization energy: Is the minimum energy required to remove an electron from the atom. First, second, and third ionization energies. Na (first is 496 kj/mol, second is 4560 kj/mol). It measures the ease with which an atom loses an electron.

*Ionization energy explains why some atoms tend to give off one electron while others give off more.


3)      Electron affinity: Is the amount of energy absorbed or released when an electron is added to the atom. It measures the ease with which an atom gains an electron.



Metals, non-metals, and metalloids





























Chapter 8:

Basic concepts of chemical bonding


Lewis symbols: The electrons in the outermost occupied (s) and (p) orbitals (valence shell) are shown as dots or dashes.


The octet rule: Atoms tend to gain, lose, or share electrons to achieve the noble gas configuration in their valence shells (8 e except H & He 2 e).


Chemical bond: are the attractive forces that hold atoms together.

1)      Ionic bond: Results from the transfer of electrons from one atom or group of atoms to another. It gives ionic compounds.

Electron configuration of ions:

Sizes of ions compared to neutral atoms:


2)      Covalent bond: results from sharing electrons between two atoms. It can be a single, double, or triple bond.

Bond polarity and electronegativity:


Electronegativity: Is the ability of an atom in a molecule to attract electrons to itself.


According to electronegativity, the covalent bond can be classified to:

A)    Nonpolar covalent: When the bonding electrons are shared equally between two atoms with the same or similar electronegativity.

B)     Polar covalent: When the bonding electrons are shared unequally between two atoms with large difference in their electronegativity.

In general: H-N, H-O, H-F, and H-Cl are polar. Nonmetal to another different nonmetal is polar. Any thing else is nonpolar.




1-     Write the skeleton of the molecule:

a)      The least electronegative atom is in the center except (H).

b)      Oxygen atoms do not bond to each other except in O2, O3, O2-2, O2-

c)      In the ternary acids, hydrogen usually bonds to oxygen atom only.

d)      For ions or molecules that have more than one central atom, the most symmetrical skeleton possible is used.


2-     Calculate (N), the number of outer (valence) shell electrons needed by all atoms to achieve Noble gas configuration (8 electrons, except H & He are 2 electrons).


3-     Calculate (A), the number of electrons Available in the outer (valence) shell of all atoms and account for ions: (Add an additional electron for each negative charge, or subtract an electron for each positive charge).


4-     Calculate (S), the total number of electrons Shared in bonds. [S = N A]

5-     Place the (S) electrons into the skeleton as shared pairs (bonds).


6-     Place the additional electrons into the skeleton as unshared electron pairs starting with the central atom to fill the octet rule for every atom.


7-     Place any extra electrons on the central atom as unshared pairs.


Examples: CH4, HBrO3, NO3-, C2H4


Resonance Structures (hybrid): Different Lewis formulas for the same molecule or ion.(CO32-).


Formal Charge: Is the charge on individual atoms in the molecule or ion.


Formal Charge = [# of valence electrons in isolated atom]-[# of electrons assigned to the atom in Lewis structure].


Oxidation State: is the algebraic sum of the values of F.C.


Formal charge and stability of molecules: The lower the absolute sum of the formal charges, the more stable the molecule is. (SO4-2, ClO3-, COCl2)




1-      SMALL ATOMS:

        Beryllium (Be) accepts only 4-electrons; Example is BeCl2, The (N)= 4 + 2(8)= 20

        Boron (B) and Aluminum (Al) accept only 6-electrons;

Example is BF3, The (N) = 6 + 3(8) = 30


2-     LARGE ATOMS: The atoms with atomic number equals to 15 and larger can accept more than 8-electrons;

Example 1: PF5

N = 8 + 5(8) = 48

A = 5 + 5(7) = 40

S = 48-40 = 8

Bonds = 8/2 = 4 bonds. Place the four bonds and just add the messing fifth bond. Then complete the octet rule for each atom in the molecule.

Example 2: XeF2

N = 8 + 2(8) = 24

A = 8 + 2(7) = 22

S = 24-22 = 2

Bonds = 2/2 = 1 bond. Place this bond then add the second messing bond and complete the octet rule for all atoms and any extra electrons are placed on the central atom as pairs.

Examples: SF6, XeF4, I3-, C2H2, CH2O


Some Properties of Covalent Bond:


Strength of covalent bond: is measured by the bond enthalpy.


Energy (enthalpy) of covalent bond: is the energy required to break the bond.


Bond length: is the distance between the centers of two atoms.


Bond order: is the number of bonds between two atoms.


Enthalpy of reaction from bond enthalpies: (8.14) DHrxn = SnDH(reactants) - SnDH(products)




































Chapter 9:

Molecular Geometry and Bonding Theories


The valence shell electron pair repulsion theory: (VSEPR)

The best arrangement of a given number of electron pairs about the central atom, is the one that minimizes the repulsion between them.


Electron pair geometry: The spatial arrangement of electron pairs (bonding and nonbonding) in a molecule.[multiple bonds are considered the same as the single bonds]

(Linear - trigonal planar - tertahedral - trigonal bipyrimdal - octahedral)


Molecular geometry: The spatial arrangement of atoms in a molecule.


Molecules with more than one central atom: (acetic acid)


Polarity of molecules

1)      Diatomic molecules: if the bond is polar, then the molecule is polar.


2)      Polyatomic molecules: -if all bonds are nonpolar, then it is nonpolar.

-If one bond is polar, then it is polar.

-If more than one bond is polar, then it depends on M.G.

        Polarity of bonds depends only on electronegativity difference between atoms involved in bonding.

        Polarity of bonds has nothing to do with EG, MG, or the formal charge.

        Polarity of molecules depends only on the polarity of bonds and MG.


Covalent Bonding and Orbital Overlap


Valence bond theory: In chemical bonding the valence atomic orbitals overlap to make molecular orbital.


Kinds of covalent bond:

1)      sigma (s) bond: It forms when two of the following orbitals overlap.(s+s, s+p, p+p head to head)

2)      Pi (p) bond: It forms only when two (p) orbitals overlap side-by-side.


Multiple Bonds:

-All single bonds are sigma bonds

-A double bond consists of one sigma and one pi, while a triple bond is consists of one sigma and two pi.







Orbital Hybridization

Mixing two or more atomic orbitals of central atom to create a new hybrid orbitals.(sp3, sp2, sp, sp3d, sp3d2).

Delocalized pi bonding: when pi electrons resonate between more than two atoms.

Molecular Orbital Theory

The atomic orbitals can combine to form molecular orbitals, where each orbital can have maximum of two electrons with opposite spin. The number of molecular orbitals formed is equal to the number of atomic orbitals that are combined.


1)      Bonding molecular orbital: has low energy and is filled first with electrons

2)      Anti-bonding molecular orbital: has high energy, and is filled next with electrons..


Energy level diagram, and molecular electron configurations for diatomic molecules.


The bond order = [bonding electrons-anti-bonding electrons]


Paramagnetism: When the molecule has unpaired electrons, which is attracted into a magnetic field (O2).


Diamagnetism: When the molecule has no unpaired electrons, which is repelled from a magnetic field (N2).

























Chapter 10:


Properties of gases:

1)      Gases can be compressed into smaller volumes.

2)      Gases exert pressure on their surroundings.

3)      Gases expand without limits.

4)      Gases diffuse into each other.

5)      The properties of gases are described in terms of Temp., pressure, volume, and amount.


The pressure: is force per unit area, P=F/A. The units are lb/in2 or psi.

Atmospheric pressure: is the mass of a column of air 1.0 m2 in cross-sectional area and extending to the top of atmosphere.

Mercury Barometer: is used to measure the atmospheric pressure.

Pressure units: atm, mmHg, Torr.

Standard atmospheric pressure: is the pressure measured at the sea level.(760 Torr)


A] Boyle Law (P-V):

At constant Temp. and mass, the volume is inversely proportional to the pressure.

V a 1/P or P.V = k where k is the proportionality constant.

P1V1 = P2V2

Example: A sample of gas occupies 12L under 1.2atm, what is (V) at 2.4atm.


B] Charless Law (T-V):

At constant pressure and mass, the volume is directly proportional to its absolute temperature.

V a T or V = k.T V1/T1 = V2/T2


Example: A sample, V=117ml at 100C, at what temperature is its volume 234ml?


Standard Temperature and pressure (STP): 0.0 C and 760 Torr.


C] Avogadros Law (n-T):

At constant temp. and pressure, the volume is directly proportional to the number of moles.

V a n or V = k.n V1/n1 = V2/n2


Example: 2 moles of hydrogen gas occupy 3 L in a balloon. What is the volume if another 4 moles were added to the balloon.


The combined gas law: [P1V1/T1] = [P2V2/T2]


Example: A deep sea diver exhales a 15.0ml bubble of air at a depth where the pressure is 12.0atm and the temperature is 8.0C. What is the volume of the bubble at the surface, where the pressure is 770 Torr and the temperature is 20.0C?


The Ideal Gas Equation

Real gas: it has molecular volume, collision between molecules, and attraction-repulsion.

Ideal gas: has none of the above.


Ideal gas equation: (PV = nRT) where R is the universal gas constant

(0.0821 atm.l/mol.K)


Calculations of unknown property of a gas:

1. Calculate the pressure needed to contain 5.29mol of an ideal gas at 45C in volume of 3.45L.

Calculation of number of molecules of a gas:

2. How many gaseous molecules are in a 1.0L container if the pressure is 1.6x10-9Torr and the temperature is 1475K


Application for Ideal gas equation (density and molar mass):


D = [P.MM]/[R.T]



Calculation of MW from density:

1) What is the MW of 1 mole of a gas at Temp. of 273 K that occupies 27.0L, and its density is 1.41g/L

Calculation of density from MW:

2) What is density of carbon tetrachloride at 714 Torr and 125 C?


Gas mixtures and partial pressure


Mole fraction (XA): is the ratio of moles one component to the total number of moles in the mixture.


XA = [moles of A]/[total moles]


Partial pressure (PA): is the pressure of one component in a mixture.


Daltons Law: XA= PA/Pt



1. A mixture of gases: 4.18g CHCl3, and 1.95g C2H6. At 375C and 50ml. What is the total pressure and the partial pressure of CHCl3?



2. What is the mole fraction (x) of each of the gases in a mixture that contains 0.267 atm He, 0.369 atm Ar, and 0.394 atm Xe?




Chapter 11

Effect of Intermolecular Forces on the Properties of Liquids and Solids


        Intramolecular forces: Are the covalent and ionic bonds within compounds.

        Intermolecular forces: are the forces between individual particles of a substance.(atoms, molecules, ions)

1)      Ionic-Ionic Forces: exist in the ionic compounds like NaCl.

2)      Dipole-Dipole Forces: exist between polar molecules like acetone.

3)      Hydrogen Bond: exist between polar molecules when there is a H attached to a strong electronegative atom like (N,O,F,Cl).

4)      London (Dispersion) Forces: exist between nonpolar molecules like CH4.


For the same size molecules: ionic>H-bond>polar>London.


Two factors affect the strength of the intermolecular forces:

        The size: as the size increases, the intermolecular forces increases.(B.P of CH4 161, CH3CH3 89, CH3CH2CH3 44)

        The symmetry: as the symmetry increases, the intermolecular forces increases. (B.P of neupentane 283K, Pentane 309K)



Intermolecular forces and the Liquid state


1)      Viscosity: is the resistance to flow.

2)      Surface tension: is the inward forces that caused by intermolecular attractions.

3)      Evaporation: is when molecules go from liquid phase to gas phase.

4)      Condensation: is when molecules go from gas phase to liquid phase.

5)      Vapor pressure: is the partial pressure of vapor molecules.

6)      Boiling point: is the temperature at which liquids boil.

7)      Boiling point and atmospheric pressure.

8)      Heat of evaporation (condensation): is the heat required to convert the liquid to vapor or the opposite.

9)      Heating curves: is the change in temperature versus heat.


Intermolecular forces and the Solid state


1)      Melting/freezing point: is the temperature at which solids melt or liquids freeze.

2)      Heat of fusion/solidification: is the amount of heat required to melt a solid or solidify a liquid.

3)      Sublimation/deposition: is the direct change of a solid to a gas or the opposite without passing through the liquid state.


Phase Diagrams: are the relationship between pressure and temperature.

Triple point: where the three phases coexist at equilibrium.

Critical point: is the point above which, the substance exists as a fluid.




Melting curve


Critical point

Triple point liquid


solid evaporation curve


sublimation curve gas Temperature





Structures of Solids


A) Crystalline solids: They have well-defined, ordered structure. (salts)


Unit Cell: is the smallest unit of volume of a crystal that shows all characteristics of the crystal pattern.

B) Amorphous solids: They dont have a well-defined, ordered structure. (glass, rubber)


Bonding in Solids

1) Metallic solids: metals are crystal solids in which metal atoms are embedded in a cloud of delocalized valence electrons.


3)      Ionic solids: most salts crystallize as ionic solids with ions occupying the unit cell of the crystal. (NaCl)


4)      Molecular solids: the unit cell of the crystal consists of one molecule or more and the attractions between molecules are (dipole, H-bond, and London forces). (CO2, H2O, CH4,..)



5)      Covalent-Network solids: the crystal consists of one molecule were the atoms bonded by a covalent bond. (Diamond that is made of carbon and quartz that is made of SiO2)










Chapter 13

Properties of Solutions

Solution: is a homogeneous mixture of one or more substances (solute) in another (solvent).

Solutes can be classified to:

a)      Electrolytes: substances that dissolve in water to give solutions that conduct electricity.

b)      Non-electrolytes: substances that dissolve in water to give solutions that don't conduct electricity.


The solubility process is a physical process and it depends on:

a)      The change in energy (enthalpy). DHsol = DH1 + DH2 + DH3

b)      The change in disorder (entropy).


1)      Solubility of solids in liquids: depends on:

        Solvent-solute attractions. (Exo)

        Solvent-solvent attractions. (Endo)

        Solute-solute attractions.(Endo)(crystal Lattice energy)(most important)


NH4NO3 (Endo), Instant ice packs

Na2SO4 (Exo), (Hand warming packs)


2)      Solubility of liquid in liquid(Miscibility): depends on:

        Solvent-solute attractions. (Exo)(most important)

        Solvent-solvent attractions. (Endo)

        Solute-solute attractions.(Endo)


"Like dissolves like"

  • Polar liquids dissolve in polar solvents.
  • Non-polar liquids dissolve in non-polar solvents.
  • Polar liquids don't dissolve in non-polar solvents.


3)      Solubility of gas in liquid: depends on "Like dissolves like"

-HCl gas is polar and dissolves in polar water.

-Non-polar gases don't dissolve in polar water with two exceptions, O2 and CO2


Saturated solutions and solubility

Saturated solution: when the solid and its dissolved ions are at equilibrium.

Super saturated solution: It contains higher than saturation concentration of solute.


Factors that affect solubility

1)      Temperature: depends whether the process is Exo or Endo.(thermal pollution)

2)      Pressure. Increasing the pressure will increase the solubility of gases.




Ways of Expressing Concentrations


1)      Mass percentage, part per million (ppm), and ppb:

% = [mass of solute/total mass of solution] x 100

PPm = [mass of solute/total mass of solution] x 106

PPb = [mass of solute/total mass of solution] x 109


Example: What is the ppm concentration of Pb+2 in a 3.5 g sample of ground water that contains 6.5 mg of Pb+2.


Example: Calculate the % of NaCl solution tha contains 1.50 g of NaCl in 50.0 g of water.


2)      Mole fraction (XA) = # of moles of A/total moles of solution


Example: if a solution contaions 2 moles NaCl, and 1 mole KCl in 3 moles water. What is the mole fraction of NaCl?


3)      Molarity(M)= moles of solute/liters of solution.


Example: What is the molarity of 5.5 g of KCl (74.5 g/mol) in 200 mL solution?


4)      Molality(m)=moles of solute/Kilograms of solvent.


Example 1: What mass of KNO3 (101 g/mol) should be added to 250 g of water to prepare a 0.200 molal KNO3 solution?


Example 2: A commercial bleach solution contains 3.62% NaOCl in water. Calculate (a) the molality and (b) the mole fraction of NaOCl in solution?


Example 3: If the density of a solution of 5.0 g of toluene and 225 g of benzene is 0.876 g/mL, Calculate the molarity.



The Relationship Between Concentrations


1) Mass % = Mass(solute) Mass(solution)

Mole=Mass(solute)/MW Mole=Mass/MW


2) Mole Fraction = Mole(solute) Mole(solution)



3) Molality = Mole(solute) Kg(solvent)



Colligative properties


When a solute is added to a solvent it affects some of its physical properties and theses properties are called the colligative properties.

-         Colligative properties depend only on the number of solute particles present in solution, and not on their chemical composition.


1)      The lowering of the vapor pressure of the solvent:

The perssure of a nonvolatile solute will lower the vapor pressure of the solvent.

Raoult's Law: The vapor pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent in the solution. P*

PA = X.(solvent).P*A


P*A = Vapor pressure of pure solvent. (Vapor

PA = Vapor pressure of solvent in solution. Pressure)



0.0 X 1.0


Example: Calculate the vapor pressure of water at 35.0 C in a solution prepared by dissolving 10.0 g of glucose (C6H12O6) in 150 g water.(P* of water is 42.2 Torr.


2)      The elevation of the boiling point of a solvent:

The presence of a nonvolatile solute will raise the boiling point of the solvent.

a) Nonelectrolyte: Elevation of the boiling point = DTb = Kb.m,

Where is the boiling point constant, (m) is the molality.

b) Electrolyte: Elevation of the boiling point = DTb = Kb.m.i,

Where (Kb) is the boiling point constant, (m) is the molality, and (i) is the number of particles.


Example1: What is the boiling point of 150 g water when it has 3.0 g of glucose.

DTb = 0.51( 0.057 C

The boiling point = 100 + 0.057 = 100.057 C


Example2: What is the boiling point of 150 g water when it has 3.0 g NaCl.

DTb = 0.51( = 0.347 C

The boiling point = 100 + 0.347 = 100.347 C

3)      The depression of the freezing point of a solvent:

The presence of a nonvolatile solute will decrease the freezing point of a solvent.

a) Nonelectrolyte: depression of the freezing point = DTf = Kf.m,

Where (Kf) is the freezing point constant, (m) is the molality.

b) Electrolyte: depression of the freezing point = DTf = Kf.m.i,


Example: What is the freezing point of 150 g of water when it has 3.0 g glucose.

DTf = 1.86( mol/kg) = 0.21 C

The freezing point = 0.0 - 0.21 = -0.21 C

4)      Osmosis: is the tendency of a solvent to flow through a semipermeable membrane into a more concentrated solution.

Osmotic pressure (P)= R.T.M.i


Example: What is the osmotic pressure of 0.15 M solution of NaCl at 20 C

Osmotic pressure (P)= R.T.M.i = (0.0821L.atm/mol.K)(293K)(0.15 mol/L)(2)= 7.22 atm

Some applications of osmosis: 1) Trees. 2) Reverse osmosis.


Determination of Molar Mass (MW):

Example#1: Camphor melts at 179.8 C, Kf = 40.0 C/m. When 0.186 g of an organic substance is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7 C. What is the MW of the solute?

Molality (m) = DTf / Kf =3.1/40=0.0775 mol of organic/kg camphor.

Moles of organic = = (0.0775 mol/kg)(0.02201kg) = 0.00171 moles

MW = g/mol = (0.186 g)(0.00171 mol) = 109 g/mol


Example#2: A sample of 2.05 g of polystyrene was dissolved in toluene to form 100 mL solution. The osmotic pressure was 1.21 kpa at 25 C. Calculate the MW.


M= P/RT = 1.21 kpa(1 atm/101.32 kpa)/0.0821)(298 K) = 4.88 x 10-4 mol/L

Moles of polystyrene = 0.488 mol/L(0.100 L) = 4.88 x 10-5

MW = 2.05 g/4.88 x 10-5 = 4.2 x 104 g/mol





Are solutions in which the size of dispersed particle of solute is 10-2000 A.(milk)


-Colloids can be distinguished from regular solution by their Tyndall effect.


Tyndall effect: scattering of light by colloidal particles.


Hydrophilic and hydrophobic colloids: (protein-fat in milk)


Hydrophobic colloids: can be stabilized by adding hydrophobic groups to their surface, like adding sodium stearate to oil suspended in water to give emulsion. Sodium stearate is called emulsifying agent. Biles from gallbladder emulsify fats.


Removing colloid particles: coagulation using; heat-salt-acids.







Chapter 15.

Chemical Equilibrium


Chemical equilibrium: The rates of the forward and reverse reactions are equal.

AA + bB cC + dD


The equilibrium constant: Keq

Is the ratio of the molar concentration of the products to the conc. Of the reactants at equilibrium raised to the power of their coefficients.

N2 + 3H2 2NH3

Keq = [NH3]2/[N2][H2]3

Keq, in terms of pressure:

2O3 3O2 Keq = (pO2)3/(pO3)2

*Keq depends only on temperature.


The magnitude of Keq:

Keq > 1 [products] > [reactants]

Keq < 1 [products] < [reactants]

Keq = 1 [products] = [reactants]


The relationship between Keq of forward and reverse:

Keq(f) = 1/Keq(r)

*Keq of a reaction that has been multiplied by a number is Keq raised to a power equal to that number. Br2 + Cl2 ↔ 2BrCl Keq = 0.42


Homogeneous equilibrium: all substances are in the same phase.

Heterogeneous equilibrium: substances are in different phases.

"If a pure solid or a pure liquid is involved in heterogeneous equilibrium, its conc. is not included in the Keq expression".


CO2(g) + H2(g) CO(g) + H2O(l)

ZnO2(s) + 2CO(g) Zn(s) + 2CO2(g)

NH3(g) + H2O(l) NH4OH(aq)

CH3COOH(aq) + CH3OH(aq) CH3COOCH3(aq) + H2O(l)


Calculating equilibrium constant: (15.7, page 589)

Applications of Keq:

1)      Predicting the direction of the reaction(reaction Quotient)

Q=Keq, Q>Keq, Q<Keq. (15.9, page 591)

2)      The calculation of equilibrium concentrations. (15.10, page 592)


Factors that affect equilibrium:

1)      Le Chatelier's Principles

2)      Change in volume and pressure. (Increase in pressure shifts the reaction in the direction that produces the smaller number of moles).

3)      Change in temp. 4) Catalyst.

Chapter 16.

Acid-base Equilibria


Bronsted-Lowry definition for acids and bases: is based on proton transfer reaction.

Acid: is proton donor.

Base: is proton acceptor.


Conjugate acid-base pairs.


Strength of acids and bases:

1-     Strong acids and bases: they ionize completely in dilute aqueous solution.

2-     Weak acids and bases: they ionize slightly in dilute aqueous solution.


The auto-ionization of water: ion product of water, Kw = [H][OH] = 1.0 x 10-14

Example: What is the [OH] if [H] is 3.0 x 10-8 M


The pH scale: is negative Logarithm [H+], pOH

PH + pOH = pKw = 14


1)      What is the pH of a sample of lemon juice where [H] = 3.8 x 10-4 M.

2)      Calculate the pH of a solution in which [OH] is 0.01M

3) What is the pH of a 0.040 M solution of HClO3

4) What is the pH of a 0.50 M solution of NaOH

5) If the pH of a solution is 3. What is the [H].


Weak acids: acid-dissociation constant, ka

HA + H2O A- + H+ Ka = [A-][H+]/[HA]


1) 3% of 0.45 M acetic acid solution is ionized. Calculate the ka.

3)      Calculate the [H+] in 0.20 M of HCN. Ka = 4.0 x 10-10


Polyprotic acids: H2SO4, H3PO4


Weak bases: base-dissociation constant, kb

A- + H2O HA + OH-

Example: Calculate the [OH] for 0.50 M aqueous ammonia. Kb = 1.8 x 10-5


The Relationship between ka and kb: ka x kb = kw. And pka + pkb = pkw.

Example: Calculate the kb and pkb for F- in HF if the ka of HF is 6.8 x 10-4.


Acid-base properties of salt solutions

When salts dissolve in water they completely ionize. Their ions react with water to produce either H+ or OH- (hydrolysis)

BA A- + H2O HA + OH- NaCl Cl- + H2O HCl + OH-

B+ + H2O BOH + H+ Na+ + H2O NaOH + H+

Salts can be classified according to the pH of their solutions to: Neutral, acidic, and basic.

Acid-base behavior and chemical structure


Factors that affect acid strength:

1- The electronegativity of the atom that is attached directly to the hydrogen.

3-     The size of the atom that is attached directly to the hydrogen.


A)    Binary acids: within a group, the size dominate; HI>HBr>HCl>HF

Within a period: the electronegativity dominate; HF>H2O>NH3>CH4


B)     Oxyacids: As the electronegativity of the nonmetal atom increases, the acidity increases; HClO3>H2SO4>H3PO4.


Lewis acids and bases

Lewis acid: is an electron pair acceptor.

Lewis base: is an electron pair donor.

































Additional aspects of aqueous equilibria

The Common-Ion effect: It describes the solution that produces the same ion by two different compounds.



CH3COONa + H2O Na+ + CH3COO-


Buffer solution: it consists of a weak acid or base and its salt. It resists the change in the pH upon addition of small amounts of acid or base.



1)      Calculate the [H+] and the pH of a solution that is 0.1 M in CH3COOH and 0.20 M in CH3COONa.

2)      Calculate the [OH-] and the pH of a solution that is 0.20 M in aqueous ammonia and 0.10 M in NH4Cl.


Henderson-Hasselbalch equation: Is a short cut to calculate the pH of buffer solutions.

A)    Acid: pH = pka + log[conjugate base or salt]/[acid]

B)     Base: pOH = pkb + log[conjugate acid or salt]/[base]


Preparing Buffer Solution:

Example: calculate the ratio of the concentrations of CO32- and HCO3- ions needed to achieve buffering at pH 9.5. Pka of H2CO3 = 10.25.


Buffer capacity: Is the quantity of added H+ or OH- that the buffer solution can tolerate without exceeding a specified pH range.


Addition of strong acids or bases to buffers:

Example: What is the pH of 250 mL buffer solution that contains 0.225 M HPO42- and 0.330 M H2PO4-. a) What is the pH if 0.40 g of solid NaOH is added? b) What is the pH after the addition of 0.040 M HCl?

H2PO4- + H2O HPO42- + H3O+ pka = 7.21


Acid-Base Titrations:(measuring the Ka & Kb)

**At half equivalence point the [HA] = [A-], and pH = pKa

1)      Strong acid-Strong base titrations:

Example: Calculate the pH when 49 mL of 0.100 M NaOH solution is added to 50.0 mL of 0.100 M HCl solutions.


2)      Weak acid-Strong base titrations:

Example: Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH. Ka = 1.8 x 10-5.


Get the moles before and after, then convert back to molarity with total volume of 95 mL

Ka = [CH3COO-][H+]/[CH3COOH]



Solubility Equilibria

Solubility product constant:

Zn(OH)2(s) Zn+2 (aq) + 2 OH(aq)


Keq = [Zn][OH]2/[Zn(OH)2(s)] since the activity of the solid is one, then;

Ksp = [Zn][OH]2 = 4.7 x 10-17

Ksp for AgCl = [Ag][Cl] = 1.8 x 10-10

Molar solubility: is the number of mole that dissolve to give one liter of saturated solution.

Gram solubility: is the number of grams that dissolve to give one liter of saturated solution.


1) Calculate the molar and gram solubility for BaSO4. (Ksp = 1.2 x 10-10)


BaSO4 Ba + SO4

x x

Ksp = [Ba][SO4]= x2 =1.2 x 10-10

X = 1.1 x 10-5 = molar solubility = [Ba] = [SO4]

Gram solubility = [1.1 x 10-5 mol/L. 233 g/mol] = 2.55 x 10-13 g/L


2) Calculate the Ksp for AgCl, if one liter of saturated solution contains 1.9 x 10-3 g of dissolved AgCl.

AgCl Ag + Cl

Convert the grams/L to moles/L = 1.9 x 10-3 g/L x mole/143 g AgCl = 1.3 x 10-5 mol/L

Ksp = [Ag][Cl] = [1.3 x 10-5]2 = 1.810-5


3)      What is the solubility in g/L for Zn(OH)2 Ksp = 4.5 x 10-17

Zn(OH)2 Zn + 2 OH

X 2X

4)      Ksp = [Zn][OH]2 = [x][2x]2 = 4x3 = 4.5 x 10-17

X = 2.24 x 10-6 M = [Zn] , [OH] = 2Zn = 4.5 x 10-6 M

Gram solubilty = [2.24 x 10-6 mol/L . 99 g/mol] = 2.2 x 10-4 g/L


Factors that affect solubility:

1-     The common ion effect; the solubility of a slightly soluble salt is decreased by the presence of a second solute that furnishes a common ion.

Example: a) Claculate the molar solubility of MgF2 in pure water. B) Calculate the molar solubility of MgF2 in 0.1 NaF solution. Ksp(MgF2) = 6.4 x 10-9

2-     The pH:

* Insoluble hydroxids can be made more soluble by reducing the pH

Zn(OH)2 Zn + 2OH

HCl Cl + H+

  • The solubility of slightly soluble salts containing anions of weak acids increases as the pH decreases.


Na2CO3 2Na+ + CO3-2 H2CO3


Precipitation and separation of ion:

The precipitation process can be predicted by comparing the concentration of ions (Qsp) at any time and their concentration at equilibrium (Ksp)

Qsp< Ksp more solid will dissolve

Qsp = Ksp at equilibrium

Qsp > Ksp precipitation occurs

Example: If 100 mL of 0.00075 M sodium sulfate, and 50 mL of BaCl2, solutions are mixed, Will a precipitate form? Ksp for barium sulfate is 1.1x10-10.


Selective precipitation of ions:

Example: Solid silver nitrate is slowly added to a solution that is 0.001 M each in NaBr, NaCl, NaI. Calculate the [Ag+] required to initiate the precipitation of each of the silver halides.Ksp(AgCl)=1.8x10-10, (AgBr)=3.3x10-13, (AgI)=1.5x10-16































Summary of Chapters 16&17


1) Strong Acid/base: Complete ionization



pH + pOH = 14


2) Weak acid/base: Slight ionization

[H+] [Acid], [OH-] [Base]


ACID: use Ka to find [H+] then pH

HA A- + H+

Ka = [A-][H+]/[HA]


BASE: use Kb to find [OH-] [H+] then pH

Z(OH)2 Z+ + 2 OH-

Kb = [Z+][OH-]2/[Z(OH)2]


3) Common ion and Buffer solution: Made of weak acid/base and their salts.

HA A- + H+


{When [salt or conjugate base] is added}


To calculate pH use Henderson-Hasselbalch Equations.


4) PH of Titrations:


a) Strong acid/strong base:

Find the moles of H+ and the OH- then find the one that exist in excess and calculate that excess amount and converted to Molarity. From the molarity find the pH or the pOH.


b) Weak acid with strong base:

Find the left over moles of the acid after neutralization with base, and then find the moles of conjugated base of the acid. Convert the moles to Molarity, then find the [H+] from the ka expression of the acid.


5) Insoluble salts: Use Ksp to find Molar solubility and Gram solubility

ZA Z+ + A-

Ksp = [Z+][A-]


ZA2 Z+ + 2A-

Ksp = [Z+][A-]2


6) Common ion and Solubility:


a)     Adding a common ion will decrease solubility by shifting the reaction to left.


ZA Z+ + A-

Z+ A-


b)    Removing a common ion will increase solubility by shifting the equilibrium to right.


Z(OH)2 Z+ + 2 OH-



c) Predication of precipitation: compare between the Ksp and the Qsp as follows;


Qsp < Ksp (more solid will dissolve)

Qsp = Ksp (reaction at equilibrium)

Qsp > Ksp (precipitation occurs)


d) Selective precipitation of ions: use the Ksp expression to calculate the concentration of the added ion that will initiate the precipitation of the desired ion.


















Chapter 19

Chemical Thermodynamics

Spontaneity of Chemical processes

Spontaneous reaction: It follows its preferred direction unless acted on by some external agent.

Non-spontaneous reaction: The product will not form under the given conditions.


Reversible and irreversible processes:

1)      Reversible process: the process that can be reversed without net change in either the system or the surroundings. Like the equilibrium state of ice/water at 0.0 C and it is non-spontaneous.

2)      Irreversible process: the process that cant be reversed without net change in either the system or the surroundings. Like the fall of a rock and it is spontaneous.



(1) original (2) expands with work (3) back to original by work


The Entropy (S): is a measure of the disorder of the system.

* Entropy increases as temperature increases. S(g)>S(l)>S(s)


Standard Molar Entropy (S): entropy of 1.0 mole of substance under standard conditions.


Second Law of thermodynamics:

Total disorder increases in any spontaneous process

Things tend to become disorder

DS (+) more disorder. DS (-) more order.


Examples: predict whether DS is + or

a) H2O(l) H2O(g)

b) Ag(aq) + Cl(aq) AgCl(s)

c) 4Fe(s) + 3O2(g) 2Fe2O3(s)


Standard reaction entropy: DS

DS(reaction) = n DSf (products) + n DSf (reactants)

Example: Calculate the entropy of the following reaction:

N2H4(g) + 2H2O2(l) N2(g) + 4H2O(g)


DS(reaction) =[1(191.5)+4(188.8)]-[1(238.5)+2(109.6)] = +489 j/mol.k


Entropy for a phase change at constant temperature:

DS = [DHvap or Dfus]/T (Fig 19.14)

Example: Calculate DS when 1 mol of water is converted into 1 mol of steam at 100 C at 1 atm. DHvap = 40.67 kj/mol

Free energy (Gibbs free energy): DG

The value of DG indicates the spontaneity of the reaction.

DG = + non-spontaneous, DG = 0 at equilibrium, DG = - spontaneous.


Example: Calculate DG for a reaction at 27 C if DH = -642.2kj/mol and DS = 605.9 j/mol.k.

Standard free energy of formation: DGf

It is the energy required to form one mole of substance under standard conditions.


Standard free energy of the reaction: DG(reaction)

DG(reaction) = n DGf (products) + n DGf (reactants)


Example: Calculate DG(reaction) for the following reaction:

2C3H6 + 2NH3 + 3O2 2C3H3N + 6H2O

DG(reaction) =[6(-237.13) + 2(208.6)]-[3(0.0) + 2(-16.45) + 2(74.62)] = -1122 kj/mol


The Relationship between DG and Keq:

DG = DG + RT LnQeq 1

At equilibrium DG = 0, and Qeq = Keq, so #1 become;

0 = DG + RT LnKeq

DG = - RT LnKeq or Keq = e-(DG/RT)


Standard free energy DG:

Indicates the extent to which a chemical reaction occurs under standard conditions.

DG = - Keq >1 poducts are favored

DG = 0 Keq =1 at equilibrium

DG = + Keq <1 reactants are favored


Example: Calculate DG for PCl3 + Cl2 PCl5 at 25 C where Keq= 1.9


Example: Calculate Keq at 25 C for N2 + O2 2NO where DG = 173.1 kj/mol