1. © Art Traynor 2011
Chemistry
Solution Stoichiometry
Solutions
Solvent
Referring to a substance possessing the propensity to dissolve another
( i.e. the Solute ) into a Solution of which the Solvent is characterized as
forming the majority constituent.
Wiki: “Solvent”
Section 4.3, (Pg. 140)
Zumdahl
Dissolution / Dilution
An interaction between two substances whereby one species – the Solvent – acts to
effect a uniform disassociation of another – the Solute – thus maximizing both
Disorder and Potential Energy in the Solute ( i.e. Solvation ) to the degree that
the Solvent can affect and forming a homogenous Solvent-Solute Complex.
Wiki: “Dissolution”
Solute
Referring to a substance possessing the propensity to become dissolute in
combination with another of which the Solute is characterized as forming the
minority constituent.
Wiki: “Solvent”
Section 4.1, (Pg. 127)
Zumdahl
Section 4.2, (Pg. 130)
Zumdahl
Section 4.2, (Pg. 130)
Zumdahl

2. © Art Traynor 2011
Chemistry
Solution Stoichiometry
Solutions
Solution
A Solvation Mixture forming a homogenous, single phase, Solvent-Solute complex
( in contra-distinction to other multi-phase Mixtures such as colloids,
suspensions, or emulsions ) which does not precipitate a chemical reaction or
change of chemical configuration in either of the constituent species , but which
nevertheless can be accompanied by changes in composition energetics and
characterized by the Concentration of its Solute.
Wiki: “Solvent”
Concentration
A characteristic of a Solution expressing the relative abundance of its Solute –
canonically denoted as a mass-volumetric quantity by a variety of disparate
measures ( e.g. Molarity , Mass Percent, Mole Fraction, Molality, Density ) .
Wiki: “Concentration”
Section 11.1, (Pg. 500)
Zumdahl
Density is the mass per unit volume of any substance
⍴ = →
m
V
mass
volume

3. © Art Traynor 2011
Chemistry
Solution Stoichiometry
Aqueous Solutions
Aqueous Solution
A solution in which the solvent is water Wiki: “ Aqueous Solution”
Section 4.1, (Pg. 127)
Zumdahl

4. © Art Traynor 2011
Chemistry
Solution Stoichiometry
Dilution / Dissolution
Dilution Formula
A Dilution can be considered as a mathematical “scaling” of a fixed ( i.e. Constant )
concentration of a substance – in this interpretation, concentration can be thought of
as “scaling” a Volume: [ X ]i · Vi , as a linear combination
Section 4.3, (Pg. 142)
Zumdahl
Vf[ X ] i
[ X ] i
Vi
Vi
For some [ X ]¬ i < [ X ]i the product [ X ]i · Vi may be equated with a
similar product Vf · [ X ]¬ i where Vf is an additive mixture of Vi and a
quantity Vsv
of a shared Solvent
Solvent = Sv
Vsv
Vf
[ X ] ¬ i
Vf = Vi + Vsv
Vi
Vsv
= Vf – Vi
The two product terms equated thus describe a proportional relationship
of the form:
[ X ]i · Vi = Vf · [ X ]¬ i
M1 · V1 = M2 · V2
or
ab = cd
or
or
=
a
c
b
d
= a· d – b · c
a b
c d
a b
c dA =
det ( A ) = a· d – b · c

5. © Art Traynor 2011
Chemistry
Solution Stoichiometry
Dilution / Dissolution
Dilution Formula ( PST )
Example: What volume of 16 M sulfuric acid must be used to
prepare 1.5 L of 0.10 M H2 SO4 solution ?
Section 4.3, (Pg. 141)
Sample Exercise 4.9
Zumdahl
Problem Solving Technique (PST)
Variables
Initial Volume Vi
[ A ]i
Final
Concentration
Vf
[ A ]f
Units
Molarity M =
Moles of Solute
Liters of Solution
Initial
ConcentrationInitial Final
[ X ]k
Vk
A = H2 SO4 (l )
[ A ]i = 16 M =
16 mol
1 L
[ A ]f = 0.10 M =
0.10 mol
1 L
Vi = UNKOWN Vf = 1.5 L
Final Volume
Formulae
Dilution [ A ]i Vi = [ A ]f Vf16 mol
1 L
Vi
1
=
0.10 mol
1 L
1.5 L
1
[ A ] i V i [ A ] f V f

6. © Art Traynor 2011
Chemistry
Solution Stoichiometry
Dilution / Dissolution
Dilution Formula ( PST )
Example: What volume of 16 M sulfuric acid must be used to
prepare 1.5 L of 0.10 M H2 SO4 solution ?
Section 4.3, (Pg. 141)
Sample Exercise 4.9
Zumdahl
Problem Solving Technique (PST)
16 mol
1 L
Vi
1
=
0.10 mol
1 L
1.5 L
1
[ A ] i V i [ A ] f V f
16 mol
1 L
Vi
1
=
0.10 mol
1 L
1.5 L
1
Solve for Vi
1 L
16 mol
[ A ] i
– 1
16 mol
1 L
Vi
1
=
0.10 mol
1 L
1.5 L
1
1 L
16 mol
1 L
16 mol
Vi =
0.10
1
1.5
1
1 L
16
[ A ] i
– 1

7. © Art Traynor 2011
Chemistry
Solution Stoichiometry
Dilution / Dissolution
Dilution Formula ( PST )
Example: What volume of 16 M sulfuric acid must be used to
prepare 1.5 L of 0.10 M H2 SO4 solution ?
Section 4.3, (Pg. 141)
Sample Exercise 4.9
Zumdahl
Problem Solving Technique (PST)
V i [ A ] f V f
Vi =
0.10
1
1.5
1
1 L
16
[ A ] i
– 1
Vi =
( 0.10 ) ( 1.5 ) L
16
Variables
Initial Volume Vi
[ A ]i
Final
Concentration
Vf
[ A ]f
Units
Molarity M =
Moles of Solute
Liters of Solution
Initial
Concentration
Final Volume
Formulae
Dilution [ A ]i Vi = [ A ]f Vf
Vi =
( 0.15 ) L
16
Vi = 0.009375 L
0.0.0.9.375 L = 9.375 x 10 – 3 L
Moving 3 positions
in the “+” directionVi =
①② ③

8. © Art Traynor 2011
Chemistry
Solution Stoichiometry
Dilution / Dissolution
Dilution Formula ( PST )
Example: What volume of 16 M sulfuric acid must be used to
prepare 1.5 L of 0.10 M H2 SO4 solution ?
Section 4.3, (Pg. 141)
Sample Exercise 4.9
Zumdahl
Problem Solving Technique (PST)
V i
Variables
Initial Volume Vi
[ A ]i
Final
Concentration
Vf
[ A ]f
Units
Molarity M =
Moles of Solute
Liters of Solution
Initial
Concentration
Final Volume
Formulae
Dilution [ A ]i V1 = [ A ]f V2
Vi =
0.0.0.9.375 L = 9.375 x 10 – 3 LVi =
9.375 m LVi =
0.0.0.9.375 L = 9.375 x 10 – 3 L
Moving 3 positions
in the “+” direction
①② ③

9. © Art Traynor 2011
Chemistry
Solution Stoichiometry
Dilution / Dissolution
Dilution Formula ( PST )
Example: What volume of 16 M sulfuric acid must be used to
prepare 1.5 L of 0.10 M H2 SO4 solution ?
Section 4.3, (Pg. 141)
Sample Exercise 4.9
Zumdahl
Variables
Initial Volume Vi = 9.375 m L
[ A ]i = 16 M
Final
Concentration
Vf = 1.5 L
[ A ]f = 0.10 M
Units
Molarity M =
Moles of Solute
Liters of Solution
Initial
Concentration
Final Volume
Formulae
Final Volume Vf = Vi + Vsv
Volume of Solvent Vsv
= Vf – Vi
Vi = 9.375 mL
Vf
[ A ] = 16M [ A ] ¬ i = 0.10 M
Vsv
= 1.490625 L
Vf = Vi + Vsv
Vsv
= 1.5 mL – 0.009375
Final Volume V f Initial Volume V i
Volume of Solvent V sv
Vf = Vi + Vsv
1.5 L = 9.375 mL + Vsv
Volume of Solvent V sv
Vsv
= Vf – Vi
Vsv
= 1.5 L – 0.009375 L
Vsv
= Vf – Vi
Solvent
Vsv
= 1.490625 L
Vf = 1.490625 L + 0.009375 L
Vf = 1.5 L

10. © Art Traynor 2011
Chemistry
Pressure
Definitions
Pressure
– n. 1. The exertion of a force upon a surface by an object, fluid, etc., in
contact with it. Physics. 2. Force per unit area ( symbol “ P ” ).
( Pressureless – adj.; Pressure, Pressured, Pressuring – v. ).
Webster’s Encyclopedic
Unabridged Dictionary
of the English Language
( Pg. 1532 )
A measure of Force,
normal to a surface,
expressed per unit area of the referent surface
over which the Force is distributed.
Variables
Pressure
(Normal) Force F
P
P =
F
A
Area
of Contact Surface
A
A scalar quantity
relating a vector surface element
(normal to the referent surface)
with the normal force which acts upon it
As a scalar quantity
pressure has no direction

11. © Art Traynor 2011
Chemistry
Pressure
Units
Pascal – SI Unit
One Newton per square meter.
Webster’s Encyclopedic
Unabridged Dictionary
of the English Language
( Pg. 1532 )
N
m2
Newton - Defined
 the amount of Net Force
One Newton ( N ) is
 that gives an acceleration of one-meter per second squared
1m
s 2
 to a body with a mass of one kilogram 1 kg
1 Newton = ( 1 kilogram ) ( 1 meter per second squared)
1 N =
1kgm
s 2
kg
m · s2
=
Wiki: “ Pressure”

12. © Art Traynor 2011
Chemistry
Pressure
Units
Standard Atmosphere ( atm ) – Legacy Unit
One Standard Atmosphere is defined as:
Webster’s Encyclopedic
Unabridged Dictionary
of the English Language
( Pg. 1532 )
1 atm = 101,325 Pa = 1.01325 x 105 Pa = 101.325 kPa
Torricelli ( torr ) – Legacy Unit
One torr is defined as:
Pa = 1.33 Pa = atm
101,325
760
1
760
Wiki: “ Torr ”
Wiki: “ Pressure”

13. © Art Traynor 2011
Chemistry
Gas Laws
Boyle’s Law
Boyle’s Law
An expression positing a constant relationship between the
product of the volume of a confined gaseous substance
and its characteristic pressure
Section 5.2, (Pg. 187)
Zumdahl
PV = k
Stated in the alternative, Boyles’ Law posits an inverse relationship
between the Volume of a gaseous substance and its characteristic
Pressure (at a particular temperature)
V = = k
k
P
1
P
This form of Boyle’s Law reveals
its consonance with the linear
form y = mx + b ( with b = 0 )
A gas strictly conforming to Boyle’s Law
is referred to as an Ideal Gas

14. © Art Traynor 2011
Chemistry
Gas Laws
Charles’s Law
Charles’s Law
An expression positing a proportional relationship between
the characteristic Volume of a Gas and its Temperature
Section 5.2, (Pg. 187)
Zumdahl
V = bT
Gaseous Volume is a
function of Temperature,
scaled by a proportionality
constant

15. © Art Traynor 2011
Chemistry
Gas Laws
Avogadro’s Law
Avogadro’s Law
An expression positing a proportional relationship between the characteristic
Volume of a Gas and the profusion of its constituent particles
Section 5.2, (Pg. 191)
Zumdahl
V = an
Gaseous Volume is a
function of particle profusion
scaled by a proportionality
constant

16. © Art Traynor 2011
Chemistry
Gas Laws
Ideal Gas Law
Ideal Gas Law ( IGL )
An expression amalgamating three of the principal relationships
characterizing gaseous medium behavior
Section 5.2, (Pg. 191)
Zumdahl
Boyle’s Law :
Gaseous Volume is a
function of particle profusion
scaled by a proportionality
constant
Charles’s Law :
Avogadro’s Law :
V =
k
P
V = bT
V = an
Ideal Gas Law : V = R
T n
P
V = R
T n
P
P
1
P · V = R ·
T n
P
P
1
→ PV = RTn
PV = nRT