Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
POINTERS TO REVIEW
Based on PRC Board Resolution no 10 series of 2013
MODULE 5 BLUEPRINT (17.5%)
PharmacistLicensure
Licensure Examination
Pharmacist
Examination
PHYSICAL PHARMACY
PHYSICAL
PHARMACY
Daryl E. Magno, RPh
University of the Philippines-Manila
College of Pharmacy
PH ARMACEUTICS
Pharmaceutical
Manufacturing
NO. OF ITEMS
NO. OF EASY
ITEMS
NO. OF
MOD ERATE
ITEMS
NO. OF
D IFFICULT
ITEMS
29
12
12
5
Dos age
Forms
29
12
12
5
Phys ical Pharmacy
25
10
10
5
17
6
6
5
Pharmaceutical
Juris prudence
and Ethics
POINTERS TO REVIEW
Based on PRC Board Resolution no 10 series of 2013
MODULE 5 BLUEPRINT (17.5%)
PH ARMACEUTICS
NO. OF ITEMS
Pharmaceutical
Manufacturing
NO. OF EASY
25
ITEMS
10
TOPICS
NO. OF
MOD ERATE
ITEMS
NO. OF
D IFFICULT
ITEMS
10
5
INTRODUCTION
NO. OF ITEMS
Phys ical Pharmacy Principle
Solubility and Dis tribution Phenomena, Buffer and Is otonic
Solutions , Interfacial Phenomena, Micrometics
Colloids , Coars e Dis pers ion, Rheology, Complexation
and
Protein Binding, Kinetics
3
3 (each)
2 (each)
WHAT IS PHYSICAL PHARMACY?
- Physical pharmacy integrates knowledge of mathematics,
physics and chemistry and applies them to the
pharmaceutical dosage form development
- Physical pharmacy is the study of the physical and
chemical properties of drugs and their dosage forms
- Physical pharmacy is the application of physical chemical
principles to various aspects of pharmacy
Lecturer: Daryl E. Magno, RPh
DEFINITION OF TERMS
Atom
- Atoms are the smallest particle into which an
element can be divided
- Basic unit of an element
Molecule
- One or more atoms joined together by chemical
bonds
- Smallest unit of a compound that has the physical
and chemical properties of that compound
1
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
DEFINITION OF TERMS
O
C
DEFINITION OF TERMS
H
H
ATOM of CARBON
MOLECULE of WATER
Electron
- A particle which has a negative charge
Proton
- A particle which has a positive charge
Neutron
- A particle with no charge
Nucleus
- At the center of the atom containing both protons
and neutrons
DEFINITION OF TERMS
DEFINITION OF TERMS
Nucleon
- Collective term for protons and neutrons
Isotopes
- Atoms of the same element with the same number
of protons, but differing numbers of neutrons
Atomic Number
Example:
- Number of protons
Mass Number
- Number of protons and neutrons
ISOTOPE
Protium
- Nucleon number
Orbital
- A region within the atom in which electrons have the
highest probability of being found
FORMULA
ATOMIC
NUMBER
MASS
NUMBER
NUMBER OF
NEUTRONS
1
1H
1
1
0
Deuterium
2
1H
(D)
1
2
1
Tritium
3
1H
(T)
1
3
2
DEFINITION OF TERMS
DEFINITION OF TERMS
Isotones
- Atoms belonging to different elements having the
same number of neutrons
Isobars
- Atoms belonging to different elements with the same
mass numbers but different atomic numbers
Example: Carbon and Nitrogen have the same number
of neutrons (7)
Example: Calcium and Argon both have the same mass
numbers (40)
13
6C
and 7N14
Lecturer: Daryl E. Magno, RPh
20Ca
40
40
18Ar
2
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
CHECKPOINT
1. Study of the physi cal and chemi cal propertie s of
drugs and their dosage forms.
a. Pharmacokinetics
b. Pharmacotherapeutics
c. Pharmaceutical Dosage Forms
d. Physical Pharmacy
2. One or more at oms j oined t ogether by che mica l
bonds.
a. Atom
b. Molecule
c. Both A and B
d. None of the above
5. Number of protons
a. Atomic Number
b. Mass Number
c. Molecular Weight
d. Atomic Weight
Lecturer: Daryl E. Magno, RPh
3. Example of an atom.
a. Carbon Dioxide
b. Nitric Acid
c. Sulfur
d. All of the above
4. Atoms of the same element with the same number
of protons, but differing numbers of neutrons
a. Isotopes
b. Isotones
c. Isobars
d. None of the above
FORCES OF ATTRACTION
PHYSICAL PROPERTIES
TYPES OF PROPERTIES
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Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
FORCES OF ATTRACTION
INTRAMOLECULAR FORCES
FORCES OF ATTRACTION
§ Forces that hold atoms together within a molecule
INTERMOLECULAR FORCES
§ Forces that exist between molecules
FORCES OF ATTRACTION
INTRAMOLECULAR FORCES
1. Ionic bond
2. Covalent bond
3. Metallic bond
INTERMOLECULAR FORCES
*Photo grabbed from www.khanacademy.org
1. Dipole-dipole
2. London forces
3. Hydrogen bond
FORCES OF ATTRACTION
FORCES OF ATTRACTION
INTRAMOLECULAR FORCES
INTRAMOLECULAR FORCES
1. Ionic bond
§ Electrostatic attraction between ions that have
opposite charges
§ Between metals and nonmetals
§ Complete transfer of valence electron(s)
between atoms
2. Covalent bond
§ Bond is formed between atoms that have
similar electronegativities—the affinity or
desire for electrons
§ Sharing of electrons in order to achieve octet
configuration
§ 2 types of Covalent bond:
i). Nonpolar
ii) Polar
Lecturer: Daryl E. Magno, RPh
4
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
FORCES OF ATTRACTION
FORCES OF ATTRACTION
INTRAMOLECULAR FORCES
INTRAMOLECULAR FORCES
2. Covalent bond
§ Nonpolar- the difference in electronegativity
between bonded atoms is less than 0.5
2. Covalent bond
§ Polar- the difference in electronegativity
between bonded atoms is between 0.5 and 1.9
FORCES OF ATTRACTION
FORCES OF ATTRACTION
INTRAMOLECULAR FORCES
INTERMOLECULAR FORCES
3. Metallic bond
§ Bond exists only in metals
§ Valence electrons are free to move through the lattice
§ The electrons act as a "glue" giving the substance a
definite structure
1. Van der Waals Forces
§ Weakest of the intermolecular forces and exist
between all types of molecules
TYPE
DESCRIPTION
OTHER TERM
KEESOM
Dipole-dipole
Orientation forces
DEBYE
Dipole-induced
dipole
Induced dipoleinduced dipole
Induction forces
LONDON
Dispersion forces
FORCES OF ATTRACTION
FORCES OF ATTRACTION
INTRAMOLECULAR FORCES
INTRAMOLECULAR FORCES
A. Keesom
B. Debye
Lecturer: Daryl E. Magno, RPh
5
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
FORCES OF ATTRACTION
FORCES OF ATTRACTION
INTRAMOLECULAR FORCES
INTRAMOLECULAR FORCES
C. London Forces
2. Hydrogen bonding
§ Special kind of dipole-dipole interaction that
occurs specifically between a hydrogen atom
bonded to either an oxygen, nitrogen, or
fluorine atom
FORCES OF ATTRACTION
FORCES OF ATTRACTION
INTRAMOLECULAR FORCES
INTRAMOLECULAR FORCES
3. Ion-dipole
§ Electrostatic attraction between an ion and a
neutral molecule that has a dipole
4. Ion-induced dipole
§ Ion induces a dipole in an atom or a molecule
with no dipole
PHYSICAL PROPERTIES
1. Additive Property
§ Depends on the sum of the individual properties of the
components present in a system
PHYSICAL PROPERTIES
§ Ex. Molecular weight
2. Constitutive Property
§
§
Depends on the type and arrangement of the
components present in a system
Ex. Optical activity
3. Colligative Property
§ Depends on the number of components present in a
system
§
Lecturer: Daryl E. Magno, RPh
Ex. Boiling point elevation
6
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
TYPES OF PROPERTIES
§ Extensive Property
- Depends on the size or the amount of material
- Ex. mass, length, volume
TYPES OF PROPERTIES
§ Intensive Property
- Does not depend on the size or amount of
material
- Ratio of two extensive properties
- Ex. Density, specific gravity
Density
Density
§ Measure of mass per unit of volume
§ unit: g/mL
§ Two types:
§ Two types:
1. Bulk Density
2. Tapped Density
1. Bulk Density
ØMeasure of interparticulate interactions
ØIndex of the ability of the powder to flow
ØDetermined by measuring the volume of a known
mass of powder sample that has been passed
through a
• Graduated cylinder (Method I)
• Volumeter (Method II)
Density
§ Two types:
2. Tapped Density
Ø achieved by mechanically tapping a measuring
cylinder containing a powder sample
Density
§ Measure of mass per unit of volume
§ unit: g/mL
Ø Methods:
– Method I – mechanical tapped density tester
(fixed drop: 14 ± 2 mm; nominal rate: 300
drops/min)
– Method II – mechanical tapped density tester
(fixed drop: 3mm (±10%); nominal rate: 250
drops/min
Lecturer: Daryl E. Magno, RPh
7
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
What is the density of a 85 mL liquid if its weight
is 38 g?
A. 0.56 g/mL
B. 0.53 g/mL
C. 0.45 g/mL
D. 0.41 g/mL
What is the weight of 60 mL of oil whose density
is 0.9645 g/mL?
A. 56.0
B. .5600
C. 57.87
D. 54.88
CORRECT ANSWER: C
CORRECT ANSWER: C
What is the density of a 3.5 L liquid if its weight
is 90 g?
A. 0.56 g/mL
B. 0.056 g/mL
C. 0.026 g/mL
D. 0.063 g/mL
Density
§ Two types:
§ Bulk density- ratio of the mass of an untapped
powder sample and its volume
§ Tapped density- increased bulk density attained
after mechanically tapping a container containing
the powder sample
CORRECT ANSWER: C
Specific gravity
Methods of Determining Specific Gravity
§ Dimensionless unit
§ Ratio of the density of a substance to the density of
water at a specified temperature
1. Measurement by Pycnometer
2. Measurement by Sprengel-Ostwald Pycnometer
3. Measurement by Hydrometer
4. Measurement by Oscillator-Type Density Meter
Lecturer: Daryl E. Magno, RPh
8
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
Calculating Specific Gravity
Calculating Specific Gravity by Weight
1. Calculating Specific Gravity by Weight
2. Calculating Specific Gravity by Density
1. Obtain the weight of the liquid in question
2. Obtain the weight of an identical volume of water
3. Calculate the ratio of the liquid’s weight to the
weight of water
Calculating Specific Gravity by Weight
Calculating Specific Gravity by Weight
1. Obtain the weight of the liquid in question
2. Obtain the weight of an identical volume of water
1. Obtain the weight of the liquid in question
2. Obtain the weight of an identical volume of water
3. Calculate the ratio of the liquid’s weight to the
weight of water
3. Calculate the ratio of the liquid’s weight to the
weight of water
Exa mple : If you weighed 10 0 mL of a cetone at 25
degrees C, it w oul d weig h 0 .17 314 pounds. W eighi ng
the sa me volume of water at the sa me te mperatur e
give you 0.22 pounds . What is the spe ci fic gravity of
acetone?
Exa mple : If you weighed 10 0 mL of a cetone at 25
degrees C, it w oul d weig h 0 .17 314 pounds. W eighi ng
the sa me volume of water at the sa me te mperatur e
give you 0.22 pounds . What is the spe ci fic gravity of
acetone? Answer: 0.17314/0.22 = 0.787
Calculating Specific Gravity by Density
Calculating Specific Gravity by Density
1. Obtain the density of the liquid in question
2. Obtain the density of an identical volume of water
1. Obtain the density of the liquid in question
2. Obtain the density of an identical volume of water
3. Calculate the ratio of the liquid’s density to the
density of water
3. Calculate the ratio of the liquid’s density to the
density of water
Exa mple : I f you had a sa mple that was eight gra ms and
nine milliliters, what its specific gravity?
Lecturer: Daryl E. Magno, RPh
9
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
Calculating Specific Gravity by Density
1. Obtain the density of the liquid in question
2. Obtain the density of an identical volume of water
3. Calculate the ratio of the liquid’s density to the
density of water
Exa mple : I f you had a sa mple that was eight gra ms and
nine milliliters, what its specific gravity?
Answer:
8g/9mL = 0.89g/mL
0.89g/mL ÷ 1g/mL = 0.89g/mL
1. Weakest type of intermolecular force of attraction.
a. London
b. Debye
c. Keesom
d. None of the above
2. Sharing of ele ctrons in order to a chieve octe t
configuration.
a. Ionic Bond
b. Covalent Bond
c. Metallic Bond
d. All of the above
5. What is the weight of 0.05 L of oil whose density is
0.8465 g/mL?
a. 42.325
b. 0.042
c. 4.233
d. 423.25
Lecturer: Daryl E. Magno, RPh
CHECKPOINT
3. Example of property where it depends on the sum
of the individual properties of the components
present in the system.
a. Additive
b. Constitutive
c. Colligative
d. All of the above
4. Ratio of untapped powder sample and its volume
a. Bulk Density
b. Tapped Density
c. Both
d. None
STATES OF MATTER
10
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
STATES OF MATTER
STATES OF MATTER
1. Solid
3. Gas
§
§
Fixed volume and shape
Rigid (particles locked into place)
§
§
Assumes the shape and volume of its container
Particles can move past one another
§
Not easily compressible
§
Compressible
§
Does not flow easily
§
Flows easily
2. Liquid
§ Assumes the shape of the container which it occupies
§ Particles can move/slide past one another
4. Plasma
§
Hot ionized gas consisting of positively and negatively
charged electrons
§ Not easily compressible
§ Flows easily
STATES OF MATTER
STATES OF MATTER
Property
Solid
Liquid
Gas
Shape
Definite
Variable
Variable
Volume
Definite
Definite
Variable
IFA Strength
Strong
Strong
Weak
Vibration
Gliding
Constant
Random Motion
Particle
Arrangement
Closely packed,
Fixed position
Fairly closely
packed, not fixed
positions
Widely spaced,
moved
independently
Particle
Arrangement
Closely packed,
Fixed position
Fairly closely
packed, not fixed
positions
Widely spaced,
moved
independently
Molecular
Motion
STATES OF MATTER
STATES OF MATTER
5. Bose-Einstein Condensate
Liquid Crystal
§
Forms within a few degrees of absolute zero
§ Mesophase
§
Formation of “SUPER ATOM”
§ Intermediate phase
§ Blends the structures and properties of the liquid
and solid states
§ ROPE (Rigid, Organic, Polarizable, Elongated)
§ Thermotropic VS Lyotropic Crystals?
Lecturer: Daryl E. Magno, RPh
11
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
STATES OF MATTER
STATES OF MATTER
Types of Liquid Crystals
Type 1: Smectic
Types of Liquid Crystals
Type 2: Nematic
§ Soap-like
§ Thread-like
§ Molecules tend to line themselves up
into layers
§ Simplest form of liquid crystal
§ More restricted movement than nematic
§ Crystal molecules have no orderly
position and are free to move
STATES OF MATTER
Types of Liquid Crystals
Type 3: Cholesteric
§ Chiral nematic
§ Molecules being aligned and at a slight
angle to one another
GASES
GASES
GAS LAWS
Kinetic Molecular Theory of Gases
Boyle’s Law
§ Mariotte Law
1. Gases are composed of a large number of particles that
behave like hard, spherical objects in a state of constant,
random motion.
2. These particles are much smaller than the distance between
particles. Most of the volume of a gas is therefore empty
space.
3. There is no force of attraction between gas particles or
between the particles and the walls of the container.
§ States that “At constant temperature, the
pressure P of a gas varies inversely with its volume V”
P1V1= P2V2
4. Collisions between gas particles or collisions with the walls of
the container are perfectly elastic.
Lecturer: Daryl E. Magno, RPh
12
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
BOYLE’S GAS LAWS
BOYLE’S GAS LAWS
Sample Problem:
Sample Problem:
Calculate the press ure in at mosphere s in an engi ne at
the e nd of the compres sion stroke. As sume t hat at the
start of the stroke, the pre ssur e of the mixt ure of
gasoline and air i n the cyli nder i s 7 45. 8 mm Hg and the
volume of ea ch cylinder is 246. 8 mL. Ass ume that the
volume of the cylinder is 24 .2 mL at the end of the
compression stroke.
Calculate the press ure in at mosphere s in an engi ne at
the e nd of the compres sion stroke. As sume t hat at the
start of the stroke, the pre ssur e of the mixt ure of
gasoline and air i n the cyli nder i s 7 45. 8 mm Hg and the
volume of ea ch cylinder is 246. 8 mL. Ass ume that the
volume of the cylinder is 24 .2 mL at the end of the
compression stroke.
Answer: 10 atm
BOYLE’S GAS LAWS
BOYLE’S GAS LAWS
Sample Problem:
Sample Problem:
A gas tank holds 27 85 L of pr opa ne, C3H8, at 83 0
mmHg. W hat is the volume of t he propane at sta ndar d
pressure?
A gas tank holds 27 85 L of pr opa ne, C3H8, at 83 0
mmHg. W hat is the volume of t he propane at sta ndar d
pressure?
Answer: 3042 L
GAS LAWS
Charles’ Law
§ States that, “At constant pressure, the volume V of a
gas is directly proportional to its absolute (Kelvin)
temperature T”
V1/T1= V2/T2
Lecturer: Daryl E. Magno, RPh
CHARLES’ GAS LAWS
Sample Problem:
Assume t hat the volume of a ball oon filled wit h H2 i s
1.00 L at 2 5 degree s C. Cal culate t he v olume of the
balloon whe n it is coole d to -78 degree s C in a lowtemperature bath made by adding dry ice to acetone.
13
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
CHARLES’ GAS LAWS
CHARLES’ GAS LAWS
Sample Problem:
Sample Problem:
Assume t hat the volume of a ball oon filled wit h H2 i s
1.00 L at 2 5 degree s C. Cal culate t he v olume of the
balloon whe n it is coole d to -78 degree s C in a lowtemperature bath made by adding dry ice to acetone.
A sample of oxyge n occupies a v olume of 160 dm3 at
91° C. W hat will be v olume of oxygen w he n the
temperature drops to 0.00° C?
Answer: 0.65 L
CHARLES’ GAS LAWS
CHARLES’ GAS LAWS
Sample Problem:
Sample Problem:
A sample of oxyge n occupies a v olume of 160 dm3 at
91° C. W hat will be v olume of oxygen w he n the
temperature drops to 0.00° C?
A sample of oxyge n occupies a v olume of 160 dm3 at
91° C. W hat will be v olume of oxygen w he n the
temperature drops to 0.00° C? How about in liters?
Answer: 120 dm 3
CHARLES’ GAS LAWS
Sample Problem:
A sample of oxyge n occupies a v olume of 160 dm3 at
91° C. W hat will be v olume of oxygen w he n the
temperature drops to 0.00° C? How about in liters?
GAS LAWS
Gay-Lussac’s Law
§ States that, “At constant volume, the pressure P of a
gas is directly proportional to its absolute (Kelvin)
temperature T”
P1/T1= P2/T2
Answer: 120 L
Lecturer: Daryl E. Magno, RPh
14
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
GAY-LUSSAC’S GAS LAWS
GAY-LUSSAC’S GAS LAWS
Sample Problem:
Sample Problem:
A gas has a pressure of 69 9.0 mm Hg at 40.0 °C. W hat
is the temperature at a pressure of 760.0 mm Hg?
A gas has a pressure of 69 9.0 mm Hg at 40.0 °C. W hat
is the temperature at a pressure of 760.0 mm Hg?
Answer: 340.3 Kelvin
GAY-LUSSAC’S GAS LAWS
GAY-LUSSAC’S GAS LAWS
Sample Problem:
Sample Problem:
If a gas at 740 mm Hg and 70. 0 ºC has its pre ssur e
lowered to 72 0 mm Hg, what will its temperature be if
volume remains constant?
If a gas at 740 mm Hg and 70. 0 ºC has its pre ssur e
lowered to 72 0 mm Hg, what will its temperature be if
volume remains constant?
Answer: 334K
GAS LAWS
GAS LAWS
Avogadro’s Law
§ States that, “Volume of a gas is directly proportional
to the amount of gas at a constant temperature and
pressure.”
Amontons’ Law
§ States that, “Given a constant number of mole of a
gas and an unchanged volume, pressure is directly
proportional to temperature.”
Lecturer: Daryl E. Magno, RPh
15
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
GAS LAWS
COMBINED GAS LAWS
Combined Gas Law
§ Combination of Boyle's Law, Charles' Law and GayLussac’s Law
Sample Problem:
COMBINED GAS LAWS
COMBINED GAS LAWS
Sample Problem:
Sample Problem:
A sample of sulfur dioxide occupies a volume of
652 mL at 40.° C and 720 mm Hg. What volume
will the sulfur dioxide occupy at STP?
The initial temperature of a 1.00 liter sample of
argon is 20.° C. The pressure is decreased from
720 mm Hg to 360 mm Hg and the volume
increases to 2.14 liters. What was the change in
temperature of the argon?
Answer: 538.75mL
A sample of sulfur dioxide occupies a volume of
652 mL at 40.° C and 720 mm Hg. What volume
will the sulfur dioxide occupy at STP?
COMBINED GAS LAWS
GAS LAWS
Sample Problem:
Ideal Gas Law
§ Combination of Boyle's Law, Charles' Law,
Amontons’ Law and Avogadro's Law
The initial temperature of a 1.00 liter sample of
argon is 20.° C. The pressure is decreased from
720 mm Hg to 360 mm Hg and the volume
increases to 2.14 liters. What was the change in
temperature of the argon?
PV=nRT
Answer: 313.51 K
Lecturer: Daryl E. Magno, RPh
16
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
GAS LAWS
Ideal Gas Law
§ Combination of Boyle's Law, Charles' Law,
Amontons’ Law and Avogadro's Law
§ n is number of moles (mass/molecular weight)
§ R is the gas constant
-
1/25/18
IDEAL GAS LAWS
Sample Problem:
How many mol ecules are t here i n 98 5 mL of nitroge n
at 0.0° C and 1.00 x 10-6 mm Hg?
R = 0.0821 liter·atm/mol·K
R = 8.3145 J/mol·K
R = 8.2057 m3·atm/mol·K
R = 62.3637 L·Torr/mol·K or L·mmHg/mol·K
IDEAL GAS LAWS
IDEAL GAS LAWS
Sample Problem:
Sample Problem:
Calculate the mass of 15.0 L of NH3 (MW = 17.04
g/mol) at 27° C and 900 mm Hg.
How many mol ecules are t here i n 98 5 mL of nitroge n at
0.0° C and 1.00 x 10-6 mm Hg?
Answer: 3.48 X 1013 N2 molecules
IDEAL GAS LAWS
GAS LAWS
Sample Problem:
Sample Problem:
Calculate the density in g/L of 478 mL of krypton
(MW=83.798 g/mol) at 47° C and 671 mm Hg.
Calculate the mas s of 15. 0 L of NH3 (MW = 1 7.0 4
g/mol) at 27° C and 900 mm Hg.
Answer: 12.3 g
Lecturer: Daryl E. Magno, RPh
17
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
GAS LAWS
GAS LAWS
Sample Problem:
Real Gas Equation
2 Factors:
Calculate the de nsity in g /L of 47 8 mL of krypton
(MW=83.798 g/mol) at 47° C and 671 mm Hg.
Answer: 2.82 g/L
§ As pressure increases, the volume of a gas becomes very
small and approaches zero. While it does approach a small
number, it will not be zero because molecules do occupy
space (i.e. have volume) and cannot be compressed
(INCOMPRESSIBI LIT Y )
§ Intermolecular forces do exist in gases
(INTERMOLECULAR FORCES)
GAS LAWS
GAS LAWS
Real Gas Equation (Van der Waals Equation)
Henry’s Law of Gas Solubility
§ States that “At a constant temperature, the amount
of a gas that dissolves in a liquid is directly
proportional to the partial pressure of that gas in
equilibrium with that liquid”
an2/v2: accounts for the internal pressure per mole
resulting from the intermolecular forces of attraction
between the molecules
nb: accounts for the incompressibility of the molecules
GAS LAWS
DALTON’S LAW OF PARTIAL PRESSURE
Dalton’s Law of Partial Pressure
§ States that “The total pressure in a mixture of gases
is equal to the sum of the partial pressure of each gas”
Sample Problem:
Let's say that we have one container with 24.0 L of
nitrogen gas at 2.00 atm, and another container with
12.0 L of oxygen gas at 2.00 atm. The temperature of
both gases is 273 K.
PTOTAL= PA + PB…
Lecturer: Daryl E. Magno, RPh
If both gases are mixed in a 10L container, what are
the partial pressures of nitrogen and oxygen in the
resulting mixture? What is the total pressure?
18
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
DALTON’S LAW OF PARTIAL PRESSURE
GAS LAWS
Sample Problem:
Graham’s Law
§ States that “The rate of effusion of a gaseous
substance is inversely proportional to the square root
of its molar mass.”
§ Empirical relationship stating that the ratio of the
rates of diffusion or effusion of two gases is the
square root of the inverse ratio of their molar masses
Let's say that we have one container with 24.0 L of nitrogen gas
at 2.00 atm, and another container with 12.0 L of oxygen gas at
2.00 atm. The temperature of both gases is 273 K.
If both gases are mixed in a 10L container, what are the partial
pressures of nitrogen and oxygen in the resulting mixture?
What is the total pressure?
Partial pressure of Nitrogen is 4.79 atm
Partial pressure of oxygen is 2.40 atm
Total Pressure is 7.19 atm
1. Description of solid except
a. Fixed volume
b. Fixed Shame
c. Flow Easily
d. Rigid
CHECKPOINT
3. Third type of liquid crystals
a. Smectic
b. Nematic
c. Cholesteric
d. A and B
2. Formation of condensate
a. Plasma
b. Liquid Crystals
c. Bose-Einstein Condensate
d. Any of the above
5. Factor(s) affecting real gas equation
a. Incompressibility
b. Intermolecular forces
c. Both
d. None
4. States that “At constant temperature, the
pressure P of a gas varies inversely with its volume V
a. Gay Lussac’s Law
b. Henry’s Law
c. Graham’s Law
d. Mariotte Law
Lecturer: Daryl E. Magno, RPh
19
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Module 5: Pharmacutics (17.5%)
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LIQUIDS
§ Less kinetic energy than do gases
§ Occupy a definite volume
§ Take the shape of container
LIQUIDS
§ Denser than gases
§ Not compressible
LIQUIDS
LIQUIDS
Vapor Pressure
§ Pressure of the saturated vapor above a liquid
resulting from the escape of surface liquid
molecules
§ A measure of the tendency of molecules to escape
from a liquid
Vapor Pressure
§ Clausius-Clapeyr on Equation
LIQUIDS
LIQUIDS
Clausius-Clapeyr on Equation
Clausius-Clapeyr on Equation
Sample Problem:
Sample Problem:
A certain liquid has a vapor pressure of 6.91 mmHg at 0
°C. If this liquid has a normal boiling point of 105 °C,
what is the liquid's heat of vaporization in kJ/mol?
Given R = 8.31147 J/mol K
A certain liquid has a vapor pressure of 6.91 mmHg at 0
°C. If this liquid has a normal boiling point of 105 °C,
what is the liquid's heat of vaporization in kJ/mol?
- shows the relationship between the
vapor pressure of a liquid and the temperature
Answer: 38445 J/mol or 38.4 kJ/mol
Lecturer: Daryl E. Magno, RPh
20
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
LIQUIDS
LIQUIDS
Clausius-Clapeyr on Equation
Clausius-Clapeyr on Equation
Sample Problem:
Sample Problem:
Determine the vapor pressure of water at 150°C. The
vapor pressure of water at 110 °C is 1.5atm and molar
heat of vaporization of 9500 cal/mole. Given R =
1.986cal/mol K
Determine the vapor pressure of water at 150°C. The
vapor pressure of water at 110 °C is 1.5atm and molar
heat of vaporization of 9500 cal/mole. Given R =
1.986cal/mol K
Answer: 4.88 atm
LIQUIDS
LIQUIDS
Dispersed Systems
§ Systems in which one substance (dispersed phase) is
distributed, in discrete units, throughout a second
substance (continuous phase or vehicle)
Dispersed Systems
1. True Solutions
§
§
§
A solution is a homogenous mixture (one phase)
composed of two or more substances
A solute is dissolved in another substance, known
as a solvent
Particle size is < 1 nm
LIQUIDS
LIQUIDS
Dispersed Systems
2. Colloidal Dispersions
Dispersed Systems
3. Coarse Dispersions
§
§
A system having a particle size intermediate
between that of a true solution and a coarse
dispersion
Particle size is 1 nm-0.5 um
Lecturer: Daryl E. Magno, RPh
§
Size of particle is larger than that of colloids
§
§
Particle size is > 0.5 um
Suspensions and emulsions
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SOLUTIONS
Solute and Solvent
Electolyte
Solute
SOLUTIONS
LIQUIDS
Nonelectrolyte
• Yields ions
• Does not yield ions
• Conduct electricity • Does not conduct
• NaCl, HCl, CaCl 2 ,
electricity
CH3 COOH
• Sucrose
Solvent: phase of the solution and the largest portion of the
solution
LIQUIDS
True Solutions
SOLUBILITY TERM
PARTS OF SOLVENT REQUIRED TO DISSOLVE
ONE PART OF SOLUTE
Very Soluble
Less than 1
Freely Soluble
1-10
Soluble
10-30
Sparingly Soluble
30-100
Slightly Soluble
100-1000
Very slightly Soluble
1000-10000
Practically Insoluble
Greater than 10000
Factors that Affect Solubility
§ Temperature
- Endothermic reactions
- Absorb heat
- Increase temperature, increase solubility
- Exothermic reactions
- Release heat
- Decrease temperature, increase solubility
LIQUIDS
Factors that Affect Solubility
§ pH
- Critical pH for weak acid and weak base
§ Presence of Salts
- Salting in (addition of salt, increased solubility)
- Salting out (addition of salt, decreased solubility)
§ Nature of solute and solvent
§ Particle size
COLLIGATIVE PROPERTY
OF SOLUTIONS
§ Pressure (for gases)
Lecturer: Daryl E. Magno, RPh
22
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
LIQUIDS
LIQUIDS
Colligative Properties of Solutions
1. Vapor Pressure Lowering
Colligative Properties of Solutions
1. Vapor Pressure Lowering
§
§
When a nonvolatile solute is added to a volatile
solvent, the solute decreases the escaping tendency
of the solvent
The vapor pressure of the solution is lowered
proportional to the relative number of solute
molecules
Psolvent = X solventPosolvent
Sample Problem #1: What is the vapor pressure of an
aqueous solution that has a solute mole fraction
of 0.1000? The vapor pressure of water is 25.756
mmHg at 25 °C.
Psolvent = X solventPosolvent
LIQUIDS
LIQUIDS
Colligative Properties of Solutions
1. Vapor Pressure Lowering
Colligative Properties of Solutions
1. Vapor Pressure Lowering
Sample Problem #1: What is the vapor pressure of an
aqueous solution that has a solute mole fraction
of 0.1000? The vapor pressure of water is 25.756
mmHg at 25 °C.
Sample Problem #2: Calculate the vapor pressure of a
solution made by dissolving 50.0 g
glucose, C6H12O6 (MW= 180g/mol), in 500 g of
water. The vapor pressure of pure water is 47.1
torr at 37°C.
Answer: 23.18 mmHg
Psolvent = X solventPosolvent
LIQUIDS
LIQUIDS
Colligative Properties of Solutions
1. Vapor Pressure Lowering
Colligative Properties of Solutions
1. Vapor Pressure Lowering
Sample Problem #2: Calculate the vapor pressure of a
solution made by dissolving 50.0 g
glucose, C6H12O6 (MW= 180g/mol), in 500 g of
water. The vapor pressure of pure water is 47.1
torr at 37°C.
Sample Problem #3: Calculate the vapor pressure
of a solution made by dissolving 50.0 g CaCl2
(MW= 111g/mol) in 500 g of water. The vapor
pressure of pure water is 47.1 torr at 37°C.
Answer: 46.63 torr
Lecturer: Daryl E. Magno, RPh
Psolvent = X solventPosolvent
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LIQUIDS
LIQUIDS
Colligative Properties of Solutions
1. Vapor Pressure Lowering
Colligative Properties of Solutions
2. Boiling Point Elevation
Sample Problem #3: Calculate the vapor pressure
of a solution made by dissolving 50.0 g CaCl2 in
500 g of water. The vapor pressure of pure
water is 47.1 torr at 37°C.
§
BP is the temperature at which the vapor pressure
of the liquid becomes equal to the external
atmospheric pressure (1 atm or 760 mmHg)
§
The BP of the solution containing a nonvolatile
solute would be higher than the pure solvent
ΔT = i K b m
Answer: 46.35 torr
LIQUIDS
LIQUIDS
Colligative Properties of Solutions
2. Boiling Point Elevation
Van’t Hoff Factor
Complete Dissociation
ΔT = i K b m
§
Ebullioscopic constant (Kb) is the molal boiling
point elevation constant and its unit is degrees
Celsius per molal (°C/molal)
§
Van’t Hoff factor (i) is the number of particles
after dissociation
§
m is the molal concentration
Substance
Van’t Hoff Factor
Non-electrolytes
1
NaCl
2
MgCl2
3
LIQUIDS
LIQUIDS
Van’t Hoff Factor
Incomplete Dissociation
Colligative Properties of Solutions
2. Boiling Point Elevation
ΔT = i K b m
Sample Problem: What is the molecular mass of an
organic compound if 16.00 g of the compound is
dissolved in 225.0 g of carbon tetrachloride raises the
boiling point to 85.36 °C?
Given: BP and Kb of CCl4 is 76.72°C and 5.03°C kg/mol
respectively
Lecturer: Daryl E. Magno, RPh
24
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
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LIQUIDS
LIQUIDS
Colligative Properties of Solutions
2. Boiling Point Elevation
Colligative Properties of Solutions
3. Freezing Point Depression
ΔT = i K b m
Sample Problem: What is the molecular mass of an
organic compound if 16.00 g of the compound is
dissolved in 225.0 g of carbon tetrachloride raises the
boiling point to 85.36 °C? Given: BP and Kb of CCl4 is
76.72°C and 5.03°C kg/mol
§
§
FP is the temperature at which a liquid turns into a
solid when cooled
The FP of the solution containing a nonvolatile
solute would be lower than the pure solvent
ΔT = i K f m
Answer: 41.4 g/mol
LIQUIDS
LIQUIDS
Colligative Properties of Solutions
3. Freezing Point Depression
Colligative Properties of Solutions
3. Freezing Point Depression
ΔT = i K f m
§
Cryoscopic constant (Kf) is the molal freezing point
depression constant and its unit is degrees Celsius
per molal (°C/molal)
§
Van’t Hoff factor (i) is the number of particles
after dissociation
§
m is the molal concentration
ΔT = i Kf m
Sample Problem #1:
When 20.0 grams of an unknown nonelectrolyte
compound are dissolved in 500.0 grams of benzene,
the freezing point of the resulting solution is 3.77 °C.
The freezing point of pure benzene is 5.444 °C and the
Kf for benzene is 5.12 °C/m. What is the molar mass of
the unknown compound?
LIQUIDS
LIQUIDS
Colligative Properties of Solutions
3. Freezing Point Depression
Colligative Properties of Solutions
3. Freezing Point Depression
Sample Problem #1:
ΔT = i Kf m
Sample Problem #2:
Lauryl alcohol is obtained from coconut oil and is used
to make detergents. A solution of 5.00 g of lauryl
alcohol in 0.100 kg of benzene freezes at 4.1 °C and
the Kf for benzene is 5.12 °C kg/m. What is the molar
mass of the lauryl alcohol? Freezing point of pure
benzene is 2.756 °C.
When 20.0 grams of an unknown nonelectrolyte compound are
dissolved in 500.0 grams of benzene, the freezing point of the
resulting solution is 3.77 °C. The freezing point of pure benzene
is 5.444 °C and the K f for benzene is 5.12 °C/m. What is the
molar mass of the unknown compound?
Answer: 122.34 g/mol
Lecturer: Daryl E. Magno, RPh
25
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
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LIQUIDS
LIQUIDS
Colligative Properties of Solutions
3. Freezing Point Depression
Colligative Properties of Solutions
4. Osmotic Pressure
ΔT = i Kf m
§
Osmosis is the spontaneous net movement of water
across a semi-permeable membrane from a region
of high to low water concentration
§
The pressure required to prevent osmosis in
solutions
Sample Problem #2:
Lauryl alcohol is obtained from coconut oil and is used
to make detergents. A solution of 5.00 g of lauryl
alcohol in 0.100 kg of benzene freezes at 4.1 °C and
the Kf for benzene is 5.12 °C/m. What is the molar
mass of the lauryl alcohol?
∏ = iMRT
Answer: 190 g/mol
LIQUIDS
LIQUIDS
Colligative Properties of Solutions
4. Osmotic Pressure
Colligative Properties of Solutions
4. Osmotic Pressure
∏ = iMRT
Where
∏ is the osmotic pressure in atm
i = van’t Hoff factor of the solute
M = molar concentration in mol/L
R = universal gas constant (0.0821 liter·atm/mol·K)
T = absolute temperature in K
∏ = iMRT
Sample Problem #1:
How much glucose (C6H12O 6) [MW = 180g/mol)
per liter should be used for an intravenous
solution to match the 7.65 atm at 37 °C
osmotic pressure of blood?
LIQUIDS
LIQUIDS
Colligative Properties of Solutions
4. Osmotic Pressure
Colligative Properties of Solutions
4. Osmotic Pressure
∏ = iMRT
∏ = iMRT
Sample Problem #1:
Sample Problem #2:
How much glucose (C6H12O6) [MW = 180g/mol) per
liter should be used for an intravenous solution to
match the 7.65 atm at 37 °C osmotic pressure of
blood?
What is the osmotic pressure of a solution prepared
by adding 13.65 g of sucrose (C12H22O11) [MW =
342g/mol) to enough water to make 250 mL of
solution at 25 °C?
Answer: 54.1 g
Lecturer: Daryl E. Magno, RPh
26
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
LIQUIDS
Colligative Properties of Solutions
4. Osmotic Pressure
∏ = iMRT
Sample Problem #2:
What is the osmotic pressure of a solution prepared by
adding 13.65 g of sucrose (C12H22O11) MW = 342g/mol)
to enough water to make 250 mL of solution at 25 °C?
Answer: 3.9 atm
ISOTONIC SOLUTIONS
LIQUIDS
LIQUIDS
Isotonic Solutions
1. Hypertonic Solutions
Isotonic Solutions
2. Hypotonic Solutions
§
§
A solution with more solute compared to cell
concentrations
Solutions that freeze lower that -0.52 °C
§
A solution with less solutes compared to cell
concentrations
Solutions that freeze higher than --0.52 °C
§
Results: crenation of the cell (shrinkage)
§
Result: swelling and lysis of the cell
§
LIQUIDS
LIQUIDS
Isotonic Solutions
3. Isotonic Solutions
Isotonic Solutions
§
§
§
Solutions for which a living cell does not gain or
lose water
Solutions with similar osmotic pressure as that
of body fluids
Similar concentration as 0.9% (w/v) NaCl
solution
Lecturer: Daryl E. Magno, RPh
27
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
LIQUIDS
Methods of Adjusting Isotonicity
Class I: Addition of a tonicity adjusting agent
a. Freezing Point Depression Method/ Cryoscopic
Method
METHODS OF
ADJUSTING ISOTONICITY
b. Sodium Chloride Equivalent Method/ E Value
Method
Class II. Addition of water and dilution with buffered
isotonic solution
a. White Vincent Method
b. Sprowl’s Method
METHODS OF ADJUSTING ISOTONICITY
METHODS OF ADJUSTING ISOTONICITY
Class 1: Addition of a tonicity adjusting agent
1. Cryoscopic Method
Class 1: Addition of a tonicity adjusting agent
1. Cryoscopic Method
– Isotonic freezing point depression: 0.52 °C
– Isotonic NaCl concentration: 0.9% (w/v) NaCl
𝑤% =
0.52 − 𝑎
𝑏
Where
W% = conc g/100mL of adjusting substance
a = freezing point depression of 1% adjusted substance multiple to
the percentage strength
B = freezing point depression of adjusting substance
𝑤% =
0.52 − 𝑎
𝑏
Sample Problem #1: How much NaCl is required to
render 100 ml of a 1% soln. of apomorphin HCL
isotonic? Fpd of 1%NaCl=0.52º, Fpd of 1%drug=0.08º
METHODS OF ADJUSTING ISOTONICITY
METHODS OF ADJUSTING ISOTONICITY
Class 1: Addition of a tonicity adjusting agent
1. Cryoscopic Method
Class 1: Addition of a tonicity adjusting agent
2. E Value Method
Sample Problem #2: Compound the precription:
Freezing point depression of 1% atropine solution is
0.07.
Rx
Atropine Sulfate
Sodium Chloride
Purified Water
Make isotonic solution.
Lecturer: Daryl E. Magno, RPh
– E value: gram of NaCl equivalent to 1 gram of
substance
– Isotonic NaCl concentration: 0.9% (w/v) NaCl
1%
q.s.
100
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METHODS OF ADJUSTING ISOTONICITY
METHODS OF ADJUSTING ISOTONICITY
Class 1: Addition of a tonicity adjusting agent
Class 1: Addition of a tonicity adjusting agent
2. E Value Method
2. E Value Method
Step 1: Calculate the amount of NaCl represented by the
ingredients in the prescription
Step 2: Calculate the amount of NaCl that would make the
volume of solution specified in the prescription
Step 3: Subtract the amount of NaCl represented by the
ingredients in prescription from the amount of NaCl that
would make the specific volume in the prescription isotonic
Step 4: if the tonicity agent is not NaCl, follow step 1-3 and
the amount of NaCl is divided with the E value of that
chemical
Sample Problem #1: How many grams of NaCl should
be used in compounding the following prescription?
Rx
Pilocarpine nitrate (E = 0.22)
NaCl
0.3g
qs
Purified Water
30mL
qs
Make isotonic solution
Sig for the eye
METHODS OF ADJUSTING ISOTONICITY
METHODS OF ADJUSTING ISOTONICITY
Class 1: Addition of a tonicity adjusting agent
2. E Value Method
Class 1: Addition of a tonicity adjusting agent
2. E Value Method
Sample Problem #2: How many grams of NaCl should
be used in compounding the following prescription to
obtain an isotonic solution?
Sample Problem #3: How many grams of boric acid
should be used in compounding the following
prescription to obtain an isotonic solution?
Rx
Rx
Cocaine HCl (E = 0.16)
NaCl
2%
qs
Purified Water
30mL
qs
Atropine sulfate (E = 0.13)
Boric Acid (E = 0.52)
1%
q.s
Purified Water
30mL
qs
METHODS OF ADJUSTING ISOTONICITY
METHODS OF ADJUSTING ISOTONICITY
Class 1: Addition of a tonicity adjusting agent
2. E Value Method
Class 2: Addition of a water and dilution with buffered
isotonic solution
Sample Problem #4: How many grams of boric acid
should be used in compounding the following
prescription to obtain an isotonic solution?
Rx
Phenacaine HCl (E = 0.17)
Chlorambutanol (E = 0.18)
0.6g
0.3g
Boric Acid (E = 0.52)
Purified Water
qs
q.s
60mL
Lecturer: Daryl E. Magno, RPh
1. White Vincent Method
v= 𝑤 𝑋 𝐸𝑣𝑎𝑙𝑢𝑒 𝑋 111 .1
Sample Problem: Using White Vincent Method,
compute for the required volume to make 0.3 g of
Atropine sulfate (E=0.13) isotonic.
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Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
METHODS OF ADJUSTING ISOTONICITY
Class 2: Addition of a water and dilution with buffered
isotonic solution
1. Sprowl’s Method
V= 0.3𝑔 𝑋 𝐸𝑣𝑎𝑙𝑢𝑒 𝑋 111 .1
1/25/18
METHODS OF ADJUSTING ISOTONICITY
Sample Problem
Rx
Phenacaine hydrochloride 0.1g
Boric acid
0.50g
Distilled water q.s.
100mL
(E=0.20)
(E=0.50)
Calculate the amount of water needed to make the
solution isotonic.
LIQUIDS
Acid-Base Equilibria
Theories on Acids and Bases
1. Arrhenius Theory
THEORIES ON ACIDS
AND BASES
§
Acids are substances which produce hydrogen
ions in solution
§
Bases are substances which produce hydroxide
ions in solution
LIQUIDS
LIQUIDS
Acid-Base Equilibria
Theories on Acids and Bases
Acid-Base Equilibria
Theories on Acids and Bases
2. Bronsted-Lowry Theory
§ An acid is a proton (hydrogen ion) donor
§ A base is a proton (hydrogen ion) acceptor
Lecturer: Daryl E. Magno, RPh
3. Lewis Theory
§ An acid is an electron pair acceptor
§ A base is an electron pair donor
30
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
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LIQUIDS
LIQUIDS
Acid-Base Equilibria
pH
Acid-Base Equilibria
Sorensen’s pH Scale
§ The negative logarithm of the H+ concentration
§ pH > 7 is basic
§ Example, [H+] = 5 X 10 -6 , what is the pH?
§ pH = 5.3
§ pH = 0 is neutral
§ pH < 7 is acidic
LIQUIDS
LIQUIDS
Acid-Base Equilibria
Ionization
Acid-Base Equilibria
pH Calculations
§ The complete separation of ions in a crystal lattice when a
salt is dissolved
§ For weak acids
HA + H2 O
§ For weak bases
B + H2 O
Strong Acids
pH = -log [H+]
H3 O+ + A-
Strong Base
pOH = -log [OH-]
pH = pKw - pOH
OH- + BH+
Weak Acids
pH = -log ( 𝒌𝒂 𝑿 𝑪𝒂 )
Weak Base
pOH = log (√𝒌𝒃𝑿 𝑪𝒃 )
§ For the ionization of water
H20 + H2 0
H3 0+ + OH-
LIQUIDS
LIQUIDS
Acid-Base Equilibria
Buffer
Acid-Base Equilibria
Henderson-Has sel bal ch Equation
•
Solutions that have the property of resisting
changes in pH when acids or bases are
added to them
•
Consists of either:
ü Weak acid and its conjugate base
ü Weak base and its conjugate acid
Lecturer: Daryl E. Magno, RPh
pH = pK a + log ([A-]/[HA])
pH = pK b + log ([B]/[OH -])
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Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
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LIQUIDS
LIQUIDS
Acid-Base Equilibria
Henderson-Has sel bal ch Equation
Acid-Base Equilibria
% Ionization
§ Acid
Sample Problem:
Calculate the pH of a buffer solution made from 0.20 M
HC2H3O2 and 0.50 M C2H3O2- that has an acid
dissociation constant for HC2H3O2 of 1.8 x 10-5
% ionization = 100/ (1+10(pKa - pH))
§ Base
% ionization = 100/ (1+10(pH - pKa))
LIQUIDS
LIQUIDS
Acid-Base Equilibria
% Ionization
Acid-Base Equilibria
% Ionization
Sample Problem:
If the pH-pKa=-1, what percentage of weak base is
nonionized?
Sample Problem:
If the pH-pKa=-1, what percentage of weak base is
nonionized?
Answer: 9.1%
LIQUIDS
Acid-Base Equilibria
Buffer Capacity
§
Buffer Efficiency
§
§
Buffer Index
Buffer Value
§
The ability of a buffer solution to resist changes in
pH upon addition of an acid/alkali
§
Bmax (Maximum Buffer Capacity) = 0.567[C]
Lecturer: Daryl E. Magno, RPh
INTERFACIAL
PHENOMENA
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LIQUIDS
LIQUIDS
Interfacial Phenomena
§ Attributed to the effects of molecules found at
the interface (boundary between 2 phases)
§ Interfacial tension if its between two liquids and
surface tension between solids and gas and
liquid and gas
Interfacial Phenomena
§ Capillary action- water “climb” upwards through
thin glass tubes (called capillary tubes) which
depends on the attraction between water
molecules and the glass walls of the tube
(adhesion), as well as on interactions between
water molecules (cohesion)
§
Surfactants – “surface active agents”, lower the
interfacial/surface tension
LIQUIDS
Interfacial Phenomena
§ Wetting phenomenon- determined by
measuring the contact angle, which the liquid
forms in contact with the solids or liquids
§ Contact Angle– angle which the liquid makes
with the solid surface
WETTABILITY CHANGE
COLLOIDAL
DISPERSION
Lecturer: Daryl E. Magno, RPh
33
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
LIQUIDS
LIQUIDS
Colloidal Dispersions
LYOPHILIC
(SOLVENTLOVING)
DISPERSED
PHASE
SOLVATION
EFFECT OF
ELECTROLYTES
Large organic
molecules
Solvated
LYOPHOBIC
ASSOCIATION
(SOLVENT-HATING) (AMPHOPHILIC)
Inorganic particles
STABLE
Little
UNSTABLE
Aggregates
(micelles)
Hydro or lipo
portion is
solvated
depending on
the medium
Salting out may
occur at high
salt conc.
Colloidal Dispersions
Properties of Colloids
1. Optical
§ Tyndall Effect
- ability to scatter light
2. Kinetic
§
Brownian Motion
- molecules in random constant motion
§ Diffusion
- governed by Fick’s Law of Diffusion
LIQUIDS
Colloidal Dispersions
Properties of Colloids
3. Electric
§
§
Nernst Potential
- electrothermodynamic potential
RHEOLOGY
- relate cell potentials
Zeta potential
- Electrokinetic potential
- results to flocculation
LIQUIDS
LIQUIDS
Rheology
§ Study of the deformation and flow properties of
matter
§
Viscosity- Measure of resistance to gradual
deformation by shear stress or tensile stress
1. Absolute Viscosity
2. Kinematic Viscosity
§
§
Lecturer: Daryl E. Magno, RPh
Elasticity-measure of stickiness or structure
Viscoelasticity- materials exhibit both elastic
behavior and viscous flow
34
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
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LIQUIDS
LIQUIDS
Rheology
Rheology
Absolute Viscosity
Kinematic Viscosity
Relative Viscosity
𝜼=
𝑭
𝑮
𝜂
𝑝
Ratio of viscosity of
solutions to viscosity of
solvent
1. Newtonian Flow
§ Constant viscosity with increasing rate
§ Has a linear relationship between shear rate and shear stress
𝜅=
LIQUIDS
LIQUIDS
Rheology
Rheology
Non-Newtonia n
2. Non-Newtonian Flow
A. Shear-dependent Viscosity
1. Plastic
2. Pseudoplastic
3. Dilatant
B. Time-dependent Viscosity
1. Thixotropy
2. Rheopexy
A. Shear-dependent Viscosity
1. Plastic
§ Bingham bodies
§ Require a yield stress before they begin to flow,
i.e., the shear stress-strain curve doesn’t pass
through origin
LIQUIDS
LIQUIDS
Rheology
Non-Newtonian
A. Shear-dependent Viscosity
2. Dilatant
§ Viscosity of the fluid increases when shear is
applied
§ Shear-thickening
Rheology
Non-Newtonian
3. Pseudoplastic
§ the more shear applied, the less viscous it becomes
§ Shear-thinning
2. Rheopexy
§ the longer the fluid undergoes shear stress, the higher its
viscosity
Lecturer: Daryl E. Magno, RPh
B. Time-dependent Viscosity
1. Thixotropy
§ show a time-dependent change in viscosity; the longer the
fluid undergoes shear stress, the lower its viscosity
(Reversible)
§ Time-dependent shear thinning
§ Time-dependent shear thickening
35
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
LIQUIDS
1/25/18
LIQUIDS
Determination of Viscosity
1. Ostwald Glass Capillary Viscometer
2. Rotational Viscometers
LIQUIDS
Determination of Viscosity
A. Single Point
1. Capillary Viscometer
2. Falling Sphere Viscometer
B. Multipoint
1. Cup & Bob Viscometer
LIQUIDS
Determination of Viscosity
A. Multipoint
1. Cup & Bob Viscometer
a. Searle Type
b. Couette Type
2. Cone & Plate
C. Others
1. Penetrometers
1. The p rocess wh ereby solute molecules mov e from the region
of higher to lower concentration until equilibrium is reached:
a. Osmosis
b. Diffusion
c. Dissolution
d. Distribution
CHECKPOINT
Lecturer: Daryl E. Magno, RPh
2. E value is defined as
a. Equivalent NaCl enough to make a normal solution
b. Equivalent of NaCl to resemble the isotonicity of blood
c. Amount of NaCl theoretic ally equivalen t to 1 gram of a
specific chemical
d. Amount of the chemical theoretic ally equivalent to 0.9
grams of NaCl
36
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
3. Shear-thickening
a. Dilatant
b. Pseudoplastic
c. Plastic
d. Newtonian
4. Ability of colloids to scatter light
a. Brownian Movement
b. Diffusion
c. Adsorption
d. Tyndall Effect
1/25/18
5. A solution with less solutes compared to cell
concentrations will result into
a. Lysis
b. Swelling
c. Shrinkage
d. A and B
SOLIDS
§
§
Fixed shapes
Nearly incompressible
§
With strong intermolecular forces
§
With little kinetic energy
SOLIDS
SOLIDS
SOLIDS
Crystalline solids
§ Molecules are arranged in repetitious 3D lattice
units
Crystalline solids
6 Common Crystalline Structures
§ Have sharp melting points
§ Have definite geometric forms
Amorphous solids
§ aka Glasses or supercooled liquids
§ No definite order of molecules
1. Cubic (Sodium Chloride)
2. Tetragonal (Urea)
3. Hexagonal (Iodoform)
4. Rhombic (Iodine)
5. Monoclinic (Sucrose)
6. Triclinic (Boric acid)
§ No definite and sharp melting points
Lecturer: Daryl E. Magno, RPh
37
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
SOLIDS
SOLIDS
Polymorphism
§ Ability of crystalline solids to exist in more than 1
crystalline form
§ Polymorphs are solids that have more than 1
crystalline form having different physical
characteristics (melting points and solubilities)
Polymorphism
§ Enantiotropic
Ø The polymorph which can be changed from
one form into another by varying temperature
or pressure (REVERSIBLE)
§ Monotropic
Ø The polymorph which can not be changed
from one form into another (IRREVERSIBLE)
SOLIDS
Polymorphism
§ Isotropic
Ø The properties of the different polymorphs
are identical
§ Anisotropic
Ø The properties of the different polymorphs
are different
MICROMERITICS
SOLIDS
SOLIDS
Micromeritics
§ Study of small particles
Micromeritics
Fundamental Properties
§ Fundamental VS Derived properties
§ Includes:
ü Particle Size
§ Particle Size
Coarse
Conventional
> 1000 um
50 to 1000 um
ü Particle Distribution
Fine
1 to 50 um
ü Particle Shape
ü Surface Area
Very Fine
Ultra Fine
0.1 to 1 um
<0.1 um
ü Derived Properties
Lecturer: Daryl E. Magno, RPh
38
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
SOLIDS
SOLIDS
Methods of Particle Size Determination
1. Optical Microscopy
Methods of Particle Size Determination
1. Use of microscope to measure individual particles
§
2. Sieving
3. Sedimentation
4. Particle Surface Area
3 Measurements
Ferret Diameter
Two tangents separated by the
longest distance
Martin Diameter
Distance that will bisect the
particles into halves
Projected Area of the Circle
Diameter of the circle that will
enclose the particle
SOLIDS
SOLIDS
Methods of Particle Size Determination
1. Use of microscope to measure individual particles
Methods of Particle Size Determination
2. Sieve Analysis
- USP Method
- Mesh (Number of openings per linear inch)
Sample Problem:
20
40
10 g
20 g
60
80
40 g
25 g
Collecting pan 5 g
Solve for the % of particles with size > mesh 40
SOLIDS
SOLIDS
Methods of Particle Size Determination
Methods of Particle Size Determination
3. Sedimentation Method
§ Based on the sedimentation rate of particles
4. Automatic Particle Counters
§
Principle is Stokes’ Law
v = rate of settling
d = diameter
ρ s = density of the particles
ρ 0 = density of the dispersion
medium
g = acceleration due to gravity
η = viscosity of the medium
Lecturer: Daryl E. Magno, RPh
Type of Counter
Principle Involved
Coulter Counter
Electronic resistance
HIAV/ Royco Light
Blockage Instrument
Light blockage
Beckman Coulter Model
Photon correlation
Spectroscopy
Microcomputerized
mercury porosimetry
Adsorption
39
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
SOLIDS
SOLIDS
Derived Properties
Porosity
Derived Properties
Particle Volume
•
Measure of total voids present in a particle
True/ Particle Volume
Granule Volume
Vp
Vg
Volume of particle
V p + intraparticle
pores/spaces/voids
Bulk Volume
Vb
V g + interparticle
pores/spaces/voids
Void Volume
v
Volume of the spaces
SOLIDS
SOLIDS
Derived Properties
Density of Particles
Porosity
True Density
ρ
M/Vp
Granule Density
ρg
M/Vg
Bulk Density
ρb
M/Vb
• ∈
• ∈
𝒊𝒏𝒕𝒆𝒓𝒔𝒑𝒂𝒄𝒆 =
• ∈
𝒕𝒐𝒕𝒂𝒍=
𝒊𝒏𝒕𝒓𝒂𝒔𝒑𝒂𝒄𝒆=
(Vb – Vg)/Vb X 100
(Vg – Vp)/Vg X 100
(Vb – Vp)/Vb X 100
Sample Problem
Given: volume of particles = 0.3
intraparticle spaces = 0.1
spaces between particles = 1.6
Solve: Vg, Vb, ∈ 𝒊𝒏𝒕𝒆𝒓𝒔𝒑𝒂𝒄𝒆 ,∈
𝒊𝒏𝒕𝒓𝒂𝒔𝒑𝒂𝒄𝒆 ,∈ 𝒕𝒐𝒕𝒂𝒍
SOLIDS
SOLIDS
Properties of Powders for Granulation
§ Fluidity or flowability
Methods for Testing Flowability and Compressibility
1. Angle of Repose
§ Compressibility
Methods for Testing Flowability and Compressibility
1. Angle of Repose
•
The angle assumed by a cone-like pile of powder
relative to the horizontal base
Angle= Arc tan (H/R)
Lecturer: Daryl E. Magno, RPh
- The angle assumed by a cone-like pile of powder
relative to the horizontal base
Angle= Arc tan (H/R)
Sample Problem: 100g of powdered sample was made to
flow from a funnel suspended at a height of of 15cm. A
powder cone 12cm tall from the surface with a diameter
of 5cm was made. Determine the angle of repose.
40
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
SOLIDS
1/25/18
SOLIDS
Methods for Testing Flowability and Compressibility
1. Carr’s Index or Compressibility Index
CI = (Vi- Vf) / Vi x 100
2. Hausner’s Ratio
HR= Vi / Vf
*Vi= Initial Volume or Untapped Volume
*Vf= Tapped Volume
1. Size and shape are ex amples of what typ e of particl e
properties:
a. Derived
b. Fundamental
c. Both
d. None
CHECKPOINT
3. Triclinic
a. NaCl
b. Urea
c. Sucrose
d. Boric Acid
4. Properties are of polymorphs are identical
a. Enantiotropic
b. Monotropic
c. Isotropic
d. Anisotropic
Lecturer: Daryl E. Magno, RPh
2. Measuremen t of particle size through op tical microscopy
which is the distance that will bisect the particle into halves
a. Ferret Diameter
b. Martin Diameter
c. Projected Area of the Circle
d. AOTA
5. The angle assumed by a cone-like pile of powder
relative to the horizontal base
a. Angle of Repose
b. Carr’s Compressibility Index
c. Hausner’s Ratio
d. NOTA
41
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
THERMODYNAMICS
THREE LAWS OF THERMODYNAMICS
1. Law of Conservation of Energy, states that
energy cannot be created or destroyed
THERMODYNAMICS
2. The second law states that the entropy of any
isolated system always increases
3. The third law of thermodynamics states that the
entropy of a system approaches zero as
the temperature approaches absolute zero
THERMODYNAMICS
THERMODYNAMICS
Phase Diagram
• Represents the states of matter that exist as
temperature and pressure are varied
Latent Heat or Molar Heat
• Heat necessary for 1 mole of gas, solid or liquid
to change to another phase
THERMODYNAMICS
GIBB’s PHASE RULE
§ Used to determine the number of
independent variables that must be set in
order to define a system
§ Degree of Freedom
F = C-P + 2 (One-Compone nt System)
F = C-P + 1 (Two-Compone nt System)
F = C-P
(Three-Component System)
Lecturer: Daryl E. Magno, RPh
Heat of Fusion
Amount of energy that must be added to a mole of
solid at constant pressure to turn it directly into a
liquid (melting) or Amount of energy that must be
added to a mole of liquid at constant pressure to turn
it directly into a solid (freezing)
Heat of Vaporization
Amount of energy that must be added to a mole of
liquid at constant pressure to turn it directly into a gas
Heat of Sublimation
Amount of energy that must be added to a mole of
solid at constant pressure to turn it directly into a gas
THERMODYNAMICS
GIBB’s PHASE RULE
§ Example: A system comprising of liquid,
water, in equilibrium with its vapor.
Determine the degree of freedom.
§ Then we cool the liquid water and its vapor
until a third phase (ice) separates out.
Compute for the dree of freedom.
42
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
THERMODYNAMICS
THERMODYNAMICS
THERMODYNAMICS
THERMODYNAMICS
One-Component System
Two-Component System containing Liquid Phases
THERMODYNAMICS
Two-Component System containing Liquid Phases
• The line bc drawn across the region
containing two phases is termed a tie line;
• All systems prepared on a tie line, at
equilibrium, will separate into phases of
constant composition.
• These phases are termed conjugate
phases.
Lecturer: Daryl E. Magno, RPh
One-Component System
Two-Component System containing Liquid Phases
•
The maximum temperature at which the twophase region exists is termed the critical
solution, or upper consolute temperature.
•
In the case of the phenol–water system, this is
66.8°C (point h ).
•
All combinations of phenol and water above this
temperature are completely miscible and yield
one-phase liquid systems.
THERMODYNAMICS
Two-Component System containing Liquid Phases
Lever rule is a tool used to determine mole
fraction of each phase of a binary equilibrium
phase diagram.
Problem: Prepare a 50-g of mixture of 24%
phenol in water solution. Determine the
proportion of component A (water) and
component B (phenol) at 50 degrees celsius.
43
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
THERMODYNAMICS
Two-Component System containing Eutectic Mixtures
1/25/18
THERMODYNAMICS
Two-Component System containing Eutectic Mixtures
• solid–liquid mixtures in which the two
components are completely miscible in the
liquid state and completely immiscible as
solids.
• Examples of such systems are:
üsalol–thymol,
üsalol–camphor,
üacetaminophen–propyphenazone.
THERMODYNAMICS
Three-Component System
CHECKPOINT
1. Maximum temperature at which the two-phase region exists
a. Critical solution temperature
b. Tie Line
c. Conjugate phase
d. Lever arm
2. Amount of energ y that must be added to a mole of liquid at
constant pressure to turn it directly into a gas
a. Heat of Fusion
b. Heat of Vaporization
c. Heat of Sublimation
d. NOTA
Lecturer: Daryl E. Magno, RPh
3. Represents the states of matter that exists as
temperature and pressure varied
a. Thermodynamics
b. Chemical Kinetics
c. Phase Diagram
d. NOTA
4. Point where 3 distinct phases co-exist
a. Triple
b. Eutectic
c. Critical
d. Polymorphic
44
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
CHEMICAL KINETICS
-
CHEMICAL KINETICS
Study of the rates of reactions and the mechanism of
which these reactions occur
Order of Reaction
- the way in which the concentration of the drug or
reactant in a chemical reaction affects the rate
CHEMICAL KINETICS
CHEMICAL KINETICS
1. Zero order Reaction
- Constant amount of drug is eliminated from the
body (independent of the concentration in the
body)
2. First Order Reaction
- Constant percentage of drug is eliminated from the
body (dependent of the concentration in the
body)
C= -kt + Co
ln C= -kt + ln Co
CHEMICAL KINETICS
CHEMICAL KINETICS
Example Problem:
Initially a drug has concentration of 50mg/mL and after
20 days, the drug concentration became 24mg/mL.
Compute for 1st order rate constant.
Half-life
-period of time required for the amount or
concentration of a drug to decrease by 50%
Zero order:
Half life = 0.5 C
First order
Half life= 0.693/K
Lecturer: Daryl E. Magno, RPh
45
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
1/25/18
CHEMICAL KINETICS
Sample Problem:
A drug has a biological half-life of 2 hours. At the end of
eight hours, what percentage of the drug’s original activity
will remain?
1. Atoms of the same element with the same number
of nucleon number, but differing numbers of atomic
number
a. Isotopes
b. Isobars
c. Isotones
d. AOTA
2. Crystal system of sucrose.
a. Cubic
b. Tetragonal
c. Monoclinic
d. Triclinic
5. A patient diagnosed with type II DM is administered an
oral dose of 0.1 mg chlorpropramide (pKa=2.0). Determine
the amount of drug that can be absorbed in the stomach
(pH=5.0)
a. 99.9mcg
b. 90mg
c. 10mcg
d. 0.10mcg
6. Forces between molecules of similar phases
a. Adhesion
b. Cohesion
c. Both
d. None
Lecturer: Daryl E. Magno, RPh
FINAL
CHECKPOINT
3. Mixing a hypertonic solution with red blood cell will
cause _________ of the red blood cell
a. Bursting
b. Crenation
c. Chelation
d. Hemolysis
4. Dispersed system possess this property
a. Optical activity
b. Solubility
c. Thixotropy
d. Tonicity
For nos. 7-8. 6.3 mg of a boron hydride is contained in a flask of 385
mL at 25.0° C and a pressure of 11 torr.
7. Determine the molar mass of the hydride.
a. 27.70
b. 2760.50
c. 2.706
d. 276.70
8. Which of the following hydrides is contained in the flask?
Given: MW B= 10.8110 g/mol
a.
b.
c.
d.
BH3
B2 H6
B4 H10
None
46
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
9. Buffer solution is composed of:
a. Strong acid and strong base
b. Strong electrolyte and strong base
c. Weak acid and its salt
d. Strong acid and its salt
10. Process whereby solvent moves from higher to lower
concentration until equilibrium is reached
a. Osmosis
b. Diffusion
c. Adsorption
d. Absroption
13. How much NaCl is needed to adjust the following prescription to
isotonicity?
Rx
Pilocarpint Nitrate 1%
E = 0.15
NaCl
Purified Water q.s.
50mL
a. 0.5g
b. 0.450g
c. 0.375g
d. 0.245g
14. Vant’ Hoff factor of incomplete dissociated CaCl 2
a. 1
b. 1.8
c. 2.6
d. 3.4
17. Formula for intraspace porosity
a. (Vb-Vg/Vb) X 100
b. (Vb + Vp/Vb) X 100
c. (Vg-Vp/Vg) X 100
d. (Vb-Vp/Vb) X 100
18. The initial plasma concentration of a drug given IV at
9:00am is 250mg/mL. If the half-life of the drug is 6 hours,
perform a calculation to predict the plasma concentration
will be at 9:00pm that same day.
a. 125mg/mL
b. 62.5mg/mL
c. 31.25mg/mL
d. 15.62mg/mL
Lecturer: Daryl E. Magno, RPh
1/25/18
11. States that an acid is a proton donor and base is a
proton acceptor
a. Arrhenius
b. Bronsted-Lowry
c. Lewis
d. AOTA
12. USP method for adjusting isotonicity
a. E value
b. D value
c. White Vincent Method
d. Sprowl Method
15. A solution of ferrous sulfate was prepared by adding
50 grams of FeSO4 (MW= 151.9g/mol) to enough water to
make 1000mL of solution. Compute for M.
a. 0.453
b. 0.329
c. 0.555
d. 0.378
16. Weakest force of attraction
a. Keesom
b. Debye
c. London
d. NOTA
For nos. 19-20. 100g powder
Mesh #
Grams
20
40
60
80
Collecting Pan
of powder that did not pass
15g
25g
10g
35g
15g
19. What is the % of powder smaller than mesh 40 and larger than mesh 80?
a. 70%
b. 45%
c. 35%
d. 15%
20. What is the % powder larger than mesh 40?
a. 10%
b. 25%
c. 35%
d. 40%
47
Pharmacist Licensure Examination
Module 5: Pharmacutics (17.5%)
21.Using White-Vincent Method, compute for
the required volume to make 0.2g of
atropine sulfate (E=0.13) isotonic. Answer
is in 2 decimal places.
a. 2.88
b. 2.89
c. 3.00
d. NOTA
1/25/18
22. Solvent loving
a. Lyophobic
b. Lyophilic
c. Amphiphilic
d. NOTA
23. Density is a derived quantity from
a. Mass and Volume
b. Mass and Length
c. Mass and Weight
d. NOTA
24. Period of time where 90% of the original concentration
is left
a. Half-life
b. Shelf-life
c. Both
d. None
25. 322 L of hydrogen occupies a volume of 197 L at STP. If
the initial temperature of the hydrogen was 37° C, what
was its initial pressure?
a. 0.69
b. 0.79
c. 0.89
d. 0.99
Lecturer: Daryl E. Magno, RPh
END OF PHYSICAL PHARMACY REVIEW
48