Class 2 - Basic calculations and physicochemical.pdf

Learning Objectives
 At the end of this session , the student is expected to be able to :
 Understand some basic concepts of concentration
 Discuss Units for Expressing Concentration
 Explain stability of drugs
 Describe Potential Adverse Effects of Instability in Pharmaceutical
Products
 Explain factors affecting Stability
 Discuss types of stability
 Explain stability testing methods

Some basic concepts of concentration
3
 Concentration is a general measurement unit stating the amount
of solute present in a known amount of solution
Or
 An expression stating the relative amount of solute per unit
volume or unit mass of solution.
 Is a measure of how much of a material (solute) is dissolved in
certain solution and
Can be expressed in different way, which are known as the units
of concentrations.

4
 Although the terms “solute” and “solution” are often associated
with liquid samples, they can be extended to gas-phase and solid-
phase samples as well.

Units for Expressing Concentration
5
 The actual Units for reporting concentration depend on how the
amounts of solute and solution are measured.
 There are physical and chemical basis of expressing concentration.
 Chemical units interested in knowing how many moles of the
material present in each unit of volume (L) of the solution
 Physical units interested in knowing how many grams (mg, μg,..)
present in unit volume (L, ml, μl,…)
 Most common physical units used in pharmaceutical analysis:
mg/ml, μg/ml, part per million (ppm) which means number of
mass units (mg, g, …) in one million of that mass units

Common Units for Reporting
Concentration
6

Percent concentration (parts per hundred)
(Weight, Volume, and Weight-to-Volume Ratios)
7
 Express concentration as units of solute per 100 units of sample.
The term % is another way of expressing concentrations
physically.
 Weight percent (% w/w)
 Grams of solute per 100 g of solution(%w/w).
 Volume percent (% v/v)
 Milliliters of solute per 100 mL of solution (% v/v).
 Weight-to-volume percent(% w/v)
 Grams of solute per 100 mL of solution (% w/v).
E. g. A solution in which a solute has a concentration of 23% w/v
contains 23 g of solute per 100 mL of solution.

8
 Volume percent is commonly used to specify the concentration
of a solution prepared by diluting a pure liquid with another
liquid
 used where both chemicals are liquids
E.g. I. If 50 mL of acetic acid diluted by adding 50 mL of
water; there is now 50 mL of acetic acid in a total volume
of 100 mL, hence the acetic acid concentration is now 50%
v/v
E.g. II. 5% aqueous solution of methanol – usually means a
solution prepared by diluting 5.0 ml of pure methanol to give 100ml
of solution with enough water.

9
 Weight /volume percent is employed to indicate the composition
of dilute aqueous solutions of solid reagents
 Used where a solid chemical is dissolved in liquid
 Example 1, If 10 g of table salt, sodium chloride dissolved to
make up a total volume of 100 mL of solution; 10% w/v solution
of sodium chloride can be made )
 Example 2, 5% aqueous silver nitrate often refers to a solution
prepared by dissolving 5g of silver nitrate in sufficient water to
give 100ml of solution.

10
 Weight per weight
 Used where the weight of each chemical is used and not the
volume
 e.g. If 10 g of fat dissolved in 90 g ethanol so the total mass of
the whole solution is 100 g, then 10% w/w solution of fat can be
made.

11
 Percent composition of a solution can be expressed as:
100
)
/
( x
tion
massofsolu
te
massofsolu
w
w
ent
weightperc 
100
)
/
( x
lution
volumeofso
lute
volumeofso
v
v
ent
volumeperc 
100
)
/
(
/ x
lution
volumeofso
lute
weightofso
v
w
ent
volumeperc
weight 

Practical examples
12
 Calculate the number of g of NaCl in 15 ml of a 15 % w/v
solution. (Ans=2.25)
 A solution is prepared by mixing 25.5g of a salt with 150 grams
of water. Calculate the mass percent of salt in the seawater
solution.
 Cow’s milk contains 4.5% lactose by mass. Calculate the mass of
lactose present in 175 grams of milk.

13
Molar concentration
 Molarity is the concentration of a particular chemical species in
solution.
 Molar solutions are the most useful in chemical reaction
calculations because they directly relate the moles of solute to the
volume of solution.
 is the number of moles of a species that is contained in one liter
of the solution (not one liter of the solvent). Or
Analytical molarity is total number of moles of a solute in one
liter of the solution.
 The unit of molar concentration is molarity, M, which has the
dimensions of mol L-1

14
 Calculate the molar concentration of ethanol in an aqueous
solution that contains 2.30g of C2H5OH (46.07 g/mol) in 3.50L of
solution.
.(ans. 0.0143M) (Divide the mass of the solute by the
molecular weight to determine the moles of solute
present in solution).

15
Parts per million & parts per billion
 Parts per million (ppm) and parts per billion (ppb) are
examples of expressing concentrations by mass.
 ppm and ppb are mass ratios of grams of solute to one
million or one billion grams of sample, respectively.

16
 For very dilute solutions, parts per
million (ppm) is convenient way to express
concentration:
Where Cppm is the concentration in parts per million.
The units of mass in the numerator & denominator
must agree.
For even more dilute solutions we use parts per
billion.
Cppm =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜n
X 106

17
Reading assignment
 Normality
 Dilutions
 Density and specific gravity (definitions and worked examples)
Cppb =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜n
X 109

pppppppppppppppppp
18
 For example, suppose a 155.3 g sample of pond water is
found to have 1.7x10-4 g of phosphates. What is the
concentration of phosphates in ppm?
 A similar procedure would be followed to calculate ppb.
 In the above example the pond water would be 1,100 ppb.

19
 Now suppose we have 400 g sample of pond water and it
has a concentration of 3.5 ppm dissolved nitrates. What is
the mass of dissolved nitrates in this sample?

20
Physical and chemical properties of drug
molecules

The ability of a chemical compound to elicit a pharmacological/
therapeutic effect is related to the influence of various physical
and chemical (physicochemical) properties of the chemical
substance on the bio molecule that it interacts with.
 The physical properties of drug molecules, along with simple
chemical derivatisation and degradation reactions, play an
important part in the design of analytical methods.
 Drug molecules can be complex, containing multiple functional
groups that in combination produce the overall properties of the
molecule.

Various Physico-Chemical Properties are,
 Solubility
 Partition Coefficient
 Ionization
 Chelation and etc…….

Partition Co-efficient
 Is one of the Physico chemical parameter which influencing the
drug transport & drug distribution., the way in which the drug
reaches the site of action from the site of application.
Partition co-efficient is defined as equilibrium constant of drug
concentration for a molecule in two phases.
The partition coefficient is a measure of the relative affinity of a
drug for the lipid and aqueous phases.
Lipid water solubility coefficient (Kp) is the ratio of dissolution
of drug in lipid as compared to water.
Kp = [drug] in lipid phase/[drug] in aqueous phase

Greater the lipid water solubility coefficient, more is the lipid
solubility of the drug and greater is the absorption.
Less the coefficient, less is the lipid solubility and less is
the absorption.
One can control the Kp by modifying the side groups on the
compound.
The more C and H on the compound, the more lipid soluble, and
thus the higher the Kp.
However, O, S and the more water-soluble the compound possess
the lower the Kp.
Drugs with benzene ring, hydrocarbon chain, steroid nucleus and
halogen groups in their structures are lipid soluble.

Chelation
The compounds that are obtained by donating electrons to a
metal Ion with the formation of a ring structure are called
chelates.
 Ligands: The compounds capable of forming a ring structure
with a metal are termed as ligands.

Introduction to acid base theory
26
 Most of the drugs used in medicine behave in solution as weak
acids, weak bases, or sometimes as both weak acids and weak
bases.
 Acidic pH favors acidic drug absorption while basic pH is better
for basic drugs.
 In order to understand and appreciate these compounds a study
must be made of simple acid–base theory.

27
In 1887, the Swedish chemist Svante August Arrhenius
suggested that solutions that conduct electricity (so-called
electrolytes) do so because they dissociate into charged
species called ions.
 Compounds of this type may be classiﬁed as strong electrolytes,
which dissociate almost completely into ions in solution, or as
weak electrolytes, which only dissociate to a small extent in
solution.
 The law of mass action can be applied to the dissociation of
water, a weak electrolyte widely used as a solvent in biological
and pharmaceutical systems.

28
 Since strong electrolytes are almost completely dissociated in
solution, measurement of the equilibrium constant for their
dissociation is very difﬁcult.
 For weak electrolytes, however, the dissociation can be expressed
by the law of mass action in terms of the equilibrium constant.
 The law of mass action states that „the rate of a chemical reaction
is proportional to the active masses of the reacting substances‟,
but for dilute solutions active mass may be replaced by
concentration, which is much easier to measure.

29
 Considering the reaction
 The equilibrium constant (K) for the reaction is given by the
product of the concentrations of the reaction products divided by
the product of the concentrations of the reactants, or

30
 When water dissociates it yields a hydrogen ion and a hydroxide.
H2O ↔ H+ + OH-
 Kw the dissociation constant in case of water dissociation
 Since the concentration of water does not change appreciably
due to dissociation, ----don‟t have an effect on the equilibrium
i.e. Kw = [H+][OH-] = 1x10-14 at 25°C

31
 Since, in pure water, [[H+]=[OH-] , the hydrogen ion concentration in
water is given by the square root of Kw.
[H+] = [OH-] = 1x10-7
 Using the concentration of hydrogen and hydroxide ions, the following
relationship results:
 Acidic Solution: [H+] > 1x10-7
 Basic Solution: [H+] < 1x10-7
The units of Kw are mole litre-1 ×mole litre -1 Or mole2 litre -2.

Calculations of pH value
32
 pH is a logarithmic measurement of the number of moles of
hydrogen ions (H+) per liter of solution.
 There are several ways to define acids and bases, but pH only
refers to hydrogen ion concentration and is only meaningful when
applied to aqueous (water-based) solutions.
 The pH of a solution is defined as the negative logarithm (to the
base 10) of the hydrogen ion concentration‟
[H+] = 10-pH


33
 pH is the measure of Solution Acidity
 In water we find an actual range of pH from about 0 to 14
 It is convenient to classify aqueous solutions according to their
pH.
 If the pH of a solution is less than 7, the solution is called acidic;
 if the pH is about 7, the solution is neutral;
 if the pH is greater than 7, the solution is called basic.

34
 Although the analogous concept of basicity is much less
commonly used,
 the pOH notation is sometimes convenient.
 Only one scale need be used, since they are related through the
ion product of water.
[H3O+][OH-] = Kw = 1 x 10-14 in molar concentration units.
 Taking the logarithm of both sides;
log [H3O+] + log [OH-] = log Kw = -14
Then;
-log[H3O+] + (-log[OH-]) = -logKw = +14 or
pH + pOH = pKw = 14

35
 In an acidic solution, then, the concentration of hydrogen ions is
greater than the concentration of hydroxide ions.
 In a neutral solution, the concentrations of hydrogen ions and
hydroxide ions are roughly equal.

36
Dissociation of Strong Acids
 A strong acid, by definition, is an acid that is completely
ionized in aqueous solution.
 If an acid is introduced into an aqueous solution the [H+]
increases.
 A strong acid is completely ionized in water and the [H+] is
equal to its molarity
Eg. 0.1M HCl contains 0.1M [H+] and has a pH of 1
 In an aqueous solution of a strong acid, acidity is essentially
determined by the concentration of strong acid alone.
 the contribution from the auto ionization of water is almost
always negligible.

37
 In the case of hydrochloric acid, the ionization equilibrium
reaction is:
HCl + H2O H3O+ + Cl-
 [H3O+] = [HCl]
 Example. To calculate the pH of 0.0010 M aqueous HCl solution;
[H3O+] = 0.0010
pH = -log[H3O+] = 3.00
 Is there any other source of protons other than the acid itself?
 The autoionization of water could give 1 x 10-7 M H3O+.
 But this additional source of protons is usually negligibly small
compared to the amount produced by the acid.

Dissociation of Strong Bases
38
 A strong base, like a strong acid, is virtually totally dissociated in
water;
 Thus 0.01 molar NaOH is actually an aqueous solution 0.01 M in
Na+ and also 0.01 M in OH-.
 Sodium ion cannot act as an acid (no protons) or as a base
(positive charge will repel a proton).
 The sodium ions have a negligible effect on the pH.
 So, the pH is determined by the hydroxide ion concentration.
 It is often more convenient to calculate the pH of basic solutions
by first calculating the pOH as is done in the following example.

39
Example-1. The pH of 0.010 M aqueous NaOH
solution is calculated as follows:
[OH-] = 0.010 M
pOH = 2.00
pH + pOH = pKw = 14.00, so
pH = 12.00
Example-2.
For strong base like 0.1M NaOH, [OH-] = 0.1M and
[H+]= and its pH is 13

40
As in the case of an aqueous solution of a strong acid, the
contribution from the autoionization of water is almost
always negligible.
In an aqueous solution of a strong base, acidity is
essentially determined by the concentration of the strong
base alone.

Dissociation of Weak acids and bases
41
 Not completely ionized in solution
 Are in equilibrium with the undissociated acid or
base
Ka for a weak acid is given by:
For instance for a 0.1M of acetic acid(Ka= )

42
CH3COOH H+ or H3O+ CH3COO-
initial 0.10 0 0
shift -x x x
final 0.10-x x x

43
The calculation of the pH of a weak base can be considered in the
same way.

Other “p” Scales
44
 The “p” in pH tells us to take the negative log of the quantity
(in this case, hydrogen ions).
Some similar examples are
pOH = −log [OH−]
pKw = −log Kw

Watch This!
45
Because
[H3O+] [OH−] = Kw = 1.0  10−14,
we know that
−log [H3O+] + −log [OH−] = −log Kw = 14.00
or, in other words,
pH + pOH = pKw = 14.00

How Do We Measure pH?
46
 There are different common methods for testing the pH of liquids, each
of which varies in its accuracy and application.
 Litmus paper is a small strip of paper that has been dipped in a
combination of dyes that change color according to the pH of the
medium in which they are contained.
 Measurement is made by briefly dipping the end of an unused strip in
the liquid and allowing it to dry.
 Acidic liquids (pH below 7) turn the paper red while alkaline liquids
(pH above 7) change it to blue or purple.
 Litmus paper is good for roughly estimating the relative pH of liquids,
but not for precise readings.

pH meter (Probe and Meter)
47
 The most accurate common means of measuring pH is through a
lab device called a probe and meter, or simply, a pH meter.
 The probe consists of a glass electrode through which a small
voltage is passed.
 The use of a specially-prepared electrode designed to allow
hydrogen ions in the solution to migrate through a selective
barrier, producing a measurable potential (voltage).
 The meter, a voltmeter, measures the electronic impedance in the
glass electrode and displays pH units instead of volts.

48
 Measurement is made by submerging the probe in the liquid until
a reading is registered by the meter.
 A pH meter typically has to be calibrated before each use with
two standard liquid solutions of known pH.

How Do We Measure pH?..........
For more accurate
measurements, one
uses a pH meter,
which measures the
voltage in the
solution.
49

Ionization and Pka
50
Most of the drugs are either weak acids or base and can exist in
either ionized or unionized state.
The ionization of the drug depends on its pKa & pH.
The rate of drug absorption is directly proportional to the
concentration of the drug at absorbable form but not the
concentration of the drug at the absorption site.
Eg: Aspirin in stomach will get readily absorbed because it is
in the unionized form(99%).

51
 The extent of ionization of a compound can have a large effect on many
biological properties, such as receptor/enzyme binding, binding to
plasma proteins, CNS penetration, solubility and absorption.
 Drugs are only absorbed passively when they are unionised.
This is because the compound has to pass through a lipophilic (‘fat
loving’) membrane and this process will be unfavourable for charged
molecules.
 In a more acidic medium, such as the stomach, the percentage ionized
for an acidic compound will be less and so more compound will have
the capacity to be passively absorbed.

52
 In comparison, a basic compound in an acidic medium will be
more ionized and so less of the compound have the capacity to
undergo passive absorption.
 This may well be the reason for the observation that acidic
compounds generally have better fraction absorbed (hence
bioavailability) than bases.
 If pKa is used as a measure of acidic or basic strength:
 For an acid, the smaller the pKa value the stronger the acid;
 For a base the larger the pKa value the stronger the base.

53
 The equilibrium between un-ionized and ionized forms is
defined by the acidity constant Ka or pKa = -log10 Ka
 By using drug pKa, the formulation can be adjusted to pH
to ensure maximum solubility in water or maximum
solubility in non-polar solvent.
 The pH of a substance can be adjusted to maintain water
solubility and complete ionization.
 Eg: Phenytoin injection must be adjusted to pH 12 with Sodium
Hydroxide to obtain 99.98% of the drug in ionized form.
 Tropicamide eye drops, an anti cholinergic drug has a pka of 5.2
and the drug has to be buffered to pH 4 to obtain more than 90%
ionisation.

Dissociation Constants/Ionization constants
The greater the value of Ka, the stronger the acid.
54

Calculating Percent Ionization
 Once the pKa value of a molecule is known, then it is possible
to calculate the proportion of ionized and neutral species at any
pH.
% ionized = 100 / 1 + 10(pKa - pH)
When an acid or base is 50% ionised: pH = pKa
55

Calculating Ka from the pH
56
 The pH of a 0.10 M solution of formic acid,
HCOOH, at 25°C is 2.38. Calculate Ka for formic
acid at this temperature.
 We know that
 To calculate Ka, we need the equilibrium
concentrations of all three things.
 We can find [H3O+], which is the same as
[HCOO−], from the pH.
[H3O+] [COO−]
[HCOOH]
Ka =

57
pH = −log [H3O+]
2.38 = −log [H3O+]
−2.38 = log [H3O+]
10−2.38 = 10log [H3O+] = [H3O+]
4.2  10−3 = [H3O+] = [HCOO−]
[4.2  10−3] [4.2  10−3]
[0.10]
Ka =
= 1.8  10−4

Calculating Percent Ionization
58
 Percent Ionization =  100
 In this example
[H3O+]eq = 4.2  10−3 M
[HCOOH]initial = 0.10 M
Percent Ionization =
 100
[H3O+] eq
[HA]initial
4.2  10−3
0.10
= 4.2%

Calculating pH from Ka
59
Calculate the pH of a 0.30 M solution of acetic acid,
HC2H3O2, at 25°C.
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2
−(aq)
Ka for acetic acid at 25°C is 1.8  10−5.

60
The equilibrium constant expression is
[H3O+] [C2H3O2
−]
[HC2H3O2]
Ka =
We next set up a table…
[C2H3O2], M [H3O+], M [C2H3O2
−], M
Initially 0.30 0 0
Change −x +x +x
At Equilibrium 0.30 − x  0.30 x x
We are assuming that x will be very small compared
to 0.30 and can, therefore, be ignored.

Calculating pH from Ka…
61
Now,
(x)2
(0.30)
1.8  10−5 =
(1.8  10−5) (0.30) = x2
5.4  10−6 = x2
2.3  10−3 = x
pH = −log [H3O+]
pH = −log (2.3  10−3)
pH = 2.64

Weak Bases
Bases react with water to produce hydroxide ion.
62

Weak Bases
63
The equilibrium constant expression for this
reaction is
[HB] [OH−]
[B−]
Kb =
where Kb is the base-dissociation constant.

Weak Bases
Kb can be used to find [OH−] and, through it, pH.
64

pH of Basic Solutions
65
What is the pH of a 0.15 M solution of NH3?
[NH4
+] [OH−]
[NH3]
Kb = = 1.8  10−5
NH3(aq) + H2O(l) NH4
+(aq) + OH−(aq)
[NH3] [NH4+], M [OH−], M
Initially 0.15 0 0
At Equilibrium 0.15 - x  0.15 x x

66
(1.8  10−5) (0.15) = x2
2.7  10−6 = x2
1.6  10−3 = x
Therefore,
[OH−] = 1.6  10−3 M
pOH = −log (1.6  10−3)
pOH = 2.80
pH = 14.00 − 2.80
pH = 11.20
2
(x)2
(0.15)
1.8  10−5 =

Ka and Kb
Ka and Kb are related in this way:
Ka  Kb = Kw
Therefore, if you know one of them, you can
calculate the other.
67

BUFFER SOLUTIONS (BUFFERS)
 A buffer solution is a solution that resists changes in pH.
“Solutions which resist changes in pH when small quantities of
acid or alkali are added ” or when dilution occurs.
 If acid is added then, within reason, the pH does not fall; if base is
added, the pH does not rise.
 Buffers are usually composed of a mixture of weak acids or weak
bases and their salts.
 Function best at a pH equal to the pKa of the acid or base
involved in the buffer.
68

 Buffer solutions contain a weak acid and its conjugate base or a
weak base and its conjugate acid.
 For weak acid/conjugate base systems, when a base is added to
the solution it reacts with the weak acid to form more conjugate
base.
 When an acid is added to the same solution it reacts with
conjugate base to form more weak acid.
 In both cases the pH remains approximately constant since the
number of OH ions in solution is approximately constant.
 A similar system is set up for weak base/ conjugate acid buffers.
69

 Together the two species (conjugate acid and conjugate base)
resist large changes in pH by absorbing the H+ ions or OH- ions
added to the system.
 When H+ ions are added to the system they will react with the
conjugate base in the buffer .
 When OH- ions are added they will react with the conjugate acid
in the buffer
70

 In buffer systems the larger the amount of weak
acid/conjugate base or weak base/conjugate acid the better
the buffer.
 In addition, the best buffers have equal amounts of weak
acid and conjugate base or weak base and conjugate acid.
71

Biological use of buffers
 In biological systems (saliva, stomach, and blood), it is essential
that the pH stays „constant‟ in order for any processes to work
properly.
 Most enzymes work best at particular pH values.
 The pH of blood is normally about 7.4, if the pH varies by 0.5 it
can lead to unconsciousness and coma.
 E.g. Carbon dioxide produced by respiration can increase
the acidity of blood by forming H+ ions in aqueous solution
74

 The presence of hydrogen carbonate ions in blood removes excess H
75

 Many household and cosmetic products need to control their
pH values.
 To maintain a pH of about 6 to prevent bacteria multiplying
in baby lotion
76

Types of buffer solution
 Acidic Buffer (pH < 7)
E.g. weak acid + its sodium or potassium salt
Ethanoic acid + sodium ethanoate
 Alkaline Buffer (pH > 7)
E.g. weak base + its chloride
ammonia + ammonium chloride
The equation that predicts the behaviour of buffers is known
as the Henderson–Hasselbalch equation.
 The Henderson-Hasselbalch equation allows us to calculate the
pH of a buffered system.
It is derived as follows, by considering a weak acid that
ionizes in solution:
77

 Since the acid in question is weak, the number of A- ions
derived from dissociation of the acid itself is very small
compared with the number derived from the fully ionized salt.
 This means that [A-] is approximately equal to total
concentration [SALT]; and similarly [HA], since the acid is
weak and predominantly unionized, is approximately equal to
the total acid concentration [ACID].
 The equation can now be rewritten as
79

 The Henderson–Hasselbalch equation can also be derived from
consideration of the ionization of a weak base, B, which ionizes
in aqueous solution as follows:
 In this case the [SALT] term can be replaced by the
concentration of the conjugate acid of the weak base,
[BH+], which, in effect, yields the same equation.
80

 An example of a buffer is a mixture of acetic acid and sodium
acetate, which will ionize as follows:
 Since the acetic acid only ionizes to a small extent, there will be
a high concentration of undissociated acid or, to put it another
way, the equilibrium for the reaction will lie predominantly to the
left hand side.
81

 Sodium acetate is a salt and will ionize completely to give high
concentrations of CH3COO- and Na+.
 If ions(hydrogen)are now added to the buffer solution, they will
react with the high concentration of CH3COO- present to give
undissociated acetic acid.
 Acetic acid is a weak acid and only dissociates to a small extent,
so the pH of the solution does not decrease.
82

 Similarly, if OH-ions are added to the buffer system, the OH-
will react with the high concentration of free acetic acid present
to give water and acetate ions:
 Neither water nor acetate is sufﬁciently basic to make the
solution alkaline, so the pH of the buffer solution will not
increase.
83

Why does a buffer resist change in pH when small
amounts of strong acid or bases is added?
The acid or base is consumed by A- or HA respectively
 Le Châtelier’s Principle states that when stress is placed on a
system in equilibrium, the system will react to relieve the
stress.
 Stress refers to changes in concentration, pressure, volume,
and temperature.
84

Buffer capacity, :
 Measure of how well solution resists change in pH when
strong acid/base is added.
 Quantitative measure of this resistance to pH changes is called
buffer capacity.
 Larger   more resistance to pH change
 The buffer capacity is greatest when there are equal amount of
weak acid and conjugate base or weak base and conjugate acid.
 In addition, as the overall amount of weak acid and conjugate
base goes up the buffering capacity of the solutions go up (same
as for weak base and conjugate acid).
85

How a Buffer Works
 Consider adding H3O+ or OH- to water and also to a buffer
 For 0.01 mol H3O+ to 1 L water:
[H3O+] = 0.01 mol/1.0 L = 0.01 M pH = -log([H3O+]) = 2.0
So, change in pH from pure water: dpH = 7.00 – 2.00 = 5.0
 For the H2CO3
- / HCO3
- system: pH of buffer = 7.38
 Addition of 0.01 mol H3O+ changes pH to 7.46
So change in pH from buffer: dpH = 7.46 – 7.38 = 0.08 !!!
87

Let‟s consider a buffer made by placing 0.25 mol of acetic acid and
0.25 mol of sodium acetate per liter of solution.
What is the pH of the buffer?
CH3COOH(aq) + H2O(l) CH3COO-
(aq) + H3O+
(aq)
Ka = or
[CH3COO-] [H3O+]
[CH3COOH]
[CH3COOH]
[CH3COO-]
[H3O+] = Ka x
88

[H3O+] = Ka x = 1.8 x 10-5 x
[CH3COOH]
[CH3COO-] (0.25)
(0.25)
= 1.8 x 10-5

 1.00 mL conc. HCl gives,
 1.00 mL x 12.0 mol/L = 0.012 mol H3O+
 If 1.00 mL of concentrated HCl added to 300.00 mL of water :
Volume of solution: 1.00 mL plus 300.00 mL ; (convert ml to liter)
then, concentration H3O+ :
0.012 mol H3O+
301.00 mL soln.
= 0.0399 M H3O+
pH = -log(0.0399 M)
pH = 1.40 Without
buffer!
E.g. What will be the pH of 300.00 mL of pure water if 1.00 mL
of concentrated HCl (12.0 M) is added?
90

 Initial moles of acid and base in buffer is:
(2.00mol/L)(0.05L) = 0.1
 Initial moles of H+ added is (6.00mol/L)(0.00200L) = 0.012
 HCl is a strong acid and dissociates 100%. The H+ ion reacts
quantitatively with the conjugate base, C2H3O2-, form of the
buffer, changing the base to acid ratio of the buffer and thus the
pH.

 Remember that the final volume is 52.00 mL not
50.00 mL

3. What is the new pH after 2.00 mL of 6.00M NaOH is added
to the original buffer?
 NaOH is a strong base and dissociates 100%. The OH ion reacts
acid.
 Again the base to acid ratio of the buffer is changed and thus the
pH.
 0.012 moles of OH is added.

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