Concentration Expression
1
Percentage Weight-In-Volume

The grams of solute or constituent in 100 ml of solution.

The volume, in milliliter, represents the weight in grams of solution or
liquid preparation as if it were pure water.

Volume of solution (ml)( representing grams ) X % ( expressed as a
decimal ) = g of solute or constituent.

Weight of solute ( g )
percentage Weight-In-Volume = ---------------------------------- X 100
Volume of solution ( ml )
2
Example I
How many grams of dextrose are required to prepared
4000 ml of a 5 % solution ?
4000 ml represent 4000 gm of solution.
5 % = 0.05
g of solute or constituent = volume (ml) X % ( expressed as
a decimal ).
= 4000 g X 0.05
= 200 g
3
Example II
How many grams of potassium permanganate should be
used in compounding the following prescription?
Rx Potassium Permanganate
0.02 %
Purified Water ad
250 ml
Sig. As directed
250 ml represent 250 gm of solution.
0.02 % = 0.0002
g of solute or constituent = volume (ml) X % ( expressed as
a decimal ).
= 250 g X 0.0002
= 0.05 g
4
Weight of solute ( g )
percentage Weight-In-Volume = ---------------------------------- X 100
Volume of solution ( ml )
Example
What is the percentage strength ( w/v ) of a solution of urea, if 80
ml contain 12 g ?
12
% w/v = -------- X 100 = 15 %
80
5
Weight of solute ( g )
percentage Weight-In-Volume = ------------------------------------ X 100
Volume of solution ( ml )
Example
How many milliliters of a 3 % w/v solution can be made from 27
g of ephedrine sulfate ?
27
3 = -------- X 100
x
27
x = --------- X 100 = 900 ml
3
Volume ( in ml ) = 900 ml
6
Percentage Volume-In-Volume

The milliliters of solute or constituent in 100 ml of solution.

Volume of solution (ml) X % ( expressed as a decimal ) = ml of
solute or constituent.

Volume of solute ( ml )
percentage Volume-In-Volume = -------------------------------- X 100
Volume of solution ( ml )
7
Example I
How many milliliters of liquide phenol be used in
compounding the following prescription?
Rx liquide phenol
2.5 %
Calamine Lotion ad
240 ml
Sig. For external use.
Volume (ml) X % ( expressed as a decimal ) = ml of solute
or constituent.
240 X 0.025 = 6 ml
8
Volume of solute ( ml )
percentage Volume-In-Volume = ---------------------------------- X 100
Volume of solution ( ml )
Example I
In preparing 240 ml of a certain lotion, a pharmacist used 4 ml of
liquefied phenol. What was the percentage ( v/v ) of liquefied
phenol in the lotion ?
4
% v/v = -------- X 100 = 1.67 %
240
9
Volume of solute ( ml )
percentage Volume-In-Volume = ---------------------------------- X 100
Volume of solution ( ml )
Example II
What is the percentage strength ( v/v ) of solution of 800 g of
liquid with a specific gravity of 0.800 in enough water to make
4000 ml ?
Volume = Weight / Specific Gravity.
Volume = 800 / 0.8
Volume = 1000 ml
1000
% v/v = -------- X 100 = 25 %
4000
10
Volume of solute ( ml )
percentage Volume-In-Volume = ---------------------------------- X 100
Volume of solution ( ml )
Example III
Peppermint spirit contains 10 % of ( v/v ) of peppermint oil. What
volume of the spirit will contain 75 ml of peppermint oil ?
75
10 = -------- X 100
x
75
x = -------- X 100 = 750 ml
10
11
Percentage Weight-In-Weight

The grams of solute or constituent in 100 grams of
solution.

Weight of solution (g) X % ( expressed as a decimal )
= g of solute or constituent.

Weight of solute ( g )
percentage Weight-In- Weight = ------------------------------- X 100
Weight of solution ( g )
12
Example I
How many grams of phenol should be used to prepare 240
g of 5 % ( w/w ) solution in water ?
Weight of solution (g) X % ( expressed as a decimal ) = g of
solute or constituent.
240 X 0.05 = 12 g
13
Example II
How many milligrams of hydrocortisone should be used in
compounding the following prescription?
Rx hydrocortisone
0.125 %
Hydrophilic Ointment ad
10 g
Sig. Apply.
Weight of solution (g) X % ( expressed as a decimal ) = g of
solute or constituent.
10 X 0.00125 = 0.0125 g
= 12.5 mg
14
Example III
How many grams of a drug substance are required to make 120
ml of a 20 % ( w/w ) solution having a specific gravity of 1.15 ?
Volume = Weight / Specific Gravity.
Weight = Volume X Specific Gravity.
Weight = 120 X 1.15
Weight = 138 g ( weight of 120 ml of solution )
Weight of solution (g) X % ( expressed as a decimal ) = g of
solute or constituent.
138 X 0.2 = g of solute or constituent.
= 27.6 g plus enough water to make 120 ml.
15
Weight of solute ( g )
percentage Weight-In- Weight = ------------------------------ X 100
Weight of solution ( g )
Example IV
If 1500 g of a solution contain 75 g of a drug substance, what is
the percentage ( w/w ) of the solution ?
75
% w/w = ---------- X 100
= 5 %
1500
16
Weight of solute ( g )
percentage Weight-In- Weight = ------------------------------ X 100
Weight of solution ( g )
Example V
If 5 g of boric acid are add to 100 ml of water, what is the
percentage strength ( w/w ) of the solution ?
100 ml of water weight 100 g
100 g + 5 g = 105 g, weight of the solution
5
% w/w = ---------- X 100
= 4.76 %
105
17
Milligram Percent

The number of milligrams of substance in 100 ml of liquid.

It is used frequently to denote the concentration of a drug or natural
substance in a biologic fluid, as in the blood.

The statement that the concentration of non-protein nitrogen in the
blood is 30 % means that each 100 ml of blood contains 30 mg of
non-protein nitrogen.
18
Example I
If a patient is determined to have a serum cholesterol level
of 200 mg/dl
(a) What is the equivalent value expressed in terms of milligrams
percent?
(a) 200 mg/dl = 200 mg / 100 ml = 200 mg%.
(b) How many milligrams of cholesterol would be present in 10
ml sample of the patient's serum ?
(b) 200 (mg)  100 ml
x (mg)  10 ml
x ( mg ) = 200 X 10 / 100 = 20 mg.
19
Example II
If a patient is determined to have a serum cholesterol level
of 200 mg/dl, what is the equivalent value expressed in term of
millimoles ( mmol ) per liter ?
Molecular weight of cholesterol = 387.
1 mol cholesterol = 387 g.
1mmol cholesterol = 387 mg.
200 mg/dl
= 2000 mg/L.
387 ( mg )  1 (millimoles )
2000 ( mg )  x (millimoles )
Therefore x = ( 1 X 2000 ) / 387
= 5.17 mmol / L
20
Part Per Million ( PPM ) & Part Per Billion ( PPB )

The strength of very dilution solution are commonly expressed in
terms of part per million ( PPM ) & part per billion ( PPB ), i.e. the
number of parts of the agent per 1 million or 1 billion parts of the
whole.

For example, fluoridated drinking water (used to reduce dental
caries) often contains 1 part of fluoride per million parts of drinking
water.

The ppm or ppb concentration of a substance may be expressed in
quantitatively equivalent value of percent strength or ratio strength.
21
Example I
Express 5 ppm of ion in water in percent strength and ratio
strength ?
5 ppm = 5 parts in 1,000,000 parts
= 1 : 200,000 ( ratio strength )
= ( 1 / 200,000 ) X 100
= 0.0005 % ( percent strength )
22
Example II
The concentration of a drug additive in an animal feed is 12.5
ppm. How many milligrams of the drug should be used in
preparing 5.2 kg of feed ?
12.5 ppm = 12.5 g ( drug ) in 1,000,000 g ( feed )
Thus
12.5 g ( drug )  1,000,000 g ( feed )
x g ( drug )  5,200 g ( feed )
x = ( 5,200 X 12.5 ) / 1,000,000
= 0.065 g
= 65 mg
23
Molarity

Concept of mole
Atoms & molecules are too small, tiny to weight so the concept of
mole which is theoretical value or unit expressing the formula
weight of substance in grams.
e.g.
gm atom of O2 M.Wt is 16 = 16 gm = 1 mol
gm ion of Cl- M.Wt is 35.5 = 35.5 gm = 1 mol
gm molecules of H2O M.Wt is 18 = 18 gm = 1 mol
24
Molar Concentration

Moles / Liter is one of the most useful units for describing
concentration.

Molarity ( M ) is the number of moles of solute / Liter of solution
( not solvent ).

M X L = moles.

M X L = Weight in grams / molecular weight.
25
Example I
How can prepare 500 ml of 0.15 M Na2CO3 solution ?
i.e. How many grams needed ?
M X L = Weight in grams / molecular weight.
0.15 X 0.5 = Weight in grams / 106
Weight in grams = 0.15 X 0.5 X 106
Weight in grams = 7.95 g
Weighing 7.95 g of Na2CO3 & placed in volumetric flask 500 ml
& diluted to 500 ml .
26
Example II
What is the molarity of NaCl solution 10 g / 100 ml ?
M X L = Weight in grams / molecular weight.
M X 0.1 =
10
/
58.5.
M = 1.71 M
Example II
What is the molar concentration of 1 % (w/v) NaCl solution?
1 % = 1 g of NaCl in 100 ml
M X L = Weight in grams / molecular weight.
M X 0.1 =
1
/
58.5.
M = 0.171 M
27
Example IV
Describe the preparation of 2 L of 0.2 M HCl solution starting
with a concentrated HCl solution ( 28 % w/w, specific gravity =
1.15 ) ?
M X L = Weight in grams / molecular weight.
0.2 X 2 =
x
/
36.5.
Weight in grams = 0.2 X 2 X 36.5
Weight in grams = 14.6 g ( need to prepare the solution )
The stock of HCl is not pure only 28 % HCl :
28 g HCl  100 g of solution.
14.6 g HCl  x g of solution.
x = ( 14.6 X 100 ) / 28
x = 52.143 g
52.143 g should be taken from a solution to take it in ml instead
of grams ( specific gravity = weight / volume )
Volume = weight / specific gravity = 52.143 / 1.15 = 45.34 ml
45.34 ml placed in a volumetric flask and completed to 2 liters.
28
Molality ( m )

number of moles of solute / kg of solvent ( not solution ).

m X kg ( solvent ) = moles.

m X kg ( solvent ) = Weight in grams / molecular weight.

Molality is useful in describing the ratio of moles of solute to
solvent.

The value of ( m ) does not change with temperature but that of
( M ) dose.

We would not even have to use a volumetric flask, because we
weight the solvent.
29
Example I
Calculate the molality HCl solution ( 28 % w/w ) ?
m X kg ( solvent ) = Weight in grams / molecular weight.
Weight of solvent = 100 – 28 = 72 g ( from 28 % )
= 0.072 kg
m X kg ( solvent ) = Weight in grams / molecular weight.
m X 0.072
=
28
/
36.5
m = 28 / ( 36.5 X 0.072 )
m = 10.65 ( m)
30
Example II
Calculate the molal concentration of 1 % ( w/v ) NaCl solution ?
Where : molecular weight = 58.5 & specific gravity = 1.0055
1 % mean 1 g NaCl in 100 ml solution
Weight of solution = volume X specific gravity
= 100 X 1.0055 = 100.55 g
Weight of solvent = weight of solution – weight of solute
= 100.55 – 1 = 99.55 g
m X kg ( solvent ) = Weight in grams / molecular weight.
m X 0.09955
=
1
/
58.5
m = 1 / ( 58.5 X 0.09955 )
m = 1.72 ( m)
31
Example III
To prepare 0.15 m solution of NaCl, to prepare this solution with
this conc, how many grams of NaCl should be dissolved in
500 gm of water ?
m
X kg ( solvent ) = Weight in grams / molecular weight.
0.15 X
0.5
= Weight in grams /
58.5
Weight in grams = 0.15 X 0.5 X 58.5
Weight in grams = 4.39 g
Thus, 4.39 g of NaCl is dissolved in 500 gm of H2O to give the
desired concentration.
32
Mole Fraction

Ratio of moles of one constituent solute or solvent of solution to
the total number of moles of all constituents ( solute & solvent ).
y1 = n1/(n1+n2)
y2 = n2 /(n1+n2)
Where
n1,n2 are the number of moles.
y1 is the mol fraction of solute.
y2 is the mol fraction of solvent.
33
Example I
What is the mole fraction of both constituent in 1 % ( w/v ) NaCl
solution ( specific gravity = 1.0053 ) and what is the molality ?
Volume = Weight / Specific Gravity.
Weight = Volume X Specific Gravity.
Weight = 100 X 1.0053
Weight = 100.53 g solution
1 g NaCl in ( 100 X 1.0053 = 100.53 g solution )
Solvent 100.53 – 1 = 99.53 g solvent.
m
m
m
m
X
kg
= Weight in grams / molecular weight.
X 99.53/1000 =
1
/
58.5.
= 1 / ( 58.5 X 0.09953)
= 0.172 ( m )
34
Cont. Example I
What is the mole fraction of both constituent in 1 % ( w/v ) NaCl
solution ( specific gravity = 1.0053 ) and what is the molality ?
No. of moles = Weight in grams / molecular weight
No. of moles of solute = 1 / 58.5 = 0.171
No. of moles of solvent = 99.53 / 18 = 5.529
y1 = n1/(n1+n2) )
y2 = n2 /(n1+n2)
Mole fraction of solute = 0.171 / ( 0.171 + 5.529 ) = 0.003
Mole fraction of solvent = 5.529 / ( 0.171 + 5.529 ) = 0.996
35
Mole Percent

Mole Percent = mole fraction X 100

In the last example :
Mole fraction of solute = 0.171 / ( 0.171 + 5.529 ) = 0.003
Mole Percent = mole fraction X 100
Mole Percent = 0.003 X 100 = 0.3 %
Mole fraction of solvent = 5.529 / ( 0.171 + 5.529 ) = 0.996
Mole Percent = mole fraction X 100
Mole Percent = 0.996 X 100 = 99.6 %
36
Normality ( N )
Equivalent weight is the weight in grams of one equivalent or the
quantities of substance that combine with 1.008 gm H+.
The EQUIVALENT WEIGHT is the amount of solute needed to be
the equivalent of one mole of hydrogen ions. Therefore, the
equivalent weight is dependent on the valence of the solute.
For solutes with a valence of one (i.e. NaCl) the molecular
weight and equivalent weight are the same. When the valence
of the solute is more than one (i.e. H3PO4, valence = 3), then
the equivalent weight is equal to the molecular weight divided
by the valence.
37

The equivalent weight ( one equivalent ) of an acid or base is
that contains 1 g atom.
Equivalent weight = molecular weight / n.
n = number of replaceable H or OH for acid or base.
n = number of electrons lost or gained in oxidation, reduction
reaction.

Equivalent weight = atomic weight / number of equivalent per
atomic weight.
e.g. F & O2
1 equivalent of F = molecular weight = 19
1 equivalent of O2 = molecular weight / Valency = 16 / 2 = 8

38

Normality ( N ) = grams of equivalent weight
of solute / Liter of solution ( not solvent )

L X N = equivalent

N = weight in gram / (equivalent weight x volume)
39
Concentration Expression :
Expression
Symbol
Definition
% Weight in Volume
%W/V
Grams of solute in 100 ml of solution.
% Volume in Volume
%V/V
Milliliters of solute in 100 ml of solution.
% Weight in Weight
%W/W
Grams of solute in 100 g of solution.
Milligram %
part per million or billion
Milligrams of solute in 100 ml of solution.
PPM,PPB the number of parts of the agent per 1
million or 1 billion parts of the whole.
Molarity
M, C
Mole ( gram molecular weights) of the
solute in 1 liter of solution.
Molality
M
Mole of the solute in 1000 g of solvent.
Mole fraction
X, N
Mole percent
Normality
Ratio of the moles of one constituent (e.g.
the solute ) of a solution to the total moles
of all constituents ( solute & solvent ).
Moles of one constituent in 100 moles of
the solution ( by multiplying mole fraction
by 100 ).
N
Gram equivalent weight of solute in 140liter
of solution.
Buffer
41
pH

pH = - log [ H+]
or
pH = - log [ H3O+ ]
Example I
What is the pH of solution with [ H+] = 32 X 10-5 M/L ?
pH = - log [ H+]
pH = - log 32 X 10-5
pH = 3.495
42
Example II
The hydronium ion concentration of 0.1 M solution was found to
the 1.32 X 10-3 M, What is the pH of the solution?
pH = - log [ H3O+ ]
pH = - log [ 1.32 X 10-3 ]
pH = 2.88
Example III
If the pH of a solution is 4.72, what is the hydrogen ion
concentration?
pH = - log [ H+ ]
- log [ H+ ] = 4.72
log [ H+ ] = - 4.72
[ H+ ] = 1.91 X 10-5
take anti-log for both side
43
H+ = 1 M/l = 1 X 100 = 1
H+ = 1 X 10-1 = 0.1 M/L
H+ = 1 X 10-2 = 0.01 M/L
H+ = 1 X 10-3 = 0.001 M/L

pH = 0
pH = 1
pH = 2
pH = 3

Note : the change in one pH unit means 10 fold change in [H+].

pH + pOH = 14



44
Introduction
When a minute trace of hydrochloric acid is added to pure
water, a significant increase in hydrogen-ion concentration
occur immediately.
 In a similar manner, when a minute trace of sodium hydroxide
is added to pure water, it cause correspondingly large increase
in the hydroxyl-ion concentration.
 These change take place because water alone cannot
neutralize even trace of acid or base, i.e. it has no ability to
resist change in hydrogen-ion concentration or pH.
 Therefore, it is said to be unbuffered.
 E.g. CO2 + H2O
H2CO3 decrease pH from 7 to 5.8
These change of pH are of great concern in pharmaceutical
preparation also NaCl solution ability to resist change of pH.
To ensure stability and solubility, we used to control pH by using
a buffer
45


Buffer
:
Compound or a mixture of compound which by presence in
solution to resist change in pH up of addition of small
quantities of acid or base or a solvent.
46
Adjustment of pH by the buffer

pH important for stability & solubility, therefore should be
adjusted pH by buffer .
Also pH play important role in :
1-Parenteral dosage form.
2-Eye drops.
3-Nasal drops.

47
Adjustment of pH by the buffer

Degree of acidity and alkalinity depends on the relative
concentration of H+ ion and OH- ion.
if H+ > OH- = acidic
H+ = OH- = neutral
H+ < OH- = alkaline
Acidity and alkali may be strong or weak:
1- weak acid, pH 3.5- 7
2- strong acid, pH 0-3.5
3- weak base, pH 7-10
4- strong base, pH 10.5-14

0
3.5
Strong acid
7
Weak acid
10.5
Weak base
14
Strong base
48

The product of H+ ion and OH- ion in the any aqueous liquid is
constant i.e. Kw = [ H+] [ OH- ]

Increase of one tend to decrease of another.
49
Some notes about buffer:
1-Buffers solution should be prepared using freshly boiled and
cooled water.
2- Buffers solution should be stored in containers of Alkali-free
glass.
3-Buffers solution should be discarded no later than three
months from the date of manufacture.
50
Selection of buffer system depends on:
1.
pH rang.
2.
Buffer capacity desired.
3.
The purpose for which it is required.
4.
Compatibility with active ingredients.
51
Method of preparation of buffer
1.
Buffer equation.
2.
Buffer table.
52

Buffer solution consist of mixture of weak acid and its
conjugate base or weak base and its conjugate acid.
Weak acid + its salt
e.g. acetic acid + sodium acetate
CH3COOH CH3COONa

Weak base + its salt
e.g. Ammonia + Ammonium chloride

NH3
+ NH4Cl
53
Henderson-Hasselbalch Equation
buffer equation for weak acid and its salt
Dissociation Constant of weak acid is given by the equation :
Where :
A- = Salt
HA = Acid
we isolate the H+ and put it on the left-hand side of the equation:
54
take the negative log of each of the three terms in the last
equation, they become:
- log [H+]  this is the pH
- log Ka  this is the pKa
- log ([HA] / [A-])  to get rid of the negative sign + log ([A-] /
[HA])
Inserting these last three items (the pH, the pKa and the
rearranged log term), we arrive at the Henderson-Hasselbalch
Equation:
55
common way the Henderson-Hasselbalch Equation is presented
in a textbook explanation:
56

Remember that, in a buffer, the two substances differ by only
a proton. The substance with the proton is the acid and the
substance without the proton is the salt.

However, remember that the salt of a weak acid is a base (and
the salt of a weak base is an acid).

Consequently, another common way to write the HendersonHasselbalch Equation is to substitute "base" for "salt form"
(sometimes you will see "conjugate base" or "base form"). This is
probably the most useful way to decribe the interactions between
the acidic form (the HA) and the basic form (the A-).

Here it is:

Remember this: the base is the one WITHOUT the proton and
the acid is the one WITH the proton.
57
Buffer equation of weak base & its salt
Dissociation Constant of weak acid is given by the equation :
( B + ) ( OH- )
Kb = -------------------( BOH)
Where : B +
= Salt
BOH = Base
[Base]
pH = pKw – pKb + log ---------[ Salt]
58
An alternate form of the HendersonHasselbalch Equation

The last discussion used pH and pKa .There is a alternate
form of the Henderson-Hasselbalch Equation using pOH and
pKb .
[Salt]
pOH = pKb + log ----------[Base]
59
Mechanism of action of buffer
1 – Acid / Its salt e.g. CH3COOH/CH3COONa.
CH3COOH + OH-  CH3COO- + H2O
This means acid will react with base (OH- ( to neutralize it.
CH3COO-/Na+ + H3O+  CH3COOH + H2O
This means salt of weak acid will react with acid ( H3O ) to
neutralize it.
60
Mechanism of action of buffer… cont
2 – Base / Its salt e.g. NH3/NH4Cl.
NH3 + H3O+  NH4+- + H2O
This means base will react with acid to neutralize it.
NH4+ + OH-  NH3 + H2O
This means salt of weak base will react with base (OH- ) to
neutralize it.
61
pH + pOH = 14
For acid & its salt :
[Salt]
pH = pKa + log ---------[Acid]
For base & its salt :
[Salt]
pOH = pKb + log ----------[Base]
or
[Base]
pH = pKw – pKb + log ---------[ Salt]
62
Example I
What is the molar ratio of salt/acid required to prepare an acetate
buffer pH = 5 ?
Ka = 1.8 X 10-5
pKa = - log Ka
pKa = - log 1.8 X 10-5 = 4.75
pH = pKa + log salt/acid
5 = 4.75 + log salt/acid
log salt/acid = 5 – 4.75 = 0.25
take anti-log for both side
salt/acid = 1.78
So, Ratio of salt/acid = 1.78/1
Mole fraction of acid = 1 / ( 1.78 + 1) =0.3597
X 100 = 35.97 %
Mole fraction of salt = 1.78 / ( 1.78 + 1) =0.6403 X 100 = 64.03 %
mole fraction multiply by 100 to get of mole %
63
Example II
Prepare 200 ml of acetate buffer pH = 6 with molar conc. = 0.4 M
& Ka = 1.8 X 10-5 ?
pKa = - log Ka
pKa = - log 1.8 X 10-5 = 4.75
pH = pKa + log salt/acid
6 = 4.75 + log salt/acid
log salt/acid = 6 – 4.75 = 1.25
salt/acid = 17.78/1
take anti-log for both side
Molecular weight of acetic acid ( CH3COOH ) = 60
Molecular weight of sodium acetate ( CH3COONa ) = 82
Weight of acid = mole fraction X Conc. X M.wt X V (L)
= 1/18.78 X 0.4 X 60 X0.2
= 0.26 g
Weight of salt = mole fraction X Conc. X M.wt X V (L)
= 17.78/18.78 X 0.4 X 82 X0.2
= 6.21 g
64
Example III
If we add 0.1 M sodium acetate to 0.09 M acetic acid .What is the
pH if you know Ka = 1.8 X 10-5 ?
pKa = - log Ka
pKa = - log 1.8 X 10-5 = 4.75
pH = pKa + log salt/acid
pH = 4.75 + log 0.1/0.09
pH = 4.796
65
Example IV
What is the pH of of a solution containing 0.1 mole of
ephedrine base and 0.01 mole of ephedrine HCl / liter
of solution ?
pKb of ephedrine = 4.64
pH = pKw - pKa + log base/salt
pH = 14 - 4.64 + log 0.1/0.01
pH = 9.36 + log 10 = 10.36
Or
pOH = pKb + log salt/base
pOH = 4.64 + log 0.01/0.1
pOH = 4.64 + log 0.1 = 3.64
pH + pOH = 14
pH = 14 – pOH
pH = 14 – 3.64 = 10.36
66
Change in pH with addition of an acid or base
Calculate the change in pH of a buffer solution with the addition
of a given amount of acid or base in the following example :
Example
Calculation the change in pH after adding 0.04 mol of sodium
hydroxide to a liter of a buffer solution containing 0.2 M conc.
of sodium acetate and acetic acid. The pKa value of acetic
acid is 4.76 at 25 OC.
The pH of the buffer solution is calculated by using the buffer
equation
pH = pKa + log salt/acid
pH = 4.76 + log 0.2/0.2
pH = 4.76
67

The addition of 0.04 mol of sodium hydroxide converts 0.04
mol of acetic acid to 0.04 mol of sodium acetate.
Consequently, the conc. Of the acetic acid is decrease and
the conc. Of the sodium acetate is increase by equal
amounts, according to the following equation :
salt + base
pH = pKa + log -----------------------acid - base
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salt + base
pH = pKa + log ----------------------acid - base
0.2 + 0.04
pH = 4.76 + log ------------------0.2 - 0.04
pH = 4.76 + 0.1761 = 4.9361 = 4.94
Because the pH before addition sodium hydroxide was 4.74, the
change in pH = 4.94 – 4.76 = 0.18 unit.
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Buffer Capacity

The ability of buffer solution to resist change in pH upon
addition of acid or base.
Ka . [H+]
Buffer Capacity “ B “ = 2.303 X C X ----------------( Ka + [H+] )2
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Example I
At hydrogen ion conc. of 1.75 X 10-5 (4.76), what is the capacity of
a buffer containing 0.1 mole each of acetic acid and sodium
acetate / liter of solution ?
Total C = [acid] + [ salt] = 0.1 + 0.1 = 0.2 mol/L
Ka . [H+]
Buffer Capacity “ B “ = 2.303 X C X ----------------( Ka + [H+] )2
(1.75 X 10-5 ) . (1.75 X 10-5 )
Buffer Capacity “ B “ = 2.303 X 0.2 X ---------------------------------------[(1.75 X 10-5 ) + (1.75 X 10-5 )]2
B = 0.115
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Thank you
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