Buffer capacity MANIK

Buffer Capacity
Prepared By: Md. Imran Nur Manik; B.Pharm; M.Pharm Page 1
manikrupharmacy@gmail.com; Lecturer; Department of Pharmacy; Northern University Bangladesh.
Buffer solution
A buffer solution is one which maintains its pH fairly constant even upon the addition of small amounts
of acid or base.
In other words, a buffer solution resists (or buffers) a change in its pH. That is, we can add a small
amount of an acid or base to a buffer solution and the pH will change very little.
Typically a mixture of a weak acid and a salt of its conjugate base or weak base and a salt of its
conjugate acid.
Types
Two common types of buffer solutions are:
 ACIDIC BUFFERS–
A weak acid together with a salt of the same acid with a strong base. These are called acidic buffers.
e.g. CH3COOH+CH3COONa
(weak acid ) ( Salt )
 BASIC BUFFERS–
A weak base and its salt with a strong acid. These are called basic buffers .
e.g. NH4OH+ NH4Cl
( Weak base) ( Salt)
Illustration of buffer action
Let us illustrate buffer action by taking example of a common buffer system consisting of solution of
acetic acid and sodium acetate (CH3COOH & CH3COONa), which will ionise as follows:
CH3COOH ↔⇌H+ + CH3COO–
CH3COO⎯Na+ ↔⇌Na+ + CH3COO⎯
Since the salt is completely ionized, it provides the common ions CH3COO– in excess. The common
ion effect suppresses the ionization of acetic acid. This reduces the concentration of H+ ions which
means that pH of the solution is raised. Thus, a 0.1M acetic acid solution has a pH of 2.87 but a
solution of 0.1M acetic acid and 0.1M sodium acetate has a pH of 4.74 (Figure). Thus 4.74 is the pH
of the buffer. On addition of 0.01 mole NaOH, the pH changes from 4.74 to 4.83, while on the
addition of 0.01 mole HCl, the pH changes from 4.74 to 4.66. Obviously the buffer solution maintains
fairly constant pH and the changes in pH could be described as marginal.
Fig.1. Illustration of buffer action.
Md.
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Manik

Buffer Capacity
Mechanism/function of a buffer
Mechanism of a buffer solution may be described by the following explanations:
A) Operation of acidic buffer
The function of an acidic buffer may be explained by considering a buffer solution composed of
CH3COONa and CH3COOH.
The pH of the buffer is governed by the equilibrium
CH3COOH⇌CH3COO–+H+ ……………………………………………………………… (1)
The buffer solution has a large excess of CH3COO– ions produced by complete ionization of sodium
acetate,
CH3COONa⇌CH3COO⎯+Na+……………………………………………………………. (2)
(1) Addition of HCl: Upon the addition of HCl, the increase of H+ ions is counteracted by association
with the excess of acetate ions to form unionized CH3COOH. Thus the added H+ ions are neutralized
and the pH of the buffer solution remains virtually unchanged. However owing to the increased
concentration of CH3COOH, the equilibrium (1) shifts slightly to the right to increase H+ ions. This
explains the marginal decrease of pH of the buffer solution on addition of HCl (Figure 2).
Fig.2. Mechanism of Buffer action of an acid buffer.
(2) Addition of NaOH: When NaOH is added to the buffer solution, the additional OH– ions combine
with H+ ions of the buffer to form water molecules. As a result the equilibrium (1) shifts to the right to
produce more and more H+ ions till practically all the excess OH– ions are neutralized and the original
buffer pH restored. However, a new equilibrium system is set up in which [CH3COOH] is lower than it
was in the original buffer. Consequently [H+] is also slightly less and pH slightly higher than the buffer
pH values (Figure 2).
Fact box: If the salt results from the reaction of a strong base and weak acid (e.g. sodium acetate from
reaction of sodium hydroxide and acetic acid), then the solution formed on hydrolysis will be basic, i.e.
CH3COOH+NaOH⇌ CH3COO⎯Na+
+ H2O
CH3COO⎯Na+
+ H2O ⇌ CH3COOH+OH⎯+Na+
Na+
does not react with water to any great extent, but CH3COO⎯is the conjugate base of the weak acid
CH3COOH and is therefore strong enough to react with water to produce OH⎯ions. The increase in
concentration of OH⎯gives a basic solution. These relationships can be summarised as follows:
Strong acid +Strong base Neutral salt
Strong acid+ Weak base Acidic salt
Weak acid + Strong base Basic salt
Weak acid + Weak base Neutral salt
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Buffer Capacity
B) Operation of a Basic buffer:
The operation of a basic buffer may be explained by considering a buffer solution composed of NH4OH
and NH4Cl on the same lines as of an acid buffer (Figure 3). Upon the addition of HCl the H+ ions
combine with OH– ions of the buffer to form water molecules. The equilibrium, NH4OH⇌NH4
+ + OH–
is shifted to the right till all the additional H+ ions are neutralized and the original buffer pH restored.
When NaOH is added to the buffer solution, OH– ions associate with excess of NH4
+ ions to form
unassociated NH4OH. Thus the pH of the buffer is maintained approximately constant.
Fig.3. Mechanism of Buffer action of a basic buffer.
Calculation of the pH of Buffer Solutions
The pH of an acid buffer can be calculated from the dissociation constant, Ka, of the weak acid and the
concentrations of the acid and the salt used.
The dissociation expression of the weak acid, HA, may be represented as
HA⇌H++A–
...(1)..............................
][A
[HA]
][Hor
[HA]
]][A[H
and





Ka
Ka
The weak acid is only slightly dissociated and its dissociation is further depressed by the addition of the
salt (Na+A–) which provides A– ions (Common ion effect). As a result the equilibrium concentration of
the unionized acid is nearly equal to the initial concentration of the acid. The equilibrium concentration
[A–] is presumed to be equal to the initial concentration of the salt added since it is completely
dissociated. Thus we can write the equation (1) as-
...(2)..............................
[salt]
[acid]
][Hor 
Ka
Where, [acid] is the initial concentration of the added acid and [salt] that of the salt used.
Taking negative logs on both sides of the equation (2), we have
)3.(........................................
[salt]
[acid]
loglog]log[H- 
Ka
But, –log [H+]= pH and –log Ka = pKa
Thus from (3) we have
equation.HendersonsimplyorequationhHasselbalc-HendersonthecalledisiprelationshThis
[acid]
[salt]
logpHHence,
[acid]
[salt]
logpHor,
[salt]
[acid]
logpH



pKa
pKa
pKa
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Manik

Buffer Capacity
In a similar way, the Henderson-Hasselbalch equation for a basic buffer can be derived. This can be
stated as:
[base]
[salt]
logHpHence,  pKbO
Problem: Find the pH of a buffer solution containing 0.20 mole per litre CH3COONa and 0.15 mole
per litre CH3COOH. Ka for acetic acid is 1.8 10–5.
Problem: Calculate the concentration of acetic acid to be added to a 0.1 M solution of sodium acetate
to give a buffer of pH 5 (pKa of acetic acid=4.66).
Solution: Acetic acid is a weak acid, so its degree of ionisation is very small and the contribution to the
total concentration of acetate anions from ionisation of the acid can be ignored. The total salt
concentration is therefore 0.1 M from the fully ionised sodium acetate.
Using the Henderson–Hasselbalch equation
0.046M[acid]
[acid]
0.1
188.2
[acid]
0.1
log34.0
[acid]
0.1
log66.40.5
[acid]
[salt]
logpH




 pKa
Problem: The Ka of propionic acid is 1.3410–5. What is the pH of a solution containing 0.5 M
propionic acid, C2H5COOH, and 0.5 sodium propionate, C2H5COONa. What happens to the pH of
this solution when volume is doubled by the addition of water?
Problem: A buffer solution contains 0.015 mole of ammonium hydroxide and 0.025 mole of
ammonium chloride. Calculate the pH value of the solution. Dissociation constant of NH4OH at the
room temperature is 1.8010–5.
Hints: Henderson-Hasselbalch equation for basic buffers is first used to determine the value of pOH.
[base]
[salt]
logHpHence,  pKbO
Where Kb is the dissociation constant of the base.
We know that,
pH + pOH = 14.00
Thus, pH = 14.00 – pOH
Problem: Estimate the pH at 25°C containing 0.10 M sodium acetate and 0.03 M acetic acid pKa for
CH3COOH = 4.57.
Md.
Imran
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Manik

Buffer Capacity
Buffer capacity
Buffer capacity is a measure of the efficiency of a buffer, in resisting changes in pH. The buffer capacity
is defined as the number of moles per litre of strong monobasic acid or base required to produce an
increase or decrease of one pH unit in the solution. It is also known as buffer efficiency, buffer index,
and buffer value.
Conventionally, the buffer capacity (β) is expressed as the amount of strong acid or base, in
gram-equivalents, that must be added to 1 liter of the solution to change its pH by one unit.
A buffer solution can resist a small amount of change of pH on adding acid or alkali to the solution.
Buffer capacities ranging from 0.01-0.1 are usually adequate for most pharmaceutical solutions. In
1922, Van Slyke first introduced an approximate equation to determine the buffer capacity by the
following equation:
)1.....(..............................
ΔpH
ΔB
β 
In which β = Buffer capacity, delta Δ=a finite change, and ΔB =the small increment in gram equivalents
(gEq)/liter of strong base added to the buffer solution to produce a pH change of ΔpH. According to
equation (1), the buffer capacity of a solution has a value of 1 when the addition of 1 g. of strong base
(or acid) to 1 liter of the buffer solution results in a change of 1 pH unit.
The higher the buffer capacity the less the buffer solution changes its pH.
A more exact equation for buffer capacity:
The buffer capacity calculated from above equation is only approximate. It gives the average buffer
capacity over the increment of base added.
Koppel and Spiro and Van Slyke developed a more exact equation,
2
3
3
])[(
][
3.2 



OHKa
OHKa
Cβ
Where, C = the total buffer concentration (i.e. the sum of the molar concentrations of acid and salt).
Buffers in pharmaceutical and biological systems
BUFFERS AND THE BODY:
Body fluids contain buffering agents and buffer systems that maintain pH at or near pH=7.4. Important
endogenous (natural) buffer systems include carbonic acid/sodium bicarbonate and sodium phosphate in
the plasma and haemoglobin, and potassium phosphate in the cells.
An in vivo value of pH < 6.9 or pH > 7.8 can be life threatening. Pharmaceutical solutions generally
have a low buffer capacity in order to prevent overwhelming the body’s own buffer systems and
significantly changing the pH of the body fluids. Buffer concentrations of between 0.05 and 0.5 M and
buffer capacities between 0.01 to 0.1 are usually sufficient for pharmaceutical solutions.
Md.
Imran
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Buffer Capacity
USEFUL BUFFER MIXTURES
Components pH range
HCl, Sodium citrate 01 – 05
Citric acid, Sodium citrate 2.5 – 5.6
Acetic acid, Sodium acetate 3.7 – 5.6
K2HPO4, KH2PO4 5.8 – 08
Na2HPO4, NaH2PO4 06 – 7.5
Borax, Sodium hydroxide 9.2 – 11
KH2PO4 is the acid, with a pH in the range of 4.4–4.7. K2HPO4 has a pH of 8.0.
H2CO3⇌ H++HCO3
–
The pH of blood is controlled by a bicarbonate (H2CO3/HCO3
–) buffer system. When the pH gets too
high (high OH concentration), the OH– reacts with carbonic acid (H2CO3) to form HCO3
–) and H2O.
When the pH gets too low (high H+ concentration), the H+ reacts with HCO3
– to form H2CO3. Because
H2CO3 is a weak acid, the H+ stays associated with the H2CO3. Since pH is an important factor in
many physiological processes, a change in the blood pH is a potentially life threatening condition
requiring immediate regulation.
The HCO3
-/H2CO3 buffer system is present in blood in greatest concentration and is very important in
maintaining the pH of blood within normal limit. The concentration of HCO3
- and H2CO3 in blood are
0.02M and 0.00125M respectively and hence the [HCO3
-]/[H2CO3] ratio is 20/1. In blood, the pKa
value for first ionization stage at body temperature is 6.1.Thus the
calculation will be as
The phosphoric and protein buffers of plasma are of relatively little
important as compared with bicarbonate buffer in regulating pH.
7.4pHOr,
1.26.1pHOr,
1
20
log6.1pHOr,
[Acid]
[Salt]
logpKapH




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Manik

Buffer Capacity
Importance of biological buffer
Buffer is very important for biological system. Some of the pictures are as follows
1. Buffer maintains constant [H+] in the body required for optimum cellular activity.
2. The pH of blood (around 7.4) is controlled by bicarbonate (H2CO3/HCO3
–) buffer system.
3. The phosphate buffer system (HPO4
2-/H2PO4
-) plays a role in plasma and erythrocytes.
H2PO4
-+H2O⇌H3O++HPO4
2-
M/A: Any acid reacts with monohydrogen phosphate to form dihydrogen phosphate
HPO4
2- + H3O+ H2PO4
- + H2O
monohydrogen phosphate dihydrogen phosphate
The base is neutralized by dihydrogen phosphate
H2PO4
- + OH- HPO4
2- + H3O+
dihydrogen phosphate monohydrogen phosphate
4. Proteins as a buffer: Proteins contain –COO- groups, which, like acetate ions (CH3COO-), can
act as proton acceptors. Proteins also contain –NH3
+ groups, which, like ammonium ions
(NH4
+), can donate protons.
M/A: If acid comes into blood, hydronium ions can be neutralized by the –COO- groups
-COO- + H3O+- COOH + H2O
If base is added, it can be neutralized by the –NH3
+ groups
-NH3
+ + OH-- NH2 + H2O
Md.
Imran
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Manik

Buffer Capacity
Preparation of Pharmaceutical buffer solutions:
To prepare a pharmaceutical buffer solution having definite pH and capacity, the pharmacist should
maintain the following steps:
1. Select a weak acid having a pKa approximately equal to the pH at which the buffer is to be
used. This will ensure maximum buffer capacity.
2. From the buffer equation, calculate the ratio of salt and weak acid required to obtain the desired
pH. (4-10)
3. Consider the individual concentrations of buffer salt and acid needed to obtain a suitable buffer
capacity.
(A concentration of 0.05 to 0.5M is usually sufficient; and a buffer capacity of 0.01 to 0.1 is
generally adequate.)
4. Other factors of some importance in the choice of a pharmaceutical buffer include-
i. Availability of chemicals
ii. Sterility of the final solution
iii. Stability of the drug and buffer on aging
iv. Cost of materials
v. Freedom from toxicity
e.g. a borate buffer, because of its toxic effects cannot be used to stabilize a solution to be administered
orally or parenterally.
5. Finally, one should determine the pH and buffer capacity of the completed buffer solution using
a reliable pH meter.
Methods for adjustment of tonicity of solutions.
The flow of the solvent through a semipermeable membrane from pure solvent to solution or from a
dilute solution to concentrated solution is termed osmosis.
Osmotic pressure (π)
Osmotic pressure (π) may be defined as the external pressure applied to the solution in order to stop the
osmosis of the solvent into the solution separated by a semipermeable membrane.
Isotonic Solutions
Solutions having the same osmotic pressure are said to be isotonic. In terms of physiological fluids, the
solutions having osmotic pressure equal to the osmotic pressure of intracellular fluid is called isotonic
solution. (πsoln = πcell).
In pharmacy and medical science, isotonic solution is that solution; which have equal tonicity with body
fluid i.e. blood, serum, plasma or lacrimal fluid. 0.9% NaCl solution is also regarded as isotonic solution.
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Buffer Capacity
Hypertonic Solutions
As compared to the blood plasma if a solution has higher osmotic pressure is said to be hypertonic
solution. (πsoln >πcell).
Physiological solutions having a greater osmotic pressure than that of body fluid or 0.9% NaCl solution
is referred to as hypertonic solution.
Hypotonic Solutions
As compared to the blood plasma if a solution has lower osmotic pressure is said to be hypotonic
solution. (πsoln <πcell).
Physiological solutions with an osmotic pressure lower than that of body fluid or 0.9% NaCl solution is
referred to as hypotonic solution.
Paratonic Solutions
The solution that is not isotonic that means both the hypertonic and hypotonic solutions are called
paratonic solution.
Effect of tonicity on body( Injection on blood)
The solutions which are not isotonic with plasma may be harmful to use. On injecting the hypotonic
solutions into blood stream, it may enter the blood cells in an attempt to produce equilibrium, the cells
swells rapidly until they burst leading to hemolysis. As this damage is irreversible may lead to serious
danger to RBC.
When hypertonic solution is injected into the blood stream, the water comes out of the membrane of
RBC in order to reach equilibrium. The cells shrink leading to crenulation which is only a temporary
damage. When the osmotic pressure of two solutions becomes equal the damaged cells will come to its
original position. Hence hypertonic solutions may therefore be administered without permanent damage
to the blood cells. They should be injected slowly to ensure rapid dilution into the blood stream and to
minimize the crenulation of blood cells.
Fig: Effect of tonicity on body
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Buffer Capacity
Calculation for the Preparation of Isotonic Solutions
For the preparation of isonotic solutions, the quantities of substances to be added may be calculated by
the following methods:
1. Based on the freezing point data (Freezing point depression).
2. Based on molecular concentration.
3. Based on Sodium Chloride equivalents.
4. Graphical method based on vapor pressure and freezing point determinations.
Method: Based on freezing point depression.
The freezing point is a colligative property often used in the calculation of the isotonic solution as it can
be measured easily and accurately. The temperature at which blood plasma and tears (Lacrhrymal
secretions) freeze is –0.52ºC which is the same value of a 0.9% solution of NaCl.All solutions which
freeze at –0.52ºC will be isotonic with blood plasma and lachrymal fluid.
The freezing points are usually expressed in terms of 1% solutions. The quantity of adjusting substance
needed for making the solutions isotonic with blood plasma may be calculated from the general formula
given below:
Percentage W/V of adjusting substance needed= b
a52.0
Where, a= freezing point depression of unadjusted solution.
b= freezing point depression of 1% W/V of the adjusting substance.
Problem: Find out the concentration of NaCl, required to render or make a 1% solution of cocaine
hydrochloride isotonic with blood plasma. The freezing point of 1% W/V solution of cocaine
hydrochloride is –0.090ºC and that of NaCl is –0.576ºC.
%746.0
576.0
090.052.0
:Hints 

Problem: Find out the concentration of NaCl, required to render or make a 1.5% solution of proocaine
hydrochloride isotonic with blood plasma. The freezing point of 1% W/V solution of proocaine
hydrochloride is –0.122ºC and that of NaCl is –0.576ºC.
%585.0
576.0
)122.05.1(52.0
:Hints 

Method: Based on Sodium Chloride Equivalents.
This method has gained popularity. NaCl method is defined as the weight of solution chloride which will
produce the same osmotic effect as 1 g of the drug to prepare an isotonic solution.
NaClofdepressionpointFreezing
AofdepressionpointFreezing
ADrugofequivalentNaCl 
Formulla: Percentage of NaCl for adjustment to isotonicity=0.9–Percent strength of drug
solutionNaCl eqivalent of the the drug.
Problem: Calculate the percentage of KNO3 required to make a 0.5% isotonic solution of AgNO3.The
NaCl equivalent of KNO3 is 0.56 and NaCl equivalent of AgNO3 is 0.33.
Hints: 0.9-(0.50.33)=0.735÷0.56=1.313%

Md.
Imran
Nur
Manik

Buffer capacity MANIK

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