Gastrointestinal agents

1. Name of Institution
Bachelor of Pharmacy
First Semester
Pharmaceutical Inorganic Chemistry
(BP 104T)
Dr. Srabanti Jana©
Associate Professor
AIP, AUMP
Amity Institute of Pharmacy

2. Name of Institution
UNIT-II
Amity Institute of Pharmacy
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3. Name of Institution
Buffer
• A buffer is a solution of a weak acid and its conjugate base that
resists changes in pH in both directions—either up or down.
• A buffer works best in the middle of its range, where the amount of
undissociated acid is about equal to the amount of the conjugate
base.
• A buffer is a pair of weak acid and its salt.
• One can soak up excess protons (acid), the other can soak up
excess hydroxide (base).
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Buffers
 Mammalian tissues in the resting state have a pH
of about 7.4
 In order to maintain the required pH in an in vitro
biochemical experiment a buffer is always used
 The pH of a buffer is given by Handerson-
Hasselbalch equation:
pH= pKa + log [A-]
[HA]
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pKa = 4.76
pH 7
_
_
_
_
_
_
_
0 equiv. of NaOH 1.0
added
Buffering range:
only small pH
changes result from
addition of base or
acid
Titration of acetic acid with
sodium hydroxide
50%
dissociation
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6. Name of Institution
Ionization of acetic acid:
Resisting changes both ways
OH- H2O
Acetic acid HAc Ac- Acetate
(CH3COOH) (CH3COO-)
H+
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The dissociation of a weak acid is given by:
H𝐴 ↔ 𝐻
+
+𝐴
−
and its magnitude is controlled by the value of the dissociation constant Ka:
The expression may be approximated by writing concentrations for activities
This equilibrium applies to a mixture of an acid HA and its salt, Say MA.
If the concentration of the acid be ca and that of the Salt be cs, then the
concentration of the undissociated portion of the acid is (ca- [CH+]).
The solution is electrically neutral, hence [A-] = cs + [H+]( the Salt is completely
dissociated). Substituting these values in the equilibrium equation, we have:
Dissociation of a weak acid
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In a mixture of a weak acid and its salt, the dissociation of
the acid is repressed by the common ion effect, and [H+]
may be taken as negligibly small by comparison with ca, and
cs. Equation then reduces to
Dissociation of a weak acid
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Dissociation of a weak base
Similarly for a mixture of a weak base of dissociation constant Kb and its
Salt with a strong acid
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Preparation of buffer
 Choose a compound whose pKa is close to the pH for the solution
 Determine what the buffer concentration should be.
Three practical methods to prepare a buffer:
1- First Method : By the Titration, in the presence of one of the two buffer
forms with strong base or acid:
 Prepare a buffer composed of an acid and its salt by adding a strong
base(e.g. NaOH) to a weak acid (e.g. Acetic acid) until the required pH is
obtained
 If the other form of buffer is available (in this case sodium acetate), a strong
acid is added (e.g. HCl) until the required pH is obtained.
CH3COONa+HCl→CH3COOH+NaCl
 So acetate buffer is formed(CH3COOH/CH3COONa)
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Second Method: Using the buffer pKa , calculate the amounts (in moles) of
acid/salt or base/salt present in the buffer at the desired pH.
If both forms (i.e., the acid and the salt) are available, convert the amount required
from moles to grams ,using the molecular weight of that component, and then
weigh out the correct amounts of both forms. Or convert moles to volume if the
stock is available in the liquid form.
Advantages:
Fast.
Easy to prepare.
Additional pH adjustment is rarely necessary, and when necessary, the adjustment
is small.
Disadvantages:
Requires the buffer pKa
and solving two equations.
Preparation of buffer
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 Find a table of the correct amounts of acid/salt or base/salt required for
different pH's
 Dissolve the components in slightly less water than is required for the final
solution volume.
 Check that the pH and correct if necessary.
 Add water to the final volume.
Advantages:
Easy to do (with appropriate table).
Convenient for frequently prepared buffers.
Disadvantages:
May be impossible to find table.
Table may be incorrect.
Requires both forms of buffer.
Component amounts from table will need to be adjusted to give the buffer
concentration and volume in your solution.
The Third Method: Using table
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13. Name of Institution
Decide on the Buffer Properties
• Before making a buffer you must know;
1. what molarity you want it to be
2. what volume to make
3. what the desired pH is.
• Most buffers work best at concentrations
between 0.1 M and 10 M.
• The pH should be within 1 pH unit of the acid/
conjugate base pKa.
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Isotonicity
Osmosis is the diffusion of solvent through a
semi-permeable membrane.
Water always flows from lower solute
concentration [dilute solution] to higher solute
concentration until a balance is produced.
Osmotic pressure is the force that cause this
diffusion.
Tonicity is a measure of the osmotic pressure of two solutions
separated by a semi-permeable membrane.
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Isotonic and Buffer solutions
• Colligative properties, From Greek word" collected together”,
depend mainly on the number of particles in solution.
– vapor pressure lowering
– freezing point depression
– boiling point elevation
– osmotic pressure
• Osmosis: 2 solutions of different concentrations are separated by
a semi-permeable membrane (only permeable to the solvent) the
solvent will move from the solution of lower conc. to that of
higher conc.
• Osmotic pressure: The pressure that must be applied to the
solution to prevent the passage of the solvent through a perfect
semipermeable membrane. 15
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Isotonic and Buffer solutions
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Types of Tonicity
Hypertonic
isotonic
Hypotonic
NaCl 2%
NaCl 0.9%
NaCl 0.2%
solute
‹ solute
Inside outside
solute
= solute
Inside outside
solute
› solute
Inside outside
shrinkage
equilibrium
swelling
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18. Name of Institution
Why using isotonic solutions?
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Isotonicity & route of administration
Subcutaneous injection:
• Not necessarily “small dose” but isotonicity reduce pain.
Hypodermoclysis
• Should be isotonic “Large volume”
Intramuscular injection
• Should be isotonic or slightly hypertonic to increase penetration
Intravenous injection
• Should be isotonic “Large volume ”
• Hypotonic cause haemolysis
• Hypertonic solution may be administered slowly into a vein
• Hypertonic large volume administered through a cannula into
large vessels.
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Isotonicity & route of administration
• Intrathecal injection
• Should be isotonic
• Eye drops
• Rapid diluted by tear, but most of it is isotonic to decrease irritation
• Eye lotions
 Preferably isotonic
• Nasal drops
 Isotonic, but not essentially
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Calculations for preparation
of isotonic solution
• Freezing point depression (colligative properties)
• - 0.52°C is the freezing point of both blood serum and lacrimal
fluids
• For nonelectrolytes (negligible dissociation) as boric acid
• Boric acid: MWt 61.8 thus if 61.8 g in 1000 g of water should produce a freezing point of -1.86°C
X = 17.3 g
• So 17.3 g of boric acid in 1000 g of water (1.73 %) should make a solution isotonic with lacrimal
fluid.
• Isotonic solutions may be calculated by using this formula
𝑔 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑒𝑟 1000 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 =
0.52 𝑋 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡
1.86 𝑋 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛 (𝑖)
• The value of i for many a medicinal salt has not been experimentally
determined. Some salts (such as zinc sulfate, with only some 40% dissociation
and an i value therefore of 1.4) 21
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Calculations
• If the number of ions is known, we may use the following
values, lacking better information:
– Nonelectrolytes and substances of slight dissociation: 1.0
– Substances that dissociate into 2 ions: 1.8
– Substances that dissociate into 3 ions: 2.6
– Substances that dissociate into 4 ions: 3.4
– Substances that dissociate into 5 ions: 4.2
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Calculations
• The Sodium Chloride Equivalent (E value) of a drug: is the amount of sodium
chloride which has the same osmotic effect as 1 gram of the drug.
• How much NaCl should be used in preparing 100 ml of 1% w/v solution of atropine
sulfate, which is to be made isotonic with lacrimal fluids?
• M.Wt of NaCl = 58.5, i = 1.8
• M.Wt of atropine sulfate = 695, i = 2.6
• X = 0.12 g of sodium chloride represented by 1 g of atropine sulfate
• Sodium chloride equivalent of atropine sulfate (E value) is = 0.12
equivalent
Chloride
Sodium
ce
ofsubs
Mwt
ce
subs
of
factor
i
NaCl
of
factor
i
NaCl
of
Mwt
=

tan
tan
( )
( )
g
X
g
1
8
.
1
5
.
58
6
.
2
695
=


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