lab 10

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Isotonic Solutions
Lab 10
Definitions
• Isotonic Solution: is a solution having the same
osmotic pressure as a body fluid. Ophthalmic (eye),
nasal(nose), and parenteral (injection) solutions
should be isotonic.
• Hypotonic solution: is a solution of lower osmotic
pressure than that of a body fluid.
• Hypertonic solution: is a solution having a higher
osmotic pressure than that of a body fluid.
Calculation of Isotonic Solutions Using
Sodium Chloride Equivalents (E values)
Rx
Pilocarpine nitrate
Sod. Chloride
Purified water
ad
Make isotonic solution.
0.3 g
Q.S.
30 ml
Calculation of Isotonic Solutions Using
Sodium Chloride Equivalents (E values)
Calculate amount in grams of NaCl represented by
the ingredients in prescription. Multiply the
amount in grams of each substance by its NaCl
equivalent “either from tables or by calculation”.
2. Calculate amount in grams of NaCl, alone, that
would be contained in an isotonic solution (0.9%)
of the total volume specified in prescription.
1.
Calculation of Isotonic Solutions Using
Sodium Chloride Equivalents (E values)
Subtract the amount of NaCl represented by the
ingredients in the prescription (step#1) from the
amount of NaCl needed to prepare isotonic
solution (step#2)
4. If an agent other than sodium chloride such as
boric acid, dextrose, sodium or potassium nitrate is
to be used to make a solution isotonic, divide the
amount of NaCl (step 3) by NaCl equivalent of the
other substance.
3.
Calculation of Sodium Chloride equivalent (E
values)
• i factor is the dissociation factor.
Calculation of Sodium Chloride equivalent
(E values)
• Problems on calculating the dissociation factor (i) of an
electrolyte:
▫ Zinc Sulfate ZnSO4 is a 2-ion electrolyte, dissociation 40% in
a certain concentration. Calculate its dissociation factor.
If we have 100 particles of ZnSo4, 40% dissociation:
40 Zinc
40 SO4
60 undissociated particles.
140 particles
So the dissociation factor (i) of ZSO4 =
= 1.4 where 100
is the starting number of particles.
Calculation of Sodium Chloride equivalent
(E values)
• In general, we may use the following tabulated
values in case of solutions of 80% or higher
concentrations:
(i)
Nonelectrolytes, slightly dissociated subs.
1
Substances that dissociate into two ions
1.8
Substances that dissociate into three ions
2.6
Substances that dissociate into four ions
3.4
Substances that dissociate into five ions
4.2
Calculation of Sodium Chloride equivalent
(E values)
• Papavarine HCl (M.wt = 376) is a 2-ion electrolyte
dissociating 80%. Calculate its E-value, where its
dissociation factor (i) = 1.8
M.Wt of NaCl = 58.5
By substituting the values in the equation:
E = 58.5/1.8 X 1.8/376 = 0.156 or 0.16
Calculation of Isotonic Solutions Using
Sodium Chloride Equivalents (E values)
• How many grams of NaCl should be used in compounding the
following prescription:
Rx
Pilocarpine nitrate
Sod. Chloride
Purified water
ad
Make isotonic solution.
1.
2.
3.
0.3 g
Q.S.
30 ml
E =0.23
The amount of NaCl represented by Pilocarpine
= 0.3 x 0.23 = 0.069g.
The amount of NaCl required to make 30 ml of isotonic (0.9%):
0.9  100
X  30 ml
X = 0.27 g of NaCl.
Amount of NaCl to be added to prepare 30 ml of isotonic solution
= 0.27 – 0.069 = 0.201 g
Calculating Isotonic Solutions Using
Freezing Point Data
• The freezing point of blood and lachrymal fluid is –
0.52 oC.
• The dissolved substances in plasma or tear depress
the solution freezing point below 0.52°C.
• Any solution that freeze at T=-0.52°C is isotonic with
blood and tear
Calculating Isotonic Solutions Using
Freezing Point Data
1.
How many grams each of NaCl and dibucaine HCl are required to
prepare 30 ml of a 1% solution of dibucaine HCl isotonic with
tears?
From tables:
1% NaCl
Tf = 0.58
1% Dibucaine HCl
Tf = 0.08
Freezing point of blood and lachrymal fluid is – 0.52 oC.
▫ We need additional freezing point depression in the amount =
0.52 – 0.08 = 0.44 oC.
▫ Concentration of NaCl needed to lower the freezing point by 0.44 oC to make
the solution isotonic:
1% NaCl
Tf = 0.58
X
0.44
X= 0.76%
▫ Actual amount of NaCl to be added:
0.76
100
X
30
X= 0.228 g NaCl
▫ Actual amount of dibucaine to be added:
1
100
X
30
X = 0.3 g
Practice problems
1. Zinc Chloride (ZnCl2) is a 3-ion electrolyte;
dissociation 80% in a certain concentration,
calculate (i).
2. How many grams of boric acid should be used in
compounding the following:
RX
Phenacaine HCl
Chlorobutanol
Boric acid
Purified water
ad
Make isotonic solution
1%
0.5%
Q.S.
60 ml
E (0.2)
E (0.24)
E (0.52)
Practice problems
3.
4.
Rx
Oxytetracyline HCl
0.5%
E = 0.12
Tetracaine HCl 2% solution 15 ml
Sodium Chloride
Q.S.
Purified water
ad
30 ml
Make isotonic solution
2% solution of Tetracaine HCl is isotonic. How many mls of 0.9%
solution of Sodium Chloride should be used?
Rx
Tetracaine HCl
0.5%
E(0.18)
Epinephrine bitartarate 1:1000 solution
10 ml
Boric acid
Q.S
E(0.52)
Purified water
ad
30
Make isotonic solution
The solution of Epinephrine bitartarate (1:1000) is already isotonic.
How many grams of boric acid should be used in compounding the
prescription?
Practice problems
5. How many grams of NaCl and Naphazoline HCl should
be used in compounding the following prescription:
Rx
Naphazoline HCl
1%
NaCl
Q.S.
Purified water
ad
30ml
Make isotonic solution.
Use the freezing point depression method.
Tf blood = 0.52
Tf 1% NaCl = 0.58
Tf 1% Naphazoline HCl = 0.16
Practice problems
6. For agent having 0.32 sodium chloride equivalent,
calculate the percentage concentration of an
isotonic solution.
7. Calculate the E value for holocaine hydrochloride, if
6.7 mL of water will produce an isotonic solution
from 0.3 g of drug substance.
Home work
1. The freezing point of 5% solution of boric acid is
-1.55 oC. How many grams of boric acid should be
used in preparing 1500 mL of an isotonic solution?
2. Rx
Dextrose, anhydrous
2.5%
Sodium chloride
q.s
Sterile water for injection ad 1200mL
Label: Isotonic Dextrose and Saline Solution.
How many grams of sodium chloride should be
used in preparing the solution?

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