college of sterile products pharmacy pht 434 · • Δt = i k f m • take a 1.0-molal solution of...
TRANSCRIPT
COLLEGE OF PHARMACY
Dr. Mohammad Javed Ansari Ph.D Contact info: [email protected]
STERILE PRODUCTS PHT 434
OBJECTIVE OF LABORATORY
• Adjustment of tonicity of parenteral and ophthalmic preparations
– Freezing point depression (FPD) “cryoscopic method”
– NaCl equivalent method
– White –Vincent method
Classes of adjustment of isotonicity
• Class I
• Adding substance to lower f.p of solution to -0.52º
• 1. Freezing point depression (FPD) “cryoscopic method”.
• 2. NaCL equivalent method.
• Class II
Adding water
3. White –Vincent method
• Freezing point of blood / tears or other body tissue is approximately -0.52
• Freezing point of water is zero.
• Whenever any drug or electrolyte is added in water, the freezing point of water will decrease.
• Therefore, this method depends freezing point depression (Δ T) because of drug (Table)
Freezing point depression method
Freezing point depression method
• FPD (Δ T) value as given in table means 1% of drug solution causes this much of depression in freezing point.
• As freezing point of blood or other isotonic solution is 0.52 so we can calculate % amount of drug which will cause FPD of 0.52 (means this solution will be isotonic)
• For example if 1% of drug cause FPD of 0.3 ((Δ T=0.3) then x % will cause a depression of 0.52
• X= (0.52/Δ T) 1% = 0.52/0.3= 1.73
• i.e 1.73 % of this drug will be isotonic.
• For dextrose Δ T=0.1, i.e. 1% dextrose causes FPD of 0.1
• % of dextrose which will be isotonic?
• X= (0.52/Δ T) 1% = 0.52/0.1= 5.2% dextrose.
Freezing point depression method
• Adjustment of tonicity by FPD method depends on freezing point depression because of drug /substances. (Table)
• This value is subtracted from 0.52 (isotonic)
• Remaining FPD is achieved by addition of adjusting substance to make the final solution isotonic with blood.
• Value of a, needs to be multiplied by the factor if drug solution is not 1% (multiply by 2 if it is 2% or multiply by 0.5 if it is 0.5%)
Freezing point depression method
b
aw
52.0%
Freezing point depression method
= conc. gm/100 ml of adjusting substance.
= FPD of 1% of unadjusted substance(table)
X percentage strength
= FPD of 1% of adjusting substance (table)
b
aw
52.0%
%w
a
b
Freezing point depression method ∆Tf 1% Sucrose = 0.06 °C
∆Tf 1% Dextrose = 0.101 °C
∆Tf 1% Ascorbic acid = 0.105 °C
∆Tf 1% Citric acid = 0.11 °C
∆Tf 1% ZnSO4= 0.15 °C
∆Tf 1% MgSO4= 0.19 °C
∆Tf 1% Sodium citrate = 0.2 °C
∆Tf 1% BaCl2 = 0.23 °C
∆Tf 1% Boric acid = 0.29 °C
∆Tf 1% Na2SO4= 0.32 °C
∆Tf 1% (NH4)2SO4= 0.33 °C
∆Tf 1% Na3PO4= 0.37 °C
∆Tf 1% NaHCO3= 0.38 °C
∆Tf 1% CaCl2= 0.44 °C
∆Tf 1% KCl= 0.46 °C
∆Tf 1% MgCl2= 0.52 °C
∆Tf 1% NaCl = 0.58 °C
∆Tf 1% (NH4)2SO4= 0.33 °C
∆Tf 1% Ascorbic acid = 0.105 °C
∆Tf 1% BaCl2 = 0.23 °C
∆Tf 1% Boric acid = 0.29 °C
∆Tf 1% CaCl2= 0.44 °C
∆Tf 1% Citric acid = 0.11 °C
∆Tf 1% Dextrose = 0.101 °C
∆Tf 1% KCl= 0.46 °C
∆Tf 1% MgCl2= 0.52 °C
∆Tf 1% MgSO4= 0.19 °C
∆Tf 1% Na2SO4= 0.32 °C
∆Tf 1% Na3PO4= 0.37 °C
∆Tf 1% NaCl = 0.58 °C
∆Tf 1% NaHCO3= 0.38 °C
∆Tf 1% Sodium citrate = 0.2 °C
∆Tf 1% Sucrose = 0.06 °C
∆Tf 1% ZnSO4= 0.15 °C
• IF FPD (Δ T) value is not provided then it can be calculated.
• ΔT = i Kf m
• i : van 't Hoff factor, unitless constant directly associated with the degree of dissociation of the solute in the solvent.
• Substances which do not ionize in solution, like DEXTROSE, GLUCOSE, have i = 1.
• Substances ionizing into two ions, like NaCl, have i = 2.
• Substances ionizing into three ions, like MgCl2, have i = 3.
• Note 1: Ion pairing effect lowers the i values for example NaCl has an actual i = 1.8 because of ion pairing.
• Note 2: Substances that ionize partially in solution will have i values between 1 and 2 usually.
Freezing point depression method
• ΔT = i Kf m
• Take a 1.0-molal solution of sugar and measure its FPD. Now examine a 1.0-molal solution of NaCl and MgCl2,. FPD will be two times and three times that of sugar as NaCl and MgCl2 have 2 ions and 3 ions respectively.
• Kf is cryoscopic constant. The Greek prefix cryo- means "cold" or "freezing." In a more generic way, it is called the "molal freezing point depression constant."
• Unit of Kf is °C kg/mol
• m: molality of solution, i.e. number of moles of solute per kg solvent.
• Note 2: Substances that ionize partially in solution will have i values between 1 and 2 usually.
Freezing point depression method
Freezing point depression method
• How much NaCl is required to render 100 ml of a
2% soln. of apomorphin HCL isotonic? (F.p.d of
1%NaCl=0.58ºC, F.p.d of 1% apomorphin HCL
=0.08º).
• W% = 0.52 - (2*0.08) / 0.58
• = 0.52 - (0.16) / 0.58
• = 0.36 / 0.58
• = 0.62
• 0.62 g of NaCl and 2 g of drug will make isotonic
solutions of 2% Apomorphine HCl.
Freezing point depression method
• How much NaCl is required to render 5%
dextrose solution isotonic? (F.p.d of
1%NaCl=0.58ºC, F.p.d of 1% dextrose =0.101º)
• W% = 0.52 - (5*0.101) / 0.58
• = 0.52 - (0.505) / 0.58
• = 0.015 / 0.58
• = 0.0258 g (25.8 mg)
• 0.0258 g of NaCl and 5 g of dextrose will make
isotonic solutions of 5% dextrose.
Freezing point depression method
• Prepare and supply 0.5% isotonic
solutions of Major body electrolytes (Na,
K, Ca Mg, Cl, & phosphate) by using
available solutes in the laboratory.
• Write procedure in brief.
• Solute used as electrolyte /drug
• Solute used as adjusting substance
• Calculation steps
• Final amount of both solutes used to
prepare isotonic solution.
NaCl equivalent method
• It depends upon adjustment of tonicity of
parenteral drug solution or electrolyte
solution by addition of NaCl.
• Sodium chloride equivalent of the parenteral
drug solution or electrolyte solution
(Amount of NaCl in gram that has same
osmotic effect as 1% of drug).
• Sodium chloride equivalent of drugs are given
in the question otherwise can be calculated.
NaCl equivalent method
Steps involved in calculations • Find out types of drug & its Liso value.
• Find out mol.wt. of drug.
• Calculate E NaCl of drug
• Put calculated E value in the formula below to find out % NaCl required to reach 0.9%.
• Add this calculated amount of NaCl and given % amount of drug in the volumetric flask.
• Add sterile water to make up the volume.
NaCl equivalent method • Whenever Sodium chloride equivalent of the
parenteral drug solution or electrolyte solution is not provided it can be calculated using following equation
• 1st calculate E NaCl
• Liso is a constant which depends on the type of drug and dissociation constant.
• If Liso value is not provided then it can be found by the types of drug as per Table 1
Liso values of substances
NaCl equivalent method • Prepare 3% solution of citric acid and adjust
isotonicity by NaCl equivalent method
• Citric acid is a weak electrolyte hence Liso = 2, MW of Citric acid is 192
• ENaCl = 17 Liso / MW
• = 17* 2/192 = 34/192 = 0.177
• W% = 0.9 - (drug% * ENaCl)
• = 0.9 - (3*0.177) = 0.9 - 0.531 = 0.369 g
• 3 g citric acid and 0.369 g of NaCl will be weighed and dissolved in 100 ml of sterile water.
NaCl equivalent method Exercise 1:
Calculate amount of NaCl needed to adjust isotonicity of 1.5% Atropine sulphate. (ENaCl of Atropine sulphate =0.12gm)
NaCl equivalent method Exercise 2:
Adjust isotonicity of procaine HCl 3% using NaCl ? (ENaCl of procaine HCl =0.21gm)
3-White – Vincent method
• Principle:
• Addition of H2O to drug to make it isotonic
• Addition of isotonic vehicle to bring solution to final volume
• Steps:
• 1. Find the weight of the drug prescribed (w), the volume prescribed (v) and its sodium chloride equivalent value (E).
• 2. Multiply the weight (w) of drug with the sodium chloride equivalent value (E).
• 3. Calculate volume VH2O of isotonic solution that can be prepared from W g of drug
• 0.9/100=W*E/ VH2O
• VH2O = W* E* 111.1
3-White – Vincent method
• How to prepare and supply 30ml of 1% drug solution isotonic with body fluid (ENaCl =0.16gm)
• 30 ml 1% drug means 0.3 g in 30 ml.
• VH2O = W* E* 111.1
• VH2O = 0.3* 0.16* 111.1
• VH2O = 5.33 ml
• i.e 0.3 g of drug should be dissolved in 5.33 ml of water which is then made to 30 ml by using isotonic solutions (normal saline or other).
3-White – Vincent method
• How to prepare isotonic solution containing Phenacaine HCl 0.06 gm (ENaCl=0.16), Boric acid 0.3 gm (ENaCl=0.5) sterile distilled H2O up to 100 ml.
• V =111.1 x [(0.06 x 0.16) + (0.3 x 0.5)] = 17.7 ml
• Both drugs should be dissolved in 17.7 ml of sterile water.
• Remaining volume should be adjusted using isotonic solutions
• i.e.100 -17.7 = 82.3 ml isotonic solution is added to make it 100 ml.
3-White – Vincent method
• Exercise 1:
• Adjust isotonicity of 1.5% Atropine sulphate using white Vincent method. (ENaCl of Atropine sulphate =0.12gm)
3-White – Vincent method
• Exercise 2:
• Adjust isotonicity of procaine HCl 3% using white Vincent method? (ENaCl of procaine HCl =0.21gm)
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GOOD LUCK ..