جامعة نزوى
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Nizwa University
Faculty of Pharmacy and Nursing
Practical Notes
for pharmaceutics III
PHCY 250
Under supervision of
Assistant Prof. Dr. Anil Philip
Suppositories
A solid or semisolid mass intended to be inserted into body cavities (rectum, vagina, urethra). It consist of active ingredients dissolved or dispersed throughout inactive base.
Once inserted, a suppository either melts at body temperature or dissolve or disintegrate into the aqueous secretions of the cavity
The bases applied for suppositories are broadly classified as into two groups:
--1-- fatty bases :- these may be natural in source such as Theobroma oil (cocoa butter) or synthetic fats such as Witepsol
--2- hydrophilic bases: The most commonly used base composed of solid glycerol/gelatin mixture.
Uses
Useful when oral administration is not suitable.
← infants.
← Debilitated individuals
← Unconscious pt. (In coma)
← Nausea, vomiting, gastrointestinal disturbance or irritation
Preparation
← Hand rolling method.
← Compression method
← Fusion method. Is the method used for commercial preparation , specially with cocoa butter , PEG, Glycerin-gelatin.
Suppositories are prepared by dissolving or dispersing an active medicament in a molten base and pouring the mixture into a suppository mould.
Suppository moulds are normally available in 1 g, 2 g and 4 g .
However, because the density of the medicament may vary considerably from that of the base, the weight of the base required to make a suppository will vary depending on the medicament used. e.g 2 g of a medicament with twice the density of theobroma oil would occupy approximately the same volume as 1 g of the suppository base.
The mold will contain a known volume or weight of the base or base mixed with the drug. In order to ensure that each suppository contains the correct amount of the drug, the bulk densities of the drug should be taken into consideration.
e.g: Drug with low density will displace more of the base than a similar weight of a more dense drug.
For many drug which would be incorporated into suppository base, the weight of the drug which would displace 1 gm of tradional suppositories base theobroma oil have been calculated and referred as displacement values
"Displacement values"
The displacement values (DVs) of medicaments are applied to calculate the actual weight of suppository base and drug required to prepare medicated suppositories.
The displacement value = number of parts, by weight of a medicament that will displace one part of suppository base (normally theobroma oil).
Displacement values for various medicaments are given in the Pharmaceutical Codex.
Displacement values for any liquid = 1
Determination of displacement bases value, using 1 gm mould.
1-Weight the suppositories of un-medicated base =" a"
2-Weight of supp. with medicated base contain certain % of active ingredients
= b
3- Weight of the base needed = C
4- Weight of base displaced by drug = weight of drug = d = a - c
"DVs" Displacement values = d / a-c
= weight of drug
amount of the base displaced by the drug
Example:
Weight of 6 un medicated supp. 6x1 = 6 gm "a"
Weight of medicated supp. containing 30% drug = 7.5 gm "b"
Weight of base = 70x 7.5/100 = 5.25 "c"
Weight of drug = 30 x 7.5 /100 = 2.25 " d"
Wight of the base displaced by 2.25 gm of drug = 6 - 5.25 = 0.75 gm
DVs = d/a-c = 2.25/ 6 - 5.25 = 3
DV = P Y…………………
100 X - (100 - p) y
P = % of active ingredient
Y = Weight of suppositories with medicaments
X = Weight of pure base suppositories
So 3 parts of the drug will displace one part of the base.
This method can be applied for determination of DV of any medicament relative to different suppository bases
Solved examples:
Use the displacement value to prepare 8 suppositories each containing 300 mg of drug.
DVs = 3 mold = 1gm Total weight of 8 supp. = 8x1 = 8gm
1- weight of drug = 8x 300 = 2.4 gm
3gm of the drug displace 1 gm of the base
So 2.4 gm drug will displace x
X= 2.4/3 = 0.8 gm
Required amount of the base for preparation 8 supp. 8 – 0.8 = 7.2gm
N.B
Note that for theobroma oil suppositories you can calculate the amount of base displaced by the active ingredient by dividing the amount of active ingredient by the displacement value
, e.g. 200 mg of a drug with a DV of 4 will displace 200/4 mg or 50 mg of base in
each mould.
Solve the following
1- - Calculate the quantities required to make 10 theobroma oil suppositories (2 g mould) each containing 400 mg of zinc oxide (Displacement value = 4.7)
Solution steps
1- Calculate the total weight of zinc oxide required. "d
2- Calculate what weight of base would be required to prepare 10 un-medicated suppositories. "a "
3- Determine what weight of base would be displaced by the medicament. = d
4- Calculate the weight of base required to prepare the medicated suppositories. "b
Medicaments included as a percentage w/w
If a medicament is present in a suppository as a percentage w/w, then its displacement value is not required when calculating the respective amounts of medicament and base required to prepare the suppository.
Solve the following Example
2- What quantities are required to prepare eight theobroma oil suppositories, in a 4 g mould, containing 1% w/w lignocaine hydrochloride?
Solution steps
Displacement value for liquid medicaments with Cocoa butter base:
Liquid extract of hamamilis 0.2 gm D.V = 1
Zinc oxide 0.2 gm D.V = 5
Cocoa butter Q.S
Fiat Supp.
Mitte : III
Signature : One supp. To be inserted at bed time
Calculation:
Procedure:
1- On a boiling water bath, melt 2/3 the amount of cocoa butter in porcelain dish.
2- Remove from the W.B, add the remaining part of C.B ( to have B stable form)
3- Add the liquid of hamamilis, zno stir.
4- Pour in a conyinous stream and make over flow ( to avoid formation of holes on cooling due to shrinkage of the base).
5- Cool in ice for 5 mins.
6- Remove the overflow by spatula
7- Open the mould and release the supp.
Use : treatment of hemorrhoids
Glycerin suppositories
Glycero-gelatin base has a density 1.2 times greater than theobroma oil.
Therefore, a 1 g suppository mould will produce a 1 g theobroma oil suppository, but a 1.2 g glycero-gelatin suppository. This factor must be taken into account in displacement value calculations.
* Calculation of correction factor
C.F = average wt of certain No. of supp.
Capacity of the mould (1 gm)
C.F of cocoa butter = 1
C.F of glycerogelatin = 1.2 = average wt of other base/ wt of Cocoa butter/
C.F of soap glycerin = 1.25
Example
Calculate the quantities required to make six glycero-gelatin suppositories (4 g mould), each containing 100 mg aminophylline (Displacement value = 1.3)
Solution steps
1-Calculate the total weight of aminophylline required.
2- Calculate what weight of glycero-gelatin base would be required to prepare 10 un-medicated suppositories.
3- Determine what weight of base would be displaced by the medicament.
4- Calculate, therefore, the weight of base required to prepare the medicated suppositories.
1- Total weight of aminophylline required = 100 mg × 6 = 600 mg or 0.6 g
2- Weight of base required for unmedicated suppositories = 4 g × 6 × 1.2
(to take account of the greater density of this base) = 28.8 g
As the displacement value of aminophylline = 1.3
This means that 1.3 g of aminophylline displaces 1 g of theobroma oil
So, 0.6 g of aminophylline displaces x
X= 0.6 g of aminophylline displace ÷ 1.3 g of theobroma oil
= 0.46 g of theobroma oil
or
DV = d/(a-c ) amount of base displaced by the drug
a-c = d/ Dv = 0.6/ 1.3 = = 0.46 g of theobroma oil
For glycerol –gelatin
This means that the aminophylline would displace 0.46 g × 1.2 of the
glycero-gelatin base = 0.55 g
Therefore, the weight of base required to make medicated
suppositories = 28.8 g – 0.55 g = 28.25 g
Answer: 28.25 grams
Solve the following problem
Prepare 12 glycero-gelatin suppositories, containing 0.5% w/w cinchocaine hydrochloride. Use a 2 g mould
Solution steps
- Calculate the total weight of the medicated suppositories, allowing for the greater density of the glycero-gelatin base.
2- Calculate the weight of the drug required.
3-Subtract the weight of the drug from the total weight of the suppositories to determine the weight of the base required.
Glycero-gelatin B.P
Rx
Gelatin 14 gm
Glycerin 70 gm sp. Gravity = 1.33
Water Q.S to 100
Calculated volume of Glycerin= mass /sp. gravity
= 70 gm/1.3 = 52.6 ml
N.B
This base is suitable base for medicated supp. containing solid medicaments or not more than 20% semisolid or liquid medicaments. With more this mass becomes too soft.
Procedure:
1-Soak gelatin in enough amount of water.
2-Put the soaked gelatin + calculated amount of glycerin.
3-Transfer to water bath + gentle stirring to avoid bubbles (zigzag way)
4- remove any skin formed before pouring
5-Pour the mass while hot, into holes lubricated with liquid paraffin, do not let the melted mass overflow.
6- Cool , remove from mould.
Medicated glycerol-gelatin supp.
Boroglycerin Supp.
Boric acid 7.5 gm
Gelatin 15 gm
Glycerin 62.5 gm
Water 15 gm
Fiat Supp. 100 gm
Mitte III
Calculation
Prepare 5 to compensate the loss
1- Total weight of Rx = 100 gm
2- Wt of Boric acid = 7.5 x 5/ 100 = 0.375 gm
3- Wt of gelatin = 15 x 5 /100 = 0.75 gm
4- volume of glycerol = (62.5 x 5 /100 )/ 1.3 = 2.35 ml.
Procedure:
1-Soak gelatin in water then add 1/2 the amount of glycerol dissolve by heating
in water bath.
2-Dissolve boric acid in glycerol by the aid of gentle heating and then transfer to dissolved glycerol-gelatin base.
3- heat on W. B till a clear solution is produced & constant weight is obtained.
4- pour into lubricated mould with soap solution or liquid paraffin using a piece of cotton & avoid overflow..
Rx
Zinc oxide - glycerin Supp.
Zinc oxide 0.1 gm
Glycero-gelatin base Q.S
Fiat : Pessaries
Mitte III
Calculation
1- Calculate the DV of zinc oxide relative official B.P formula of glycerogelatin base.
2- Prepare the required pessaries
Prepare 5 to compensate the loss
Wt of drug = 0.1 x 5 = 0.5 gm ZnO
Wt of the base = 5x 1.2 = 6 gm un-medicated base " a"
Wt of glycero –gelation = 6- 0.5 = 5.5 gm
Prepare using glycerol-gelatin B.P
Gelatin 14 gm x 5.5 /100 =
Glycerin gm sp. Gravity = 1.33 70 gm x 5.5 / 100 =
Water Q.S to 100 ( equal about 16 gm x 5.5 / 100 =
Procedure:
1-Soak gelatin in water then add 1/2 the amount of glycerol dissolve by heating in water bath.
2-Dissolve boric acid in glycerol by the aid of gentle heating and then transfer to dissolved glycerol-gelatin base.
3- heat on W. B till a clear solution is produced & constant weight is obtained.
4- pour into lubricated mould avoiding overflow..
Soap glycerin Supp
Stearic acid + Sod. Carbonate ----- sod. Stearate + CO2 ↑
Stearic Soap
Adv. Over glycerogelatin
1- larger quantity of glycerin can be used (up to 50%)
2- Soap helps the action glycerin (laxative action) while gelatin dose not.
Disadv.
Hygroscopic so they must be wrapped in waxed paper or foil and protected from moisture.
Rx
Glycerin 90 gm
Sod. Bicarbonate 4.5 gm
Stearic acid 7.5 gm
Fiat Supp.
Send III
Sig. To be taken before bed time.
Calculation
Total wt of Rx = 100 gm
Prepare 5 supp.
1- Volume of glycerin = (90x 5/100) /1.3 = 3.5 ml
2- Wt. of sod. Bicarb = 4.5 x 5/ 100 = 0.225 gm
3- Wt. of stearic acid = 7.5 x 5 / 100 = 0.375 gm
Procedure:
1-Dissolve powdered sod. Bicarb. in glycerin with the aid of gentle heating
2- Add stearic acid and heat carefully until effervescence ceases and complete solution is obtained.
4- pour into lubricated mould avoiding overflow.
Uses : laxative
Calculate the correction factor C.F =
average wt. of one supp. of other base than cocoa butter
1 gm (capacity of mould)
Water soluble Supp
Macrogols bases (carbowaxes bases)
Carbowaxes (polyethylene glycol ) are synthetic polymer of ethylene oxide and have different molecular weights ranging from 200 – 20,000.
At room temp. low mol. Wt are liquids
Mol. Wt 1000- 1540 are soft solids
Higher mol. Wt are solids
Adv:
- less liable to microbial contamination
- No laxative effect & can be used as a base for medicaments
Rx
Paracetamol 0.25 % (about
Polyethtylene glycol 4000 33 %
Polyethtylene glycol 6000 47 %
Water 20 %
Mitte :III
Sig. To be used as directed.
Calculation:
Total about 100 %
No. of supp. = 3 + 2 xss = 5
Wt. of PEG 4000 = 33 x 5/ 100 = 1.65gm
Wt. of PEG 6000 = 47 x 5 / 100 = 2.35gm
Wt. of water = 20 x 5/ 100 = 1 gm = 1 ml
Wt of paracetamol = 0 .25 x 5 / 100 = 0.0125 gm
Procedure:
1-Dissolve powdered paracetamol in H2O with the aid of gentle heating
2- Melt PEG 4000 & 6000 in a porcelain dish in water bath ,
then add water & drug
3- pour into mould with out lubrication.
Solve the following problems:
1- A prescriber requests 5 glycero- gelatin suppositories be made containing 1 % w/w hydrocortisone. If a 4 mould is used , what quantities of base and medicament are needed.
2- Prepare 8 glycero- gelatin suppositories (in a 2 gm mould) each containing 2 mg of morphine hydrochloride (DV = 1.6).
3- How would you prepare 10 theobromo oil suppositories (1 gm mold) containing 2.5% w/w bismuth subgallate ?
4- How would you prepare 6 theobromo oil suppositories (2 gm mold) containing 10% w/w zinc oxide ?
5- Six theobromo oil suppositories (2 gm mold) each containing 125mg of paracetamol, are to be prepared. The displacement value of paracetamol is 1.5. What quantities of base and medicament are required?
6- A pharmacist has to prepare 6 suppositories for a patient, each suppository to be made in 1 gm mould and to contain 0.4 gm of bismuth subgallate. Calculate the amount of the base (Theobroma oil) that the pharmacist will need to use.
7-A pharmacist has the prepare twenty 2 gm of glycerol-gelatin pessaries each containing 150 mg miconazole nitrate. DV = 1.6
Assume that the pharmacist makes sufficient for 25 pessaries
Then give the amount required for each substance of the glycerol-gelatin base.
Electrolytes solutions
Milliequivalent and Millimoles calculations
From Moles to equivalent or Milliequivalents
Chemical substances in solution may remain intact or dissociate into particles known as ions, which carry an electric charge.
--- Substance that are not dissociate in solution are called " nonelectrolytes"
Such as: Urea and dextrose
--- Substances that dissociate in solution are called " electrolytes"
Such as: sodium chloride , potassium chloride.
Electrolytes ions in the blood plasma include the cations e.g Na +, K+, Ca++, Mg++, and anions Cl-, HCO3- , HPO4 -- - , SO4 - -. Organic acid - , Protein - .
Electrolytes in body fluids play an important role in maintaining the acid-base balance in the body. They play a part in controlling body water volumes and they help to regulate the body metabolism.
Application of dosage forms
Electrolytes preparations are used for the treatment of disturbance of the electrolytes and fluid balance in the body.
They are applied clinically in the form of oral solutions and syrups, as dry granules intended to be dissolved in water or juice to make an oral solution, as oral tablets and capsules and when necessary as intravenous infusions.
Mole , Millimole or micromole = is used in international systems (SI ) used in European countries to express the electrolytes concentrations.
A mole : is the molecular weight of a substances in grams
A millimole: is 1/1000 of the molecular weight in grams
= Mol. Wt
1000
Molar Solutions
A 1 molar solution is a solution in which 1 mole of a compound is dissolved in a total volume of 1 litre.
mol /L mmol/L , μmol/L
For example:
The molecular weight of sodium chloride (NaCl) is 58.44, so one gram molecular weight (= 1 mole) is 58.44g.
If you dissolve 58.44g of NaCl in a final volume of 1 litre, you have made a 1M NaCl solution.
Solve the following:
1- How many gm of sod. Chloride is needed to prepare 0.1 mole
2- How many millimoles of monobasic sodium phosphate (m.w = 138) are present in 100 gm of the substance?
3- How many milligrams would 1 mmol of monobasic sodium phosphate weigh.
4- What is the weigh, in milligrams, of 1 mmole of HPO4 - ?
Atomic weight = 95.98
5- How many millimole of calcium chloride (m.w = 147) are represented in 147 ml of a 10% w/v calcium chloride solution.
6- How many grams of anhydrous dextrose should be used to in preparing 500 ml of 0.5 mole of dextrose (M. W = 180)
Equivalent and milliequivalent
Eq. or mEq
Many important substances in the body are measured in equivalents or milli equivalent.
Eq. or mEq. Are chemical units applied in USA by clinicians, physicians, pharmacist to express the concentration of electrolytes in solutions.
This unit of measure is related to the total numbers of ionic charge in solution (takes note of the valence of the ions)
, or
It is the unit of measurement of the amount of chemical activity of an electrolytes
Note:
-- Use of equivalents and milliequivalents is valid only for those substances that have fixed ionic valences (e.g. sodium, potassium, calcium, chlorine, magnesium bromine, etc).
--- For substances with variable ionic valences (e.g. phosphorous), a reliable equivalent value cannot be determined
The technical definition of an equivalent is the amount of substance it takes to combine with 1 mole of hydrogen ions.
An example
we can look at hydrochloric acid (HCl). It takes approximately 35 grams of chloride (1 mole), to combine with 1 gram of hydrogen (1 mole) to make 1 mole of HCl (which weighs approximately 36 grams).
Since both of these elements are monovalent (carrying a valence charge with a magnitude of 1), they combine in a one-to-one ratio.
Therefore the amount of chloride that is needed to combine with one mole of hydrogen is = 1 mole or 1 equilvalent (eq).
Equivalents (Eq) = number of univalent counter ions needed to react with each molecule of substance. HCl has 1 equivalent per mole in that one mole of H+ reacts with one mole of Cl
Equivalents can also be defined in terms of metric weight. The weight of one equivalent can be determined by dividing the gram-atomic weight of the ion by its valence.
1 equivalent = gm mol. Or atomic wt
valence
Remember
Most solutes in the body are not measured in grams and moles, but milligrams and millimoles instead.
-- milliequivalent (mEq) = 1/1000 of it gm equivalent weight.
Ex = Mol.wt of KCl = 74.5
Equivalent wt = 74.5/1= 74.5 gm
MEq= 1/1000 x 74.5 = 0.0745 gm
So for monovalent 1 equivalent = 1 mol
---- For a divalent ion (Ca+2) combining with Cl- (a monovalent ion) to form calcium chloride ( CaCl2 ).
The combining ratio of calcium to chlorine is one-to-two, meaning it takes half a mole of Ca+2 to combine with one mole (or 1 eq) of Cl-. So, half a mole of calcium equals 1 equivalent.
So for divalent 1 equivalent = 0.5 mol
Another way to look at it is to say one equivalent is the amount of ion required to cancel out the electrical charge of an oppositely charged monovalent ion. You can also say that the valence charge of the ion is the number of equivalents there are in one mole of that ion.
For example, nitrogen (N-3) is a trivalent ion; therefore one mole of nitrogen equals 3 equivalents.
Using the calcium example from above, we know that one mole of calcium equals 2 equivalents.
Here are some things to keep in mind when converting to milliequivalents:
1 mEq = 10-3 eq
For monovalent ions, 1 mEq = 1 mmol
For divalent ions, 1 mEq = 0.5 mmol
For trivalent ions, 1 mEq = 0.333 mmol
Rules for calculations of milliequivalents
To convert the concentration of electrolytes in solution expressed as milliequivalent per unit volume to weight per unit volume and vice versa, use the following
mEq = mg x Valence
Atomic, molecular, formula weight
mg = mEq x Atomic, molecular, formula weight
Valence
Exercise
1. How many equivalents are present in 80 grams of calcium (molecular weight = 40 g)?
2. How many equivalents of sodium are present in 116 g of NaCl (molecular weight of Na ≈23; molecular weight of Cl ≈ 35)?
3. How many millimoles of Mg+2 would be present in a solution containing 0.8 milliequivalents?
4. How many milliequivalents of P-3 are present in a solution containing 6 millimoles?
5- Calcium has gram atomic weight of 40.08 , what is the m Eq weight
6- How many m Eq of Na+ and Cl – has a solution containing 409.5 mg of NaCl / 100 ml. ( Mol. Wt of NaCl= 58.5 )
7- What is the con. In gm/ml of a solution containing 4 mEq of calcium chloride (Ca. Cl2. 2 H2O) per ml.
Mol. Wt of calcium chloride = 147
8- A solution contains 298 mg of KCl per ml. Express this concentration in term of mEq of KCl.
Mol. Wt of KCl = 74.5
9-A – 10 ml ampoule of KCl contains 2.98 gm. What is the concentration of the solution in term of mEq/ml.
10- How many grams sod. Chloride should be used to prepare a solution containing 154 mEq per liter.
Mol. Wt of NaCl= 58.5
11- Sterile solutions of KCl containing 5 m Eq/ml are available in 20- ml containers. Calculate the amount , in grams of KCl in the container.
Mol. Wt of KCl = 74.5
12- How many ml of a solution of 2 m Eq of KCl per ml should be used to obtain 2.98 gm of KCl.
Mol. Wt of KCl= 74.5
13- A solution containing 10 mg % of K+. Express this concentration in term of mEq/L (Atomic wt of K+ = 39)
14- How many milliequivalents of potassium chloride are represented in a 15-ml dose of a 10% (w/v) potassium chloride elixir?
Osmolality
Or Osmolarity
Osmotic pressure is important to biological process that involve diffusion of or the transfer of fluids through semipermeable membranes.
Measurement of osmolar concentrations of parenteral fluids is important
The Labels of pharmacopeial solutions that provide intravenous replenishment of fluid, nutrients or electrolytes and the osmotic diuretic mannitol are required to state the osmolar concentration.
The measurement of osmolar concentration indicate whether the solution is hypotonic, iso-osmotic, hypertonic, with regard to biologic fluids and membranes.
Electrolytes have a role in controlling body water volume by establishing osmotic pressure.
This osmotic pressure is proportional to the total number of particles in solution where ether this particles are ions or molecules.
Plasma osmolality : is a measure of the concentration of substances such as sodium, chloride, potassium, urea, glucose, and other ions in human blood.
It is calculated as the osmoles of solute per kilogram of solvent.
Normal reference range of osmolality in plasma is about 280 - 303 milli-osmoles per kilogram. It is affected by changes in water content.
Osmolarity vs Osmolality Measures of osmotic concentration
Osmolarity: "millimoles" concentration of solute per liter of solution
• Osmolality: "millimoles" concentration of solute per kilogram of solvent
• Osmolarity is NOT ALWAYS equivalent to Osmolality (beware of terminology!)
The unit used to measure osmotic activity is milliosmol, & abbreviated mosm.
osmole (osmol) is a standard unit for measurement of osmotic pressure of one mole concentration of an ion or particle in a solution
Or = unit of measurement that defines the number of moles of a chemical compound that contribute to a solution's osmotic pressure. Dissolving table salt in water In chemistry, a solution is a homogeneous mixture composed of one or more substances, known as solutes, dissolved in another substance, known as a solvent. ...
e.g.,
1 mole of glucose, which is not ionizable, represent 1 osmole of solute,
1 mole of sodium chloride represented 2 osmoles of solute., as NaCl dissociate into two ions
1mole CaCl2 represent 3 osmole , dissociate into 3 ions
I mole sodium citrate represent 4 osmole, dissociate into 4 ions
1 milliosmol = 1/ 1000 x osmole
One Osmole = The molecular weight of a substance, in grams, divided by the number of ions or particles into which it dissociates in solution ( i.e its valence).
The milliosmolar value of separate ions of an electrolytes may be obtained by dividing the concentration in mg/L on the ion atomic weight .
Calculation of mosmole of separate ions of electrolytes =
= Concentration of the ion (mg/L)
Ion atomic weight
The milliosmole of the whole electrolytes in solution is equal to the sum of milliosmolar values of separate ions
Or
Total milliosmole of whole electrolytes in solution
mOsmole = weight of substances (g/L) X NO. of species X 1000
Molecular weight (gm)
mOsmole = weight of substances (mg/L) X NO. of species
Molecular weight (mgm)
Example
- calculate the number of milliosmole for a solution of 2 L containing 0.9% sod. Chloride. NaCl Mol. Wt = 58.5
For the above solution :
0.9 gm 100 ml
X 2000 ml
X= 0.9x2000 / 100 = 18 gm = 18, 000 mg
No of msomole =
Concentration of the ion (mg/L) 18,000 mg x 2 = 615.4 mosmole
Ion atomic weight 58.5 mg
Solve the following problems.
1- A solution contains 5 % of anhydrous dextrose in water for injection.
- How many milliosmole/ litre are represented by this concentration.
Mol. Wt of dectrose = 180
2- A solution contains 156 mg of potassium (K+) ions per 100 ml. How many milliosmoles are represented in a liter of the solution.
Atomic wt of K+ = 39
3- A certain electrolytes solution contains 0.9% of NaCl in 10% dextrose solution.
A- Express the conc. Of NaCl in terms of mEq/L and millisomol.
b- How many milliosmoles of dextrose are I one liter of solution.
Mol. Wt. of NaCl = 58.5
Of dextrose = 180
How many milliosmoles of dextrose are one liter of solution.
8- How many milliosmol are represented in a liter of a hypotonic (1/5 normal ) NaCl solution?
Isotonic solution (normal) contains 0.9% NaCL. Assume complete dissociation. NaCl mol. Wt = 58.5
Solve the following problems:
1- A solution contains 10 mg % of Ca 2+ ions, How many milliosmoles are represented in 1 liter of solution? (m.w. 40)
2-A solution contains 322 mg of Na+ ions per liter. How many milliosmoles are represented in the solution? (m.w. 23)
3-A solution of sodium chloride contains 77 mEq/L. Calculate its osmolar strength in terms of milliosmoles per liter. Assume complete dissociation? (m.w. 58.5)
4- Calculate the osmolar concentration , in terms of milliosmoles, represented by 1 liter of a 10 % w/v solution of anhydrous dextrose (m.w = 180) in water.
5-Calculate what is the osmolarity of an 8.4% w/v solution of sodium bicarbonate (m.w. 84).
6-Calculate the osmolarity, in milliosmoles per milliliter, of a parenteral solution containing 2 mEq/ml of potassium acetate (KC2H3O2---- m.w. 98).
7-Calculate:- a – total mEq, b- mEq/ml c- the osmolarity of a 500-ml parenteral fluid containing 5 % w/v of sodium bicarbonate. (m.w. 84).
8- A hospital medication order calls for the administration of 100 g of mannitol to a patient as an osmotic diuretic over a 24-hr period. Calculate :
(a)- how many milliliters of 15% w/v mannitol injection should be administered per hour, and (b) how many milliosmoles of mannitol (m.w. 182) would be represented in the prescribed dosage.
9-A hospital pharmacist fills a medication order calling for an intravenous fluid of dextrose 5% in 0.9% injection and 40 mEq of potassium chloride in a total volume of 1000 ml. The intravenous infusion is administered through an IV set that delivers 15 drops per milliliter. The infusion has been running at rate of 12 drops per minutes fot 15 hours.
During the 15- hour period:
How many mEq of KCl have been administered?
How many grams of Kcl have been administered?
How many millimoles of KCl have been administered?
What is the total osmolarity of the intravenous fluid?
Calculation of isotonic solutions
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Osmosis = is the phenomenon when solvent molecules pass through semi permeable membrane from dilute solution to concentrated one to be equalized.
Osmotic pressure = force that allow passage of solvent molecules
Osmotic pressure depends on the total number of particles present in solution where ether this particles are molecules or ions
So osmotic pressure will depend on:-
■ the solute concentration
■ degree of dissociation of solute molecules
N.B
Substances that dissociate have relatively greater umber of particles in solution & should exert a greater osmotic pressure than undissociated particles.
Iso-osmotic = two solution have the same osmotic pressure
Iso-tonic = a solution that have osmotic pressure as body fluids
Or = Having the same concentration of solutes as the blood:
Hypotonic = solution with osmotic pressure lower than that of body fluids
Hypertonic = solution with osmotic pressure higher than that of body fluids.
Blood, and the fluids of the eye, nose and bowel are of principle concern to the pharmacist in the manufacture and use of preparations to be mixed with those biologic fluids, which include:-
1- Ophthalmic (eye)
2- Nasal (Nose)
3- Parenteral (By injection)
4- Enema (rectal)
Ex of isotonic solutions
1- Ringer solution, Ringer's solution - an aqueous solution containing the chlorides of sodium and potassium and calcium that is isotonic to animal tissues; used to correct dehydration and (in physiological experiments) as a medium for in vitro preparations
2- saline, saline solution - an isotonic solution of 0.9% sodium chloride and distilled water.
It is generally agreed that many solutions designed to be mixed with body fluids should have the same O.P as body fluid (isotonic) for greater comfort, efficacy & safety.
Clinical consideration of tonicity
It is generally accepted that for ophthalmic and parenteral administration, isotonic solutions are better tolerated by the patient than those at extreme of hypo-hyper tonicity.
Only hyper tonic solution can be used to "draw" fluids out of edematous tissue into the administered solution.
1- Isotonic ophthalmic solution are formulated isotonic or approximately isotonic to duplicate ophthalmic tears for the comfort of the patient. (Artificial tears). They are also buffered to reduce irritation to the eye's tissues and to maintain the stability of the preparations.
2- Injection that are not isotonic should be administered
Slowly and in small quantities to minimize tissue irritation, pain, and cell fluid imbalance.
3- Tonicity of small-volume injections can be adjusted when added to large –volume infusions, because of the presence of tonic substances, e.g sod. Chloride, or dextrose in large-volume infusion which serve to adjust the tonicity of smaller added volume.
4-Large volumes of hyper tonic infusions containing dextrose , can result in hyperglycemia, osmotic diuresis and excess loss of electrolytes .
5- Excess infusions of hypo-tonic fluids can result in the osmotic hemolysis of red blood cells (breaking open of red blood cells and the release of hemoglobin into the surrounding fluid (plasma, in vivo).And surpass the upper limits of the body's capacity to safely absorb excessive fluids.
[pic]
Hemolysis. Red blood cells without (left and middle) and with (right) hemolysis.
Note that the hemolyzed sample is transparent, because there are no cells to scatter light.
Physical and chemical considerations in preparation of isotonic solutions
Osmotic pressure depends on the number than the kind of particles and the type of dissolved substances.
So substance that dissociate have a tonic effect that depend on the degree of dissociation ( ionization)
The greater the dissociation, the smaller the quantity required to produce a given osmotic pressure.
1- For non electrolytes e.g Boric acid:
100 molecules will remain as 100 molecules
2- For electrolytes e.g sod. Chloride
If we assume that sod. Chloride in weak solutions will dissociate by about 80%.
So each 100 molecules will yield 180 particles, Bec
80 Na+ + 80 Cl- + 20 Na Cl = 180 particles
So the increase in No. of particles will be 1.8 times the No. of particles produced non electrolytes
Calculation of dissociation factor : "I" factor of sod. chloride
I = total No. of particles in the solution = 180 = 1.8
No. of molecules if not un-dissociated molecules 100
Problems on calculating the dissociation factor (i) of an electrolyte:
1- Zinc Sulfate ZnSO4 is a 2-ion electrolyte, dissociation 40% in a certain concentration. Calculate its dissociation factor.
2- Zinc Chloride (ZnCl2) is a 3-ion electrolyte; dissociation 80% in a certain concentration, calculate (i).
general, we may use the following tabulated values in case of solutions of 80% or higher concentrations:
|substance |(i) dissociation factor |
|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 |
Procedure for Calculation of Isotonic Solutions Using Sodium Chloride Equivalents ( E – value):
When a prescription directs us to make a solution isotonic by adding the proper amount of some substance other than the active ingredients.
Ex 1:-
A given 0.5% solution of sod. Chloride, to make it isotonic
1- isotonic solution contain 0.9% NaCl
So to make such solution isotonic
0.9- 0.5 = 0.4 gm NaCl is needed to be added to make the solution isotonic and the final volume will be 100 ml.
Ex 2:
For a 100 ml solution of 1% w/v atropine sulfate, which is to be made isotonic with lachrimal fluids?
Solution
1- firstly we should determine how much of sod. Chloride is in effect represented by atropine sulfate.* i.e 1% atropine sulfate will be equivalent to how much of NaCl.
We can use the rule that
The quantity of one that is the equivalent in tonic effects to a given quantity of the other may be calculated by:=
Quantity of one having a certain effect in a specified quantity of solvent
quantity of the other having the same effect in the same quantity of solvent
Quantities of two substances that are tonicic equivalents are proportional to the molecular weights of each multiplied by the "i" value of the other.
Mol. Wt of atropine x I value of Nacl === mol.wt of NaCl x I value of atropine
So for 1 gm of atropine ==== x NacL
X E- value NaCl =
i.e. Calculation of Sodium Chloride equivalent factor (E value) of a substance (if it is not tabulated):
Sodium Chloride equivalent = * E-Value =
mol. wt of sod. chloride x I factor of substances
I factor of sod. chloride Mol. Wt of substances
Mol. Wt of NaCl = 58.5 i= 1.8
Mol. Wt of atropine = 695 i= 2.6
E = 58.5 x 2.6 = 152.1 / 1252 = 0.12 gm
695 x 1.8
So 1 % atropine sulfate ---=== 0.12 gm NaCl
Problem on calculating the E-value of a substance:
1- Papavarine HCl (mwt = 376) is a 2-ion electrolyte dissociating 80%. Calculate its E-value (Sodium Chloride equivalent), where
- its dissociation factor (i) = 1.8
- NaCl mol. Wt = 58.5
2- Procaine HCl (mol. Wt 273) is a 2-ion electrolytes, dissociating 80% in a certain concentration .
A – calculate its dissociation factor
b- Calculate its sodium chloride equivalent.
3- Calculate the sodium chloride equivalent for fluorescein sodium , which dissociate into three ions and has a molecular weight of 376. & i factor = 2.6
Calculation of tonic agent required use the E-value of a substance
The values of sod. Chloride equivalent (E values) are given in the text books. Example that given in the following table
The amount of NaCl represented by any substance in the prescription is given by
= number of grams of the substance x its sod. Chloride equivalent
So the following procedure should be followed for calculation iso tonic solution with sod. Chloride.
1- 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 of the total volume specified in prescription.
i.e. 0.9 à 100 because 0.9% solution of NaCl is isotonic.
X à total of the prescription.
3- 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) i.e. step # 2 – step # 1.
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 substance used to adjust Isotonicity
Problems on isotonic solutions calculations using the NaCl E-value:
1- How many grams of NaCl should be used in compounding the following prescription:
Rx Pilocarpine nitrate 0.3 g E =0.23
Sod. Chloride Q.S.
Purified water ad 30 ml
Make isotonic solution.
2- Rx Ephedrine sulfate 0.26 g (E= 0.2)
Sodium Chloride Q.S
Distilled water aq to 30 ml
Sig: Use as directed
How many mg of sodium chloride should be used in compounding the prescription.
3- Rx Atropine sulfate 1 % (E= 0.12)
Boric acid Q.S (E= 0.52)
Distilled water aq to 30 ml
Sig: one drop in each eye
How many mg of boric acid should be used in compounding the prescription.
Amount of atropine needed for 30 ml = 1x 30 /100 = 0.3 gm
4- How many grams of boric acid should be used in compounding the following prescription?
Rx
Phenacaine HCL 1.0% (NaCl eq. = 0.17)
Chlorobutanol 0.5% (NaCl eq. = 0.18)
Boric acid qs. (NaCl eq. = 0.52)
Purified H20 ad 30 mL
Make isotonic solution
Sig: One drop in each eye
5- How many grams of sod. chloride should be used in compounding the following prescription?
Rx
Dextrose, anhydrous 2.5 % (NaCl eq. = 0.18)
Sodium chloride Q.S
Sterile water for injection 1000 ml
Purified H20 ad 30 mL
Label: Isotonic dextrose and saline solution
6- How many grams of sod. chloride should be used in compounding the following prescription?
Rx
Oxytetracyline HCl 0.5% E = 0.12
Tetracaine HCl 2% solution 15 ml
Sodium Chloride Q.S.
Purified water ad to 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?
7- How many grams of dextrose monohydrate should be used in compounding the following prescription?
Rx
Ephedrine Hydrochloride 0.5 gm
Chlorobutanol 0.25 gm
Dextrose monohydrate Q.S
Rose water ad 50 ml
Sig. Nose Drops
8- How many grams of Boric acid should be used in compounding the following prescription?
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?
Buffers and Buffer solutions
Buffer solution :- is an aqueous solution, that contain certain substances or combination of substances that imparts the system the ability to maintain a desired pH and possesses the property of resisting changes in pH with the addition of small amounts of a strong acid or base.
Buffer action: is referred to the ability of buffers to resist changes in pH
Buffer capacity : is the ability to withstand rapid pH fluctuation.
Buffering capacity refers to solution's ability to keep the pH stable as
acids or bases are added.
The greater the buffering capacity, the greater the quantity of acid or base which must be incorporated with a material to alter the pH.
Buffering capacity= Moles of acid or base added in buffers solution
change in solution's pH.
B = ∆B / ∆pH
∆B = is the minimal increment of strong base or strong acid
∆pH = the change in pH
Why buffered are used:-
1- establish and maintain an ion activity within rather narrow limits.
2- In pharmacy it used to:-
a- preparation of such dosage forms as injections & ophthalmic solutions which are placed directly into pH- sensitive body fluids
b- manufacture of formulations in which the pH must be maintained at a relatively constant level to ensure maximum product stability
c- pharmaceutical tests and assays requiring adjustment to or maintenance of a specific pH for analytic purposes.
What is the composition of buffer solutions:
They are composed of a weak acid and its salt or weak base and it salt
|Acidic buffers |acetic acid |sodium acetate |
| |boric acid |sodium borate |
| |sodium acid phosphate |di-sodium phosphate |
|Basic buffer |Ammonium hydroxide |Ammonium chloride |
Selection of buffers
The selection of buffer must be given to the dissociation constant of weak acid or base to ensure maximum buffer capacity.
The dissociation constant, in the case of an acid, is a measure of the strength of the acid.
The more readily the acid dissociation, the higher its dissociation constant and the stronger is the acid.
|ACID |FORMULA |kA |pkA |
|acetic acid |H(C2H3O2) |1.74 E-5 |4.76 |
|ascorbic acid (1) |H2(C6H6O6) |7.94 E-5 |4.10 |
|ascorbic acid (2) |(HC6H6O6)- |1.62 E-12 |11.79 |
|boric acid (1) |H3BO3 |5.37 E-10 |9.27 |
|boric acid (2) |(H2BO3)- |1.8 E-13 |12.7 |
|boric acid (3) |(HBO3)= |1.6 E-14 |13.8 |
|butanoic acid |H(C4H7O2) |1.48 E-5 |4.83 |
|carbonic acid (1) |H2CO3 |4.47 E-7 |6.35 |
|carbonic acid (2) |(HCO3)- |4.68 E-11 |10.33 |
|chromic acid (1) |H2CrO4 |1.82 E-1 |0.74 |
|chromic acid (2) |(HCrO4)- |3.24 E-7 |6.49 |
|citric acid (1) |H3(C6H5O7) |7.24 E-4 |3.14 |
|citric acid (2) |(H2C6H5O7)- |1.70 E-5 |4.77 |
|citric acid (3) |(HC6H5O7)= |4.07 E-7 |6.39 |
|formic acid |H(CHO2) |1.78 E-4 |3.75 |
|heptanoic acid |H(C7H13O2) |1.29 E-5 |4.89 |
|hexanoic acid |H(C6H11O2) |1.41 E-5 |4.84 |
|hydrocyanic acid |HCN |6.17 E-10 |9.21 |
|hydrofluoric acid |HF |6.31 E-4 |3.20 |
|lactic acid |H(C3H5O3) |8.32 E-4 |3.08 |
|nitrous acid |HNO2 |5.62 E-4 |3.25 |
|octanoic acid |H(C8H15O2) |1.29 E-4 |4.89 |
|oxalic acid (1) |H2(C204) |5.89 E-2 |1.23 |
|oxalic acid (2) |(HC2O4)- |6.46 E-5 |4.19 |
|pentanoic acid |H(C5H9O2) |3.31 E-5 |4.84 |
|phosphoric acid (1) |H3PO4 |6.92 E-3 |2.16 |
|phosphoric acid (2) |(H2PO4)- |6.17 E-8 |7.21 |
|phosphoric acid (3) |(HPO4)= |2.09 E-12 |12.32 |
The dissociation constant " Ka "value of an weak acid
Ka = ( H +)( A-) where A- = salt
(HA) HA = acid
Because the numeric values of most dissociation constants are small and may vary over many powers of 10, it is more convenient to express it as negative logarithms
P Ka = - log Ka
So pKa = - log ( H +) _ log salt
acid
PH = - log (H+)
So For acid
pKa = pH _ log salt
acid
and pH = pKa + log salt
acid
The dissociation constant " Ka "value of an weak base
Ka = ( B +)( OH- ) where B + = salt
(BOH) BOH = base
Buffer equation for the base
and pH = pKw - pKb + log base
salt
Buffer equations is known as
Henderson-Hasselbalch equation, which describes pH in terms of pKa
Importance of buffer equation:-
1- calculate the pH of the buffer system if its composition is known
2- determine the molar ratio of the buffer components required to give a solution of certain pH
3- calculate the change in pH of a buffer solution with the addition of a given amount of acid or base
Solve the following problems
Calculate the pKa value of the following weak acids
|Acid |Dissociation constant at 25 C |
|acetic acid |1.75 x 10 – 5 |
|boric acid |6.4 x 10 -10 |
|lactic acid |1.38 x 10 -4 |
|formic acid |1.76 x 10 -4 |
Calculate the pH value
1- What is the pH of buffer solution prepared with 0.05 M sodium borate and 0.005 M boric acid? The pKa value of boric acid = 9.24 at 25C
2- What is the pH of a buffer solution prepared with 0.055 M sodium acetate and 0.01 M acetic acid? The pKa value of acetic acid is 4.76 at 25 C.
3- What is the pH of a buffer solution prepared with 0.5 M disodium phosphate and 1 M sodium acid phosphate ? The pKa sodium acid phosphate is 7.21 at 25 C.
4- What is the pH of a buffer solution prepared with 0.05 M ammonia and 0.05 M ammonium chloride? The pKb of ammonia is 1.8 x 10 -5 at 25 C. The pKw of water is 1 x 10 -14 at 25 C.
5- What is the molar ratio of salt/ acid is required to prepare a sodium acetate-acetic acid buffer solution with a pH of 5.76 ? The pKa value of acetic acid is 4.76 at 25 C
6- What molar ratio of salt / acid would be required to prepare a buffer solution with a pH of 4.5 ? pKa of the acid is 4.05 at 25 C
Reaction kinetics
Chemical kinetics
Is the study of the rate of chemical changes taking place during chemical reaction.
The rate processes are of fundamental concern to everyone connected with pharmaceutics :-
1- pharmacist :- who should aware by the potential of instability problems related to the drug he handle.
2- Manufacturer:- who must demonstrate the drug or dosage form with sufficient stability under storage condition for reasonable length of time (self-life).
Pharmaceutical manufacturer are responsible to design, test, produce dosage forms that contain exact quantities and qualities of drugs and are acceptable, reproducible, convenient and elegant.
3- Patient :- who must be sure that the prescribed drugs reach them in sufficient concentration to give a desired action.
The reaction rate studies is important in studying :-
1- drug stability & incompatibility, study the rate of process by which the drug become inactive either by decomposition, or loss of activity due to physically and chemically changes.
2- Dissolution:- which concern with the rate of drug dissolution from it solid dosage form to molecular solution.
3- Absorption, distribution, metabolism, concerned by the rate of drug absorption, distribution, metabolism that affect the drug therapeutic activity.
Rate of reaction:
Is the velocity or the speed by which a reactant or reactants undergo chemical changes. The rate is measured by measuring the change in concentration of reactant or product in a period of time.
According to the law of mass,
The rate of chemical reaction is proportional to the product of molar concentration of the reactants each raised to the power equal to the number of molecules undergo the reaction.
aA + bB ………………→ product
so the rate of reaction is given by = ± dc/dt
dc = change in concentration
dt = change in time
+ or – sign indicate increase or decrease in concentration dc within a time interval dt.
So rate of the above reaction = [A]a [B]b
The order of reaction = a +b = n
If n= zero ……..zero order reaction
n=1..............first order reaction
n=2 .............second order reaction
1- Zero-order reaction kinetics
If the rate of reaction is independent of the concentration of reacting species, the reaction
is called zero-order.
This means that it is not possible to increase the rate of reaction by increasing the concentration of reactant.
This means that the rate of reaction is constant
- dA/dt = constant = k
this means that there is a constant decrease in concentration of substance A with change in time t.
By integration of the above equation between the initial drug concentration (Ao) and the concentration at time t (At)
∫ At dA = -k ∫ t dt.
Ao o
At-A0 = -kt
At = Ao –kt
This the equation of straight line (Y = m x + b)
Where m = the slope, b = the intercept ,
So, if plot Concentration vs. Time will give straight line
Slope = k = m Intercept = Ao = b
-
- K can be obtained from :-
1- the curve
2- K = A0 At = A0 At
t-0 t
by calculation = C1-C2
t2-t1
[pic]
Constant rate of change
The half-life (t 1/2)
Is the time required to for the initial concentration of reactant to get reduced to the half
At = 1/2 Ao
At = Ao –kt
Ao /2 = Ao –kt1/2 = Ao /2 - Ao = – kt1/2
- Ao /2 = –kt1/2
So t1/2 = 1/2 Ao
K
It should be noted that t1/2 of zero-order reaction is proportional to the initial concentration of the drug
Shelf –life for zero-order reaction.
Is the time required for the drug to decompose by 10% (loss 10 % of its activity).
i.e decrease to 90% of its original concentration.
At = 0.9 Ao
At = Ao –kt
0.9Ao = Ao –kt0.9 = 0.9 Ao - Ao = – kt0.9
- 0.1Ao = –kt0.9
So t0.9 = 0.1 Ao
K
Draw the following data in a Cartesian or rectangular coordinate graph paper, then predicate the following by both graph and calculation methods:
- The order of reaction
- The rate reaction constant
- Half-life
- Shelf- life
Write the equation for each required item.
1-
| |Time (sec) |[S2O8-2] |
|1 |0 |0.048 |
|2 |60 |0.043 |
|3 |120 |0.037 |
|4 |180 |0.032 |
|5 |240 |0.027 |
|6 |300 |0.022 |
2-
| |Time (sec) |[S2O8-2] |
|1 |0 |0.096 |
|2 |30 |0.090 |
|3 |60 |0.084 |
|4 |90 |0.077 |
|5 |120 |0.071 |
|6 |150 |0.065 |
2- A pharmacist dissolved a few milligrams of a new antibiotic drug into exactly 100 ml of distilled water and placed the solution in a refrigerator (5 C). At various time intervals, the pharmacist removed a 10 ml aliquot from the solution and measured the amount of drug contained in each aliquot.
The following data were obtained:
Time (hr) Concentration mg /ml
0.5 84.5
1 81.2
2 74.5
4 61.0
6 48.0
8 35.0
12 8.7
a. What is the order of the decomposition process of this
antibiotic?
b. What is the rate of decomposition of this antibiotic?
c. How many milligrams of antibiotic were in the original
solution prepared by the pharmacist?
d. Give the equation for the line fits the experimental data.
3) The degradation of a multi-sulpha preparation is a zero order process. If the concentration was 0.47 mole/liter when freshly prepared & after 473 days ,its absorbance reaches 0.225 mole/liter. Calculate k & tl/2.
4) Drug suspension conc. :=125 mg/ml decay by zero order kinetics, k=0.5 mg/ml/hr. What is the conc. of the intact drug remaining after 3 days?
2- First-order reaction kinetics.
The rate of reaction is directly proportional to the first power of the concentration of single reactant.
The reaction depend on the concentration of single reactant.
So dc/dt C (remaining concentration)
dc/dt = - kc
dc/c = -K dt
by integration between Co and C at t=0 and t =t
∫ Ct dC/C = -k ∫ t dt
Co o
Ln C – Ln Co = - K (t-0)
Ln C = Ln Co - Kt
It is known that Ln = 2.303 log
2.303 log C - 2.303 log Co = -Kt
Divide both side of equation on 2.303
Log C – Log Co = -Kt/2.303
Log C = Log Co - Kt/2.303 (first order equation).
Or , in exponent form, the equation becomes
C = Co e –KT/2.303
[pic]
Plot of concentration vs. time not give straight line
But if we plot log C Vs. time it will give straight line
The slope = - K/2.303
[pic]
K = rate constant = - Slope/2.303
Or can be calculated from the equation:-
Log C – Log Co = -Kt/2.303
Log C = Log Co -Kt/2.303
K = Log Co x 2.303
C t
Or
K = 2.303 log a (initial conc)
T (a - x) (remaining con)
A = initial concentration
x = decrease in concentration
a-x = remaining concentration at time t
K = is reaction rate constant of first order is sec -1 (time -1)
2. Semi log graph paper: available with one, two, three or more cycles per sheet, each cycle representing a 10-fold increase in the numbers or a single log10 unit.
Remember semi-log graph paper has a normal x- axis scaling but the y- axis scaling is proportional to the log of the number (not the number itself).
It saves you from taking the log of each number before you plot it.
Note: Once you take the log of a number you loose the units.
[pic]
Example use of semi-log graph paper: Plot the data, draw a line "through the data", and calculate the slope of the line.
Example Cp versus Time Data
Time (hr) 1 2 4 8 12
Cp (mg/L) 20 15 6.8 3.2 1.3
[pic]
From the Figure Value
|Cp1 |22.8 |
|Cp2 |1.0 |
|t1 |0 |
|t2 |12.8 |
-- A best-fit line is drawn through the data points and extended to the extremes of the graph paper (for better accuracy).
-- Values for Cp1 and Cp2 are read from the y-axis and t1 and t2 values are read from the x-axis.
-- These values can be used to estimate a value for the slope of the line and also the rate constant for the drug elimination.
--- From the Y-axis the first point is 0, 22.8 and from the X-axis the second point is 12.8, 1 (in the format x-value, y-value).
--- Always try to use the extremes of the line for the best accuracy.
From the Figure Value
Cp1 22.8
Cp2 1.0
t1 0
t2 12.8
The slope can be calculated using the equation of the line is: [log (1.0) - log (22.8)]/ (12.8 - 0)
= (0.000 - 1.358)/12.8 = - 0.106 hr-1
The half-life (t 1/2)
Is the time required to for the initial concentration of reactant to get reduced to the half
K = Log Co x 2.303
Log C t
So
t1/2 = 2.303 log Co
K 1/2 Co
t1/2 = 2.303 log 2
K
t1/2 = 0.693/K
It should be noted that t1/2 of first -order reaction is not proportional to the initial concentration of the drug. (independent on the drug concentration)
Shelf –life for first-order reaction.
Is the time required for the drug to decompose by 10% (loss 10% of its activity). i.e decrease to 90% of its original concentration.
T0.9 = 2.303 log Co
K 0.9 Co
T0.9 = 2.303 x 0.0457
K
T0.9 = 0.1052
K
Methods for determining the order of reaction
1- substitution method: in this method, the data obtained from a kinetic experiment is substituted in the relevant integrated rate equation. The equation that yields a fairly constant value of "K" indicate the order of reaction.
2- Graphical: in this method, the data obtained from a kinetic experiment is plotted in the relevant form to determine the order of a particular reaction.
- if straight line is obtained by plotting the concentration "C" against time " t", the reaction is zero-order
- if straight line is obtained by plotting the log remaining concentration "a-x" against time " t",
the reaction is first-order
3- Half – life method:
n = log t1/2 (1) / log t1/2 (2) + 1
Log (a2/a1)
n= is the order of reaction
Half-life is determined graphically by plotting " a" vs. "t"
at two different initial concentrations (a1) (a2).
The half-life times are then read 1/2 a1 & 1/2 a2 respectively from the graph.
The values of half-life and the initial concentration are then substituted in the above equation and the order of reaction, n is obtained directly.
Tutorial 1
1- Plot the following data on both semi-log graph paper and
standard rectangle coordinate paper.
Time (min) 10 20 40 60 90 120 130
Drug A (mg) 96 89 73 57 34 10 2.5
a. Does the decrease in the amount of drug A appear to be a Zero order or First-order process?
b. What is the rate constant k and the half life?
c. Does the amount of the drug A extrapolate to zero on the x axis?
d. Write the equation for the line produced on the graph?
[pic]
2- It was found that the concentration of hydrogen peroxide remaining after4>5 minutes expressed as the volumes in ml of gas evolved was 9.6 from the initial concentration of 57.90.Calculate k using equation of first-order. How much hydrogen peroxide remained undecomposed after 25 minutes?
3- A solution of drug contained 500 unites/ml when prepared. It was analyzed after 40 days, and was found to contain 300 unites/ml. Assume that the decomposition follow first order, at what time the drug decomposed to one- half of its original concentration.
3) At 25 °C, the half-life period for the decomposition of N205 is 5.7 hours and is independent of the initial pressure of N205, calculate:
a-The time for the reaction.
b-The specific rate constant for the reaction.
4 - A solution of a drug was freshly prepared at a concentration of 300 mg/ml after 30 days the drug concentration in the solution was 75 mg/ml.
a. Assuming first order kinetics. When will the drug decline to one-half of the original concentration?
b. Assuming Zero-order kinetics. When wilt the drug decline to one-half of the orgna1 concentration?
5- . If the half-life for decomposition of a drug is 12 hr., how long will it takes for 125 mg of the drug to decompose 30%? Assume First- order kinetics and constant temperature.
6.- Two different drugs were administered on different occasions in different doses to the same subject by IV administration and the half-lives of the two drugs were determined
Dose (mg) 40 60 80
t ½ of drug A 10 15 20
t ½ of drug B 3.5 3.5 3.5
a. Which one of the two drugs is eliminated by a first order
process? Explain.
b. If 10 mg of the drug that is eliminated by first order process was administered to the same patient, how much time would be required to eliminate 7 mg of the drug.
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3-Second order reaction:
The rate of these reactions depend on the concentration of
2 reactants each raised to a power equals 1 or a single reactant raised to power 2.
1- if A+ B →→ products
-dA/ dt = -d B/dt = K2 [A]1[B]1
2- if 2A →→ products
-dA/ dt = K2 [A]2
K2 = is the specific rate constant of a second order reaction.
If a, b are the initial concentration of A & B
X = No. of each A or B remaining at time t, the reaction rate is given by:
-dx/ dt = K [a-x] [b-x]
a-x & b-x = remaining concentrations of A & B
if Initial concentration of A and B are equal
a = b
dx/dt = K (a-x)2
By integration between x = dx at time = t
& x = 0 at time = zero
x t
∫0 dx = K2 ∫0 dt
(a-x)2
X = Kt
a (a-x)
S0
Unit of K = 1 mole/ liter-1 x l/sec = liter mole-1 sec -1
T 1//2 =1 /a k
Problems on Second order Reactions
1) The saponification of ethyl acetate was done at 25°c. The initial concentration of both ethyl acetate & NaOH in the mixture were 0.01 M. The change in cone (x) of alkali during 20 min. was a 0.0566 .
mole/liter.
Compute:
a-The rate constant.
b-The half-life of the reaction
Pseudo-first order reaction:
In a second order reaction, if the concentration of one reactant is in such large excess that it virtually remain constant. Then the rate of change of concentration follows first order. "known as pseudo-first order"
Ex: hydrolysis reaction with acid or base catalyst are common example of pseudo-first –order reaction.
Pseudo zero- order reaction:
A compound decomposing in solution exhibits a first-order reaction , but if the compound present in very large concentration (insoluble excess ), then the equilibrium is maintained between the compound in solution & that in solid form.
Reaction takes place in pharmaceutical suspensions are generally of pseudo-zero order type
Aspirin suspension the reaction appears to be zero-order till aspirin is present in solid form and only when all of aspirin gets dissolved the reaction becomes first order.
Aspirin ↔ aspirin
(solid) (in solution)
Factors affecting the rate of a reaction,
Some of these factors are;
1-Temprature.
2-Ionic strength.
3-Polarity of the solvent.
4-Catalysis.
1-Temprature
According to Arrhenius, the rate of a reaction increases 2-3 times by a rise 10 °C in temp. Arrhenius equation:
Logk =log A - Ea x 1
2.303R T
where:
k= Reaction rate constant (sec- 1).
T= Temperature in Kelvin (°C +273)
A= Frequency factor(Collision factor or Arrhenius factor) [sec-1]
Ea = Energy of activation. (Cal./mole).
R = Gas constant. (=1.987 or 2 Cal/mol.°K.
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NB.: A, Ea are evaluated by determining k at several temperatures & plotting l/T against log k.
Also Ea can be obtained from determining k1& K2 at 2 different temperatures T1 &T2 using the following equation:
Log K2 = Ea x (T2- T1)
K1 2.303 R T2 T1
Problems on Arrhenius' Equation
1- the rate constant K for the decomposition of 5- hydroxymethyl furfural at 120 °C is 1.173 hr-1 or 3.258 x 10 -4 sec-1 & K2 at 140 °C is 4.86 hr -1. What is the activation energy (Ea) in K cal/mole & frequency factor A in sec-1 for the breakdown of 5 HMF within this temp range?
2-Prepare an Arrhenius plot of the following data for the degradation of methenamine at various temp.
Calculate the activation energy & Arrhenius factor.
|K (hr-1) |0.0836 |0.111 |0.233 |0.427 |
|Temp (°C) |37.5 |47 |57 |67 |
3- A product at its specified packing system (final form) was put at different elevate temperature. Data obtained was tabulated.
|Time (week) | Conc. at 50 °C |Conc. at 60 °C |Conc. at 70 °C |
|0 |100 |100 |100 |
|7 |97 |96 |94 |
|14 |96 |93 |91 |
|21 |94 |90 |88 |
| | | | |
If degradation is first order, calculate:
- K at 50 °C, 60 °C and 70 °C . (theoretical & graphical).
- K at at 25 °C
- shelf-life
- Ea (theoretical & graphical)
2- Light
Light energy may be absorbed by certain molecules which become sufficiently activated to undergo reaction. Visible & U.V. light cause photo-chemical reaction.
Examples of pharmaceutical compounds which undergo photochemical decomposition include " riboflavin, phenothiazine, nifedipine ".
3- Solvent polarity
The polar solvents tend to accelerate reactions in which the products formed are more polar than the reactants.
On the other hand, if products formed are less polar than reactants, then reaction proceeds better in solvent of relatively low polarity.
Commonly used non-aqueous solvents are, ethanol, glycerol, propylene glycol, polyethylene glycols and vegetable oils.
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Slope= -K
x/(a-x)
t
Slope = K
Log K
1/ T (Kelvin) = C +273
Slope = - Ea / 2.303 R
Intercept = Log A
Equation for calculation of K if the initial concentration of A & B are equal
K = 1 * x
t a (a-x)
K = 2.303 * Log b(a- x)
t(a-b ) a (b -x)
Equation for calculation of KÞ[pic]r˜t˜Î™Ð™[?]ššFšGšHšIšJštšušvš”š•šòååØË¿³® if the initial concentration of A & B are Not equal
Straight line if a= b
b(a-x) /a(b-x)
t
Straight line if a not equal= b
K= slope
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