MODULE 2. nSOFT nAND ASEPTIC nDOSAGE FORMS. PHARMACEUTICAL INCOMPATIBILITIES.
CONTENT MODULE 4. MEDICINAL nFORMS THAT REQUIRE ASEPTIC MANUFACTURING CONDITIONS. nPHARMACEUTICAL INCOMPATIBILITIES.
LESSON 22. ISOTONIC SOLUTIONS.
Isotonic solution
~ Colligativepropertiessimilar nto those of body fluids
~ Solute nare not capable of diffusing through biological membrane
Isoosmotic solution
Isotonic -a nbiological compatibility
isoosmotic-similarity nof chemicaland/or physical composition
Sodium chloride nsolution 0.9% w/v
~A nFreezing point depression (△tf)=0.52℃
Hypotonic solution
~-0.52℃< tf<0 ℃
~Swellin gand bursting of the cells (ex. Hemolysis)
Hypertonic solution
~tf<-0.52℃
~Shrinkageof nthe living cells
Isoosmotic solution but not isotonic solutions
~Swelling nand bursting of the cells
~The nsolute molecules -> The cells
Ex. Boric nacid(1.9%w/v)
~Hemolysisto nred blood
~Isotonic nin the eye
Sodium Chloride Equivalent Values(E)1g drug ?
g sodium chlorideE ≅17(Liso/MW)Ex7. E value? MW=340g/mole Liso=3.4E nvalue=17(3.4/340)=0.17g
Preparation of Isotonic Solutions
Freezing Point Depression Method
Amount of NaClneeded in 100mL of solution = n[0.9*(0.52-tf’)]/0.52
Ex10. 100ml of a drug solution , How much nNaCl=? Fp=-0.18℃
Amount of NaClneeded = [0.9*(0.52-0.18)]/0.52=0.59g
Sodium Chloride Equivalent Method
Step1: E * (amount of drug in solution) = X
Step2: (0.9g/100mL) * (volume of solution in mL) n= Y
Step3: Y–X = amount of NaClneeded
Ex11. 3g/500ml solution of drug(E=0.21), How nmuch NaCl?
To render the solution isotonic.
S1: 0.21 * 3g = 0.63g
S2: (0.9g/100mL) * (500mL) = 4.5g
S3: 4.5 -0.63 =3.87g
White-Vincent nMethod
Wi= weight in grams of the i thsolute in the formulation
Ei= sodium chloride equivalent of the ithsolute ithe formulav = volume of sodium chloride solution (0.9%) that contains 1g of nNaCl(this volume is 111.1 mL)
Sprowls’Method
The Sprowls’value
The volume of an isotonic solution that
can be prepared by the addition of enough
water to 0.3 g drug
“0.3g”represents the amount of drug in a
fluid ounce of a 1% solution
1 fluid ounce ≒0.0294 L
Ex15.
The Sprowls’value of Glycerin = 11.7
The E value of glycerin = ?
11.7 = 0.3 g * E * 111.1
E = 0.35
Ex15.
How much silver nitrate is needed to
prepare 100mL of isotonic solution?
An isotonic silver nitrate solution prepared
by the Sprowls’method => 13mL
0.3 : 13mL = x : 100mL
0.3g * 100mL / 13mL = 2.3g of silver nitrate
ISOTONIC nSOLUTIONS
1. nAseptic conditions.
2. nAlgorithm of preparation injectiosolutions.
3. nClassification of solutions for ninjections.
4. nMethods of calculating the nisotonic concentration:
· nbased on equatioMendeleev-Klapeyron or law Van’t Hoff
· nbased on a law Raul (for ncryoscopy constants)
· nusing isotonic equivalents by nsodium chloride.
Terms to nRemember
· nosmotic pressure
· nisotonic solution
· nhypertonic solution
· nhypotonic solution
Injectiodosage forms –
specific group nof drugs that require
Ø special conditions of preparation,
Ø the strictest adherence to aseptic,
Ø technological discipline,
Ø full responsibility for the preparation,
Ø quality control and design to dispensing drugs.
Injection solutions prepares in aseptic conditions
Aseptic conditions – defined conditions, and complex institutional narrangements required to enable to save the drugs from getting into these nmicroorganisms.
Technology process of preparation solutions for injection consists of nthe following stages:
Ø Preparation of naseptic unit and the organization of work in aseptic conditions.
Ø Preparation of ntable-wear and auxiliary materials.
Ø Preparation of nsolvents and drugs.
Ø Dissolution of ndrugs.
Ø Stabilizatioor isotones solutions
Ø Quality control nof solutions.
Ø Filtering solutions ninto bottles for dispense, checking the absence of mechanical inclusions.
Ø Closing, leak ncheck, preparation for sterilization (marking).
Ø Sterilization.
Ø Quality control nand design of drugs to dispense.
Preparation of naseptic unit and the organization of work in aseptic conditions
Preparation of table-wear and auxiliary materials
Preparation of solvents and drugs
Dissolution of drugs
Stabilization or isotones solutions
Quality control of solutions
Filtering solutions into bottles for dispense
Corking bottles
Sterilization
CLASSIFICATION OF SOLUTIONS FOR INJECTIONS
Isotonic nsolutions are solutions, which have an osmolality, equal to the osmolality of nliquids of organism (blood, plasma, lymph, tear liquid).
Methods of Adjusting nTonicity
Isotonic equivalent (E) of nsodium chloride shows the amount of sodium chloride, which creates conditions nidentical osmotic pressure, osmotic pressure equal to 1.0 g of drug.
The isotonic concentration of solution sodium chloride nis 0,9%.
Solutions of medicinal substances in concentrations, nwhich create osmotic pressure, even such to 0,9% solution of sodium chloride, nalso are isotonic.
THE CALCULATION OF ISOTONIC CONCENTRATIONS OF nSOLUTIONS
|
The name of medicinal substances |
The equivalent of substances by NaCl |
Isotonic concentration, % |
|
Sodium chloride (NaCl) |
– |
0.9 |
|
Sodium nitrite (NaNО2) |
0.66 |
1.3 |
|
Sodium sulphate (Na2SО4) |
0.23 |
3.9 |
|
Glucose (anhydrous) (C6H12O6) |
0.18 |
5.2 |
|
Boric acid (H3BO3) |
0.53 |
1.7 |
n USING THE SODIUM CHLORIDE ISOTONIC nEQUIVALENTS
Rp.: Sol. Glucosi 200 ml isotonicae
Sterilisa!
D.S. nFor intravenous introduction
Equivalent glucose/sodium chloride – 0,18;
depression of temperature of freezing 1% solution – n0,1;
М – 180,0
NaCl Equivalent Method
E = amount of NaCl equivalent in P to 1 g of drug
Ø The sodium chloride equivalent is also known as n“tonicic equivalent”.
Ø The sodium nchloride equivalent of a drug is the amount of sodium chloride that is nequivalent to (i.e., has the same osmotic effect as) 1 gram, or other weight nunit, of the drug.
Example:
Equivalent glucose/sodium chloride – 0,18
1,0 Glucose = 0,18 Sodium nchloride
Ø The sodium chloride equivalent values of many drugs nare listed in tables.
Equivalent glucose/sodium chloride – 0,18
I variant of calculation
0,18 sodium nchloride – 1,0 glucose
0,9 sodium chloride – Х
Х = (0,9 · 1,0) : 0,18 = 5,0
By the prescription glucose for 200 ml:
5,0 – 100 ml
Х – 200 ml Х n= 10,0 Glucose
II variant of calculation
100 ml isoton. sol. – 0,9 sodium chloride
200 ml – Х
X = 1,8 NaCl
0,18 sodium nchloride – 1,0 glucose
1,8 sodium chloride – Х
X = 10,0 Glucose
Sample ncalculation:
Calculate the amount of NaCl required to make the nfollowing ophthalmic solution isotonic.
Atropine nSulfate 2%
nNaCl q.s.
nAqua. pur. q.s. ad. 30 ml
nM.ft. isotonic solution
1. Determine the amount of NaCl to make 30 ml of aisotonic solution
2. Calculate the contribution of atropine sulfate to nthe NaCl equivalent
3. Determine the amount of NaCl to add to make the nsolution isotonic by subtracting (2) from (1)
Rp. Sol. Dicaini 0,3% 100 ml
n Natrii chloridi q. s. ut. f. sol. nisotonica
n D.S.
Е Dicaini /sodium nchloride = 0,18
0,18 sodium nchloride – 1,0 Dicaini
Х sodium of chloride – 0,3 Dicaini
Х = (0,18 · 0,3) : 1,0 = 0,054 NaCl
100 ml isotonic sol. NaCl ̶ n0,9 NaCl
0,9 – 0,054 = 0,846 ≈ 0,85 NaCl
Cryoscopy Method
DTf nblood and tears = – 0.52 ˚C
So for any solution to be isotonic with blood, it must nalso have a depression of 0.52 C.
For a number of drugs the freezing point depression caused nby a 1% solution is given in tables in literature.
Steps:
We find the nfreezing point depression caused by the given amount of the drug
In the nprescription in the given volume of water.
We subtract it nfrom 0.52.
For the nremaining depression in freezing point, we add sufficient sodium chloride, nknowing that 1% sodium chloride has a freezing npoint lowering of 0.58 C.
Calculation after depression of freezing temperature nof solution:
Δt – depression of freezing temperature of solution shows how many degrees nCelsius reduces the freezing temperature of 1% solution compared to the nfreezing point of pure solvent.
Freezing point depression of serum 0,52.
1% – Δt
X – 0,52 o nC
X= (0,52/ Δt) 1%
For Glucose n(Δt 1% solution 0,1 °C):
х=0,52/0,1=5,2%.
For Sodium chloride (Δt 1% solution 0,576 °C):
х=0,52/0,576=0,903%
For Magnesia sulphate (Δt 1% solution 0,08 °C):
х=0,52/0,08=6,5%
Calculation after equation of Van’t Hoff:
Calculation after equation of Mendeleyev- Klapeyron:
When calculating isotonic concentrations of nelectrolytes as the law of Van’t Hoff and by Mendeleev-Clapeyron equation, nshould make corrections, i.e. the value (0.29 x M) must be divided into nisotonic coefficient “and,” which shows how many times an increasing nnumber of particle dissociation (compared with nedysotsiyuyuchoyu matter), and nnumerically equal to:
i = 1 + a (n-1),
where
i- isotonic coefficient;
a – the degree of electrolytic dissociation;
n – number of particles formed from one molecule of nthe substance at dissociation.
|
Substance |
Isotonic coefficient |
|
Acid boric |
1,06 |
|
Glucose |
1,00 |
|
Novocain |
1,57 |
|
Sodium chloride, Sodium iodide, Sodium nitrate, Potassium chloride, Potassium iodide, Potassium nitrate |
1,87 |
|
Zinc sulphate |
1,12 |
Rp.: Sol. Glucosi 200 ml isotonicae
Sterilisa!
D.S. nFor intravenous introduction
m = 0,29 · n180,0 = 52,2 (1 l of solution)
100 ml – 5,22
200 ml – X X = n10,44
Thus, for 200 nml of solution it is necessary to take a 10,44 glucose (waterless).
m = 0,29 · n180,0 / 1,00 = 52,2 (1 l of solution),
100 ml – 5,22
200 ml – X X = n10,44
For preparation of injection solutions and eye drops nuse glucose taking into account it actual humidity.
A calculation is conducted after a formula:
Х = (100 · а) : (100 – в), where
and is an amount of glucose after the sample of nwriting;
in is humidity of glucose.
If humidity of glucose of 10%, then:
Х = (100 · 10,44) : (100 – 10) = 11,6
Some nof the many ways;
1. Freezing npoint depression method
2. NaCl nequivalent method
3. White Vincent nmethod
4. Sprowl method
5. Molecules nconcentration method
6. Graphical nmethod on vapor pressure and freezing point determination
1. FREEZING nPOINT DEPRESSION METHOD
The nplasma and blood freezing point temperature = -0.52°C
The ndissolved substances in plasma or tear depress the solution freezing point nbelow 0.52°C
Any nsolution that freeze at T=-0.52°C is isotonic with blood and tear
weight nof substance that need to be adjusted to make it hypotonic given by the nfollowing
formula;
Isotonic Solutions
• Definitions
1. Isotonic Solution: is a solution having the same osmotic npressure as a body fluid. Ophthalmic, nasal, and parenteral solutions should be nisotonic.
2. Hypotonic solution: is a solution of lower osmotic npressure than that of a body fluid. These solutions lead to swelling and nbursting of RBCs, which in turn leads to hymolysis.
3. Hypertonic solution: is a solution having a higher nosmotic pressure than that of a body fluid. These solutions lead to shrinkage nof the RBCs.
Isotonic solutions Calculations
I. Procedure for nCalculation of Isotonic Solutions Using Sodium Chloride Equivalents:
a. Calculate amount in grams of NaCl represented by the ningredients in prescription. Multiply the amount in grams of each substance by nits NaCl equivalent “either from tables or by calculation”.
b. Calculate amount in grams of NaCl, alone, that would be ncontained in an isotonic solution of the total volume specified iprescription. i.e. 0.9 ? 100 because 0.9% solution of NaCl is nisotonic. X ? total of the prescription.
c. Subtract the amount of NaCl represented by the ingredients ithe prescription (step#1) from the amount of NaCl needed to prepare isotonic nsolution (step#2) i.e. step # 2 – step # 1.
d. If an agent other than sodium chloride such as boric acid, ndextrose, sodium or potassium nitrate is to be used to make a solutioisotonic, divide the amount
of nNaCl (step 3) by NaCl equivalent of the substance used to adjust Isotonicity ni.e. valueENaCl−3 # step from .
Calculatioof Sodium Chloride equivalent of a substance (if it is not tabulated):
Sodium nChloride equivalent = NaCloffactor NaCl ofM.Wt i * substanceofM.Wt nsubstance offactor i
Problems non calculating the dissociation factor (i) of an electrolyte:
1. nZinc Sulfate ZnSO4 is a 2-ion electrolyte, dissociation 40% in a certaiconcentration. Calculate its dissociation factor.
2. nZinc Chloride (ZnCl2) is a 3-ion electrolyte; dissociation 80% in a certaiconcentration, calculate (i).
|
I 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 |
Problem on calculating the E-value of a substance:
Papavarine nHCl (mwt = 376) is a 2-ion electrolyte dissociating 80%. Calculate its E-value n(Sodium Chloride equivalent), where its dissociation factor (i) = 1.8
Problems on isotonic solutions calculations using the NaCl nE-value:
1. How many grams of NaCl should be used in compounding the nfollowing prescription: Pilocarpine nitrate 0.3 g E =0.23 Sod. Chloride Q.S. nPurified water ad 30 ml Make isotonic solution.
2. nOxytetracyline HCl 0.5% E = 0.12 Tetracaine HCl 2% solution 15 ml Sodium nChloride Q.S. Purified water ad 30 ml Make isotonic solution 2% solution of nTetracaine HCl is isotonic. How many mls of 0.9% solution of Sodium Chloride nshould be used?
3. nTetracaine HCl 0.5% E(0.18) Epinephrine bitartarate 1:1000 solution 10 ml Boric nacid Q.S E(0.52) Purified water ad 30 Make isotonic solution The solution of nEpinephrine bitartarate (1:1000) is already isotonic. How many grams of boric nacid should be used in compounding the prescription?
II. nIsotonic Solutions Calculations using Freezing nPoint Data:
The calculations involved in preparing isotonic nsolutions may be made in terms of data relating to the colligative nproperties “freezing point depression, osmotic pressure, elevation of nboiling point” of solutions. A comparison of freezing points for example may be nused for this purpose.
1. nHow many mg each of NaCl and dibucaine HCl are required to prepare 30 ml of a n1% solution of dibucaine HCl isotonic with tears? From tables: 1% NaCl Tf = 0.58 n1% Dibucaine HCl Tf = 0.08 Freezing point of blood and lachrymal fluid is – n0.52 oC. n
2. nHow many grams of NaCl should be used in compounding the following nprescription:
Rx Naphazoline Hcl 1% nNaCl Q.S. Purified water ad 30ml Make isotonic solution. Use the freezing point ndepression method.
Tf blood n= 0.52 Tf 1% nNaCl = 0.58 Tf n1% Naphazoline HCl = 0.16
III. nIsotonic Solutions Calculations using White-Vincent Method:
Computing tonicity involves the addition of water to na specified amount of drug to make isotonic solution, followed by the additioof an isotonic buffered diluting vehicle to bring the solution to the final nvolume.
V n= W * E * 111.1
Where: n
W n= grams of drug.
E n= sodium chloride equivalent (tabulated).
V n= volume in ml of isotonic solution that may be prepared by mixing drug with nwater.
Example: n
Make nthe following solution isotonic with respect to membrane. Provided that you nhave isotonic diluting solution:
Phenacaine nHCl 0.06 g (E 0.2)
Boric nacid 0.3 g (E 0.52)
Sterilized nwater to 100 ml
Iso-osmoticity and Isotonicity
If na semi-permeable membrane (one that is permeable only to solvent molecules) is nused to separate solutions of different solute concentrations, a phenomenoknown as osmosis occurs in which solvent molecules cross the membrane nfrom lower to higher concentration to establish a concentration equilibrium. nThe pressure driving this movement is called osmotic pressure and is ngoverned by the number of “particles” of solute in solution. If the nsolute is a nonelectrolyte, the number of particles is determined solely by the nsolute concentration. If the solute is an electrolyte, the number of particles nwill be governed by both the concentration and degree of dissociation of the nsubstance.
Solutions ncontaining the same concentration of particles and thus exerting equal osmotic npressures are called iso-osmotic. A 0.9% solution of NaCl (Normal nSaline) is iso-osmotic with blood and tears. The term isotonic, meaning nequal tone, is sometimes used interchangeably with the term iso-osmotic. The ndistinction between these terms comes with the realization that red blood cell nmembranes are not perfect semipermeable membranes, but allow passage of some nsolutes, such as alcohol, boric acid, ammonium chloride, glycerin, ascorbic nacid, and lactic acid. Hence, a 2% solution of boric acid while physically nmeasured to be iso-osmotic (containing same number of particles) with blood, nwill not be isotonic (exerting equal pressure or tone) with blood but is nisotonic with tears. Practically speaking, this differentiation is rarely of nsignificance and isotonicity values calculated on the basis of the number of nparticles in solution is usually sufficient.
The nclinical significance of all this is to insure that isotonic or iso-osmotic nsolutions do not damage tissue or produce pain when administered. Solutions nwhich contain fewer particles and exert a lower osmotic pressure than 0.9% nsaline are called hypotonic and those exerting higher osmotic pressures nare referred to as hypertonic. Administration of a hypotonic solutioproduces painful swelling of tissues as water passes from the administratiosite into the tissues or blood cells. Hypertonic solutions produce shrinking of ntissues as water is pulled from the biological cells in an attempt to dilute nthe hypertonic solution. The effects of administering a hypotonic solution are ngenerally more severe than with hypertonic solutions, since ruptured cells canever be repaired. The eye can tolerate a range of tonicities as low as 0.6% nand as high as 1.8% sodium chloride solution.
Several nmethods are used to adjust isotonicity of pharmaceutical solutions. One of the nmost widely used method is the sodium chloride equivalent method. The NaCl nequivalent (E) is the amount of NaCl which has the same osmotic effect (based noumber of particles) as 1 gm of the drug.
sample ncalculation: Calculate the amount nof NaCl required to make the following ophthalmic solution isotonic.
|
Rx Atropine Sulfate 2% |
1. nDetermine the amount of NaCl to make 30 ml of an isotonic solution
n
n
2. nCalculate the contribution of atropine sulfate to the NaCl equivalent
3. nDetermine the amount of NaCl to add to make the solution isotonic by nsubtracting (2) from (1)
Other nsubstances may be used, in addition to or in place of NaCl, to render solutions nisotonic. This is done by taking the process one step further and calculating nthe amount of the substance that is equivalent to the amount of NaCl calculated nin step 3.
For nexample, boric acid is often used to adjust isotonicity in ophthalmic solutions nbecause of its buffering and anti-infective properties. If E for boric acid is n0.50, then the amount of boric acid needed to replace the NaCl in step 3 can be ncalculated:
or
or, nmore simply:
Thus, n0.38 g or 380 mg of boric acid would be required to render the previous nophthalmic solution isotonic.
Isotonic Buffers
The addition of any compound to a nsolution will affect the isotonicity since isotonicity is a property of the nnumber of particles in solution. So the osmotic pressure of a solution will be naffected not only by the drug but also by any buffer compounds that are nincluded in the formulation. But after these compounds have been added, it is nstill possible that the solution will not be isotonic. It may be necessary to nadd additional sodium chloride to bring the solution to isotonicity, but that nwould require doing the calculations as shown above.
An alternative to this approach is to nuse an isotonic buffer. There are two approaches to using isotonic buffers.
Approach 1
In the first approach, the drug is ndissolved in an appropriate volume of water (V-value) to make the solutioisotonic. Then the remaining volume needed in the formulation is supplied by aisotonic buffer
An example:
|
Rx |
The formulation requires 0.3 g of nProcaine HCl. V-value tables can been found in standard references and are ntabulated to tell how many ml of water, when added to 0.3 g of drug, will nresult in an isotonic solution. For Procaine HCl, 7 ml of water added to 0.3 g nof drug will make an isotonic solution. Therefore 0.3 g Procaine HCl is ndissolved in 7 ml water, and then sufficient buffered, isotonic vehicle of nappropriate pH is added to make 15 ml.
The pH of an isotonic Procaine HCl nsolution is 5.6. Therefore, an isotonic buffer of approximately that pH would be nused. One commonly used isotonic buffer is the Sorenson’s Modified Phosphate nBuffer.
|
Sorensen’s Phosphate Vehicle |
|||
|
ml of 0.0667 M |
ml of 0.0667 M |
Resulting pH |
NaCl required for |
|
90 |
10 |
5.9 |
0.52 |
|
80 |
20 |
6.2 |
0.51 |
|
70 |
30 |
6.5 |
9.5. |
|
60 |
40 |
6.6 |
0.49 |
|
50 |
50 |
6.8 |
0.48 |
|
40 |
60 |
7.0 |
0.46 |
|
30 |
70 |
7.2 |
0.45 |
|
20 |
80 |
7.4 |
0.44 |
|
10 |
90 |
7.7 |
0.43 |
|
5 |
95 |
8.0 |
0.42 |
(from USP XXI, p. 1338)
The closest Sorenson’s buffer to pH 5.6 nwould be pH 5.9. So to complete the formulation, 8 ml of pH 5.9 Sorensen’s nbuffer would be added to the 7 ml of Procaine HCl solution. The individual namounts of the Sorenson’s buffer to add can be determined:
Therefore, the compounding procedure nwould be to weigh 0.3 g Procaine HCl and 0.04 g NaCl, add 7 ml of H2O, 7.2 ml nof 0.0667 M NaH2PO4, and 0.8 ml of 0.0667 M Na2HPO4. Filtration sterilize the nsolution and package in a sterile final container.
The limitation to this approach is that nthe final formulation pH may be different than the desired pH. The final pH nwill depend on the two pHs and buffer capacities of the Sorensen’s buffer and nthe aqueous drug solution.
Approach 2
The second method is to use the nSorensen’s buffer as the entire solvent of the formulation. In this situation, nthe sodium chloride equivalent of the active drug is subtracted from the n”NaCl required for isotonicity” listed in the table. This method has nthe advantage that the pH of the final solution will be the pH of the selected nSorensen’s buffer.
An example:
|
Ampicilli Sodium 30 mg/ml |
1. A ratio calculation will show that n0.45 g of Ampicillin Sodium is needed for this formulation.
2. The sodium chloride equivalent for nAmpicillin Sodium is 0.16. Therefore, the drug will contribute osmotic pressure nas if it was 0.072 g of sodium chloride.
3. To have a pH of 6.6, 9.0 ml of nmonobasic sodium phosphate solution and 6.0 ml of dibasic sodium phosphate nsolution are needed. Then to adjust the isotonicity, 0.0735 g of sodium nchloride is needed, but the Ampicillin Sodium equivalent will account for 0.072 ng. So an additional 0.0015 g of sodium chloride must be added.
Therefore, the compounding procedure nwould be to weigh 0.45 g of Ampicillin Sodium and 0.0015 g of sodium chloride. nAdd 9.0 ml of monobasic sodium phosphate solution and 6.0 ml of dibasic sodium nphosphate solution. Filtration sterilize the solution and package in a sterile nfinal container.
· n1. CALCULATION FOR SOLUTION ISOTONIC WITH BLOOD AND nTEARSSome of the many ways; 1. Freezing point depression method 2. NaCl nequivalent method 3. White Vincent method 4. Sprowl method 5. Molecules nconcentration method 6. Graphical method on vapor pressure and freezing point ndetermination 1. FREEZING POINT DEPRESSION METHOD • The plasma and blood nfreezing point temperature = -0.52°C • The dissolved substances in plasma or ntear depress the solution freezing point below 0.52°C • Any solution that nfreeze at T=-0.52°C is isotonic with blood and tear • weight of substance that nneed to be adjusted to make it hypotonic given by the following formula; 0.52 − n= Where, W=the weight, in g, of the added substance in 100ml of the final nsolution a =the depression of the freezing point of water produced by the nmedicament already in the solution (calculated by multiplying the value for b nfor the medicament by the strength of the solution express as % w/v) b=the ndepression of freezing of water produced by 1 per cent w/v of the added nsubstance ~mlk~ 1Freezing Point Depression Method
· n2. Example 1.1 Rx ∆Tf 1% NaCl 900mg 0.576 nDextrose q.s (20.5g) 0.101 Ft. isotonic soulution500ml Or; 0.9% ⁄ = nSolution; 500 100 . . . ( × . ) = 0.18% ⁄ = = . = 4.12 100 For n500ml; 500 = (4.12 ) × 100 = 20.6Example 1.2 Rx Ephedrine HCl 1g nChlorobutol 0.5g NaCl (1.4g) q.s Distilled water q.s ad 200ml Solution; 0.52 − n0.52 − [(0.5 × 0.165) − (0.25 × 0.138)] = = 0.576 = 0.7 n100 200 ; 200 = (0.7 ) × 100 = 1.4 ~mlk~ 2Freezing Point DepressioMethod
· n3. EXercise 1.3Prepare 500ml NaHCO3 (∆T.f n1%=0.38) so that when it is dilute with the same amount of water, it wouldbe nisotonic.Solution; Rx ∆T.f 1% NaHCO3 Xg(13.7) 0.38 Water qs ad 500mlThe nsolution is hypertonic When diluted NaHCO3 Xg Water qs ad 1000mlThis solutiois isotonic 0.52 − 0.52 − 0 = = = 1.37 ℎ ℎ 0.38NaHCO3 required to make the solution isotonic upon dilution is;1.37 nχ =100 10001.37 X 10 = 13.7g (to be dissolved to 500ml solution) ~mlk~ n3Freezing Point Depression Method
· n4. 2. NaCl EQUIVALENT METHOD (E) • Based on the factor ncalled the sodium chloride equivalent which can be used to convert a specified nconcentration of medicament to the concentration of medicament to the nconcentration of sodium chloride that will produce the same osmotic effect. • nStandard → 0.9 % w/v NaCl at isotonic • Method to calculate % at nisotonic, ∆Tf 1% =0.576 ~ NaCl 0.52 − 0.52 − 0 = = = 0.9 0.9% n⁄ 0.576 Known as normal saline • Sodium chloride equivalents (E1%) can be ncalculated from the following formula; ( , . 927) = 0.576 ( ℎ ℎ )Formula & Method 1 •Look up in the table the sodium chloride nequivalent for the srenght of solutioearest to the strenght = 0.9% – ([E1%X n%w/v] + …….) of medicament in the preparation. 2 •Multiply this by the strenght nof the medicament. 3 •substract the result from 0.9 per cent; the difference is nstrenght of sodium chloride necessary to adjust the solution to iso-osmoticity. n~mlk~ 4 aCl Equivalent Method
· n5. Example 2.1 ∆Tf 1% ascorbic acid =0.105°C ∆Tf n1% NaCl = 0.576°CWhat is the E for 1% ascorbic acid?Solution . ° E 1% ascorbic nacid= . ° = 0.18Example 2.2 Rx NaCl 0.2%w/v Dextrose q.s Ft. isotonic solutio500mlSolution; ∆Tf 1% NaCl = 0.576°C ∆Tf 1% dextrose = 0.101 °C . n°For dextrose 1 % (E1%) = . ° = 0.18Conclusion → 1% dextrose has osmotic npressure equals to 0.18% NaClNaCl in the solution = 0.2 %∴Remaining amount of NaCl to be added nfor isotonic 0.9% – 0.2% =0.7%1% dextrose equivalent to 0.18% NaCl Convert namount of NaCl requiredχ% dextrose equivalent to 0.7% NaCl to amount of ndextrose required (0.7%)(1%)χ= = 3.89% 0.18%For 500ml;3.89 χ =100 n500χ = 19.45 g ~mlk~ 5 aCl Equivalent Method
· n6. Example 2.3 Rx E 1% Ephedrine HC 1g 0.3 Chlorobutol n0.5g 0.24 NaCl q.s Distilled water 200ml isotonic Solution; For ephedrine HCl n× 100 % × 0.3 = 0.15% . For chlorobutol × 100 % × 0.24 n= 0.06% Amount of NaCl for isotonic; 0.9% – (0.15% + 0.06%) = 0.69% So for n200ml……………….= 1.38 Example 2.4 Rx E 1% Let adjust isotonicity using KCl instead nEphedrine HCl 1g 0.3 of NaCl Chlorobutol 0.5g 0.24 KCl q.s 0.4 Distilled water n200ml isotonic Solution; 1% KCl equivalent to 0.4% NaCl Convert amount of NaCl nrequired to χ% ← 0.69% NaCl (refer to example 3) amount of KCl nrequired 0.69 χ= = 1.725% 0.4 So for 200ml …..= 3.45g KCl ~mlk~ 6 aCl nEquivalent Method
· n7. Example 2.5 Calculate the percentage of sodium nchloride required to render a 0.5 per cent solution of potassium chloride niso-osmotic with blood plasma. Solution Sodium chloride equivalent of 0.5 per ncent potassium chloride = 0.76 ∴ Percentage nof sodium chloride for adjustment = 0.9 − (0.5 Χ 0.76) = 0.9 − n0.38 = 0.52 Example 2.6 Calculate the percentage of anhydrous dextrose required nto render a 1 per cent solution of ephedrine hydrochloride iso-osmotic with nbody fluid. ∴ Percentage nof sodium chloride for adjustment = 0.9 − (1 Χ 0.3) = 0.6 Equivalent npercentage of anhydrous dextrose = 0.6/0.18 = 3.33 Example 2.7 Select a nsuitable substance for an eye lotion 0.5 per cent of silver nitrate and ncalculate the percentage required to render the lotion iso-osmotic with nlachrymal secretion. Sodium chloride is unsuitable because silver nitrate is nincompatible with chloride. Potassium nitrate will be used Sodium chloride nequivalent of 0.5 per cent silver nitrate = 0.33 = 0.9 − (0.5 Χ n0.33) = 0.9 − 0.165 = 0.753 . Equivalent percentage of potassium nitrate n= = 1.3 . ~mlk~ 7aCl Equivalent Method
· n8. 3. WHITE VINCENT METHOD Principle: A Isotonic solutio+ B isotonic solution = C isotonic solution→This method involves the naddition of water to medicament to obtain an isotonic solution. Thisfollowed by nthe addition of isotonic buffer solution or preservatives isotonic solution to nthe requiredvolume. V = W X E1% X 111.1Where;V= V (ml) of isotonic solutiothat could be obtain in wg of drug in water (the amount of water to addedto nform isotonic solution)W= amount of drug in the formulaE1%=NaCl equivalent of nthe drug111.1= constant that could be find from volume for 1% isotonic NaCl n0.9% NaCl → 1% NaCl 0.9g NaCl → 100 ml 1 g NaCl → 111.1 ml n~mlk~ 8White Vincent & Sprowl Method
· n9. Example 3.1 Rx E 1% Ephedrine HCl 1g 0.3 nChlorobutol 0.5g 0.24 NaCl q.s Distilled water 200ml isotonicSolution; This nsolution is isotonic (A)For ephedrine HCl, V=1 X 0.3 X 111.1 = 33.33ml (final nvolume) So the overallFor chlorobutol, V=0.5 X 0.24 X 111.1 = 13.33ml (final nvolume) This solution is isotonic (B) solution is isotonic200ml – (33.33 + n13.33) ml = 153.34ml∴ Amount of NaCl nrequired to adjust 153.34ml to isotonic; This solution is isotonic (C) 0.9= n× 153.34 = 1.38 100 4. SPROWL METHOD • Using white Vincent method but w nis set to constant, 0.3 V = W X E1% X 111.1 or V = 33.33E1% ~mlk~ 9White nVincent & Sprowl Method
· n10. 5. MILIEQUIVALENT (mEq) • Definition; the gram nequivalent weight of an ion is the ionic weight (the sum of atomic weights of nthe element in an ion) in gram divided by the valence of that ion. • A nmililequivalent is one thousandth part of the gram equivalent weight, the same nfigure expressed in milligram. Table 5.1 makes this clear Gram equivalent nweight Ion Ionic weight ℎ Weight of n1mEq (mg) ( ) Sodium Na+ 23 23 23 Potassium K+ 39.1 39 39.1 Calcium Ca2+ 40 20 n20 Chloride Cl− 35.5 35.5 35.5 Bicarbonate HCO3− 61 61 61 Phosphate nHPO4− 96 48 48 • ∴ Gram nequivalent weight i) For ion; ℎ ℎ = Example 1 Eq sodium → 23/1 n= 23g 1Eq chloride → 35.5/1 =35.5 • The weight of a salt containing 1mEq nof a particular ion is obtained by dividing the molecular weight of the salt by nthe valency of that ion multiply by the number of such ion in the molecules ℎ , , 1 ℎ = × ℎ ℎ ~mlk~ 10Miliequivalent (mEq)
· n11. Table 5.2 makes this clear Weight of salt M. of nValency of Number of containing 1mEq Ion Salt used the salt the ion ion in salt nof the ion Na+ Sodium chloride 58.5 1 1 58.5mg Na+ Sodium phosphate 358† 1 2 n179.0mg Ca2+ Calcium chloride 147† 2 1 73.5mg Cl- Calcium chloride 147 1 2 n73.5mg († N.B. the water of crystallization must not be ignored) ii) For salt; ℎ ℎ ℎ = Example (monovalent) 1 Eq → n58.5/1= 58.5g Example (not monovalent) 1Eq CaCl2.H2O → 147/2 n=73.5Miliequivalent (mEq) → 1/1000 Gram Eq weightExample1mEq sodium →23mg1mEq nchloride → 35.5g1mEq NaCl → 58.5 mg1mEq CaCl2.H2O →73.5 mg n~mlk~ 11Miliequivalent (mEq)
· n12. Conversion Equations I. To convert mEq/litre to nmg/litre; × II. To convert mEq/litre to g/litre; × 1000 III. To nconvert mEq/litre to %w/v; × 10000• The BPC includes table showing the nweight s of salt that contain 1 mEq of specified ion; Mg of salt containing 1 nIon Miliequivalent (mEq) mg Salt mEq of specified ion Na+ 23.0 Sodium chloride n58.5 Sodium bicarbonate 84 K+ 39.1 Potassium chloride 74.5 Ca2+ 20.0 Calcium nchloride 73.5 Mg2+ 12.5 Magnesium sulphate 123 Magnesium chloride 101.5 Cl- n35.5 Sodium chloride 58.5 HCO3- 61.0 Sodium bicarbonate 84 HPO4- 48 Sodium nphosphate 179 ~mlk~ 12Miliequivalent (mEq)
· n13. Example 5.1 Calculate the quantities of salt nrequired for the following electrolyte solution; Sodium ion 20 mEq Potassium nion 30 mEq Magnesium ion 5 mEq Phosphate ion (HPO4-) 10 mEq Chloride ion 45 mEq nWater for injection, to 1 litre From table 5.3:- 1 mEq of both potassium and nchloride ion are contained in 74.5 mg of potassium chloride. 1 mEq of both nsodium and phosphate ion are contained in 179 mg of sodium phosphate. 1 mEq of nboth magnesium and chloride ion are contained in 101.5 mg of magnesium nchloride. 1 mEq of both sodium and chloride ion are contained in 58.5 mg of nsodium chloride. Therefore; 30 mEq of potassium ion is provided by 30 X 74.5 mg nof potassium chloride which will also supply 30 mEq of chloride ion. 10 mEq of nphosphate ion is provided by 10 X 179 mg of sodium phosphate which will also nsupply 10 mEq of sodium ion. 5 mEq of magnesium ion is provided by 5 X 101.5 mg nof magnesium chloride which will also supply 5 mEq of chloride ion. There nremains a deficiency of 10mEq of each sodium and chloride ions which cn be nprovided by 10 X 58.5 mg of sodium chloride. The formula becomes; ~mlk~ n13Miliequivalent (mEq)
· n14. mg g mEq × × K+ Na+ Mg2+ Cl- HPO4- n1000 Potassium chloride 74.5 X 30 2.235 30 30 Sodium phosphate 179.0 X 10 1.790 n10 10 Magnesium chloride 101.5 X 5 0.508 5 5 Sodium chloride 58.5 X 10 0.585 10 n10 Water for injection, to 1 litre to 1 litre 30 20 5 45 10 55 55 The fact that ncation and anion balance confirm that the formula has been worked out correctly nExample 5.2 Express the following formula as percentage w/v. Sodium ion 147 mEq nPotassium -ion 4 mEq Calcium ion 4 mEq Chloride ion 155 mEq Water for nInjections, to 1 liter. From Table 20.6 and the appropriate conversioequation, the required percentages are— Sodium Chloride 5.85 X 147 ÷ 10 n000 = 0.860% W/v Potassium Chloride 74.5 X 4 ÷ 10 000 = 0.030% w/v nCalcium Chloride 73.5 X 4 ÷ 10 000 = 0.029% w/v Water for injections, to n1 liter ~mlk~ 14Miliequivalent (mEq)
· n15. Example 5.3 Prepare 500 ml of an Intravenous nsolution containing 70 mEq of sodium, 2 mEq of potassium, 4 mEq of calcium and n76 mEq of chloride. The number of milligrams of the various chlorides which ncontain I mEq of the required ions is obtained from Table 20.6 and the formula nbecomes— Sodium Chloride 70 X 58.5 ÷ 1000 = 4.0Q5 g Potassium Chloride 2 nX 74.5 ÷ 1000 = 0.149 g Calcium Chloride 4 X 73.5 ÷ 1000 = 0.294 ng Water for Injections, to 500 ml It is not unusual for students to nmiscalculate the amounts for this type of formula due to failing to appreciate nthat mEq is a unit of weight and not an abbreviation for mEq per litre. This nleads to an incorrect halving of the final quantities in the above example. nExample 5.4 Express 0.9 per cent sodium chloride solution to mililequivalent nper litre. 0.9 × 10000 = 154 / 58.5 To Convert Percentage w/v to mEq The nnumber of grams (C) per 100 milliliters is converted to mg/liter by multiplying nby 10 000. This, divided by the weight (W) in mg of salt containing 1mEq, will ngive the number of mEq/liter. × 10000 = / ~mlk~ 15Miliequivalent (mEq)
· n16. Adjustment to Iso-osmoticity with Blood Plasma nBased on Miliequivalent The equation used is:- = 310 − Where; a = number nof mililequivalent per liter of medicament present and b = number of nmililequivalent per liter of adjusting substance required Example 5.5 Calculate nthe amount of sodium chloride required to adjust a solution containing 40 mEq nof each potassium and chloride ion to iso-osmoticity with blood plasma. A nsolution containing 40 mEq of chloride ion provide a total of 80 mEq of anioand cation. ∴ = 310 − n80 = 230 230 mEq will be provided by 115 mEq of sodium ions and 115 mEq nchloride ions. The formula of the solution will be:- Potassium chloride 74.5 mg nX 40 = 2.98 g Sodium chloride 58.5 mg X 115 = 6.73 g Water for injection, to 1 nlitre ~mlk~ 16Miliequivalent (mEq)
· n17. 6. MILIMOLES In the SI, the unit for chemical nquantity is mole and the term equivalent and mililequivalent becomes absolute. nConsequently the method of expressing the composition of body and infusiofluid is changing from the mililequivalent to mole notation. By analogy with natoms and molecules, a mole of an ion is its ionic weight in grams but the nnumber of moles of each of the ions of a salt in solution depends on the number nof each ion in the molecule of the salt. It follows that the quantity of salt, nin mg, containing 1 mmol of a particular ion can be found by dividing the nmolecular weight of the salt by the number of that ion contained in the salt. nFor example— Salt Ion Quantity of salt (In mg) containing 1mmol NaCl Na+ M.Wt n/1 = 58.5 CI- M.Wt / 1 = 58.5 CaCl2 Ca2+ M.Wt / 1 = 147 Cl- M.Wt /2 = 73.5 nNa2HPO4 Na+ M.Wt /2 = 179 HP042- M.Wt /1 = 258 NaH2PO4 Na+ M.Wt /1 = 156 H2P04- nM.Wt /1 = 156 Conversion Equations To convert quantities expressed in mmol per nlitre into weighable amounts. The following formulae may be used:- = × = n× ÷ 1000 ⁄ = × ÷ 10000 Where W is the number of nmg of salt containing 1 mmol of the required ion and M is the number of mmol nper litre ~mlk~ 17Milimole(mmol)
· n18. Example 6.1 Calculate the quantities of salts nrequired for the following electrolyte solution. Sodium 60 mmol Potassium 5 nmmol Magnesium 4 mmol Calcium 4mmol Chloride 81 mmol Water for injection, to 1 nlitre From Table 20.8— 5 mmol of potassium ion is provided by 5 x 745 mg of npotassium chloride which also provides 5 mmol of chloride ion. 4 mmol of nmagnesium ion is provided by 4 x 203 mg of magnesium chloride which also nprovides 2 x 4 mmol = 8 mmol of chloride ion, since there are two chloride ions nin the molecule. 4 mmol of calcium ion are provided by 4 x 147 mg of calcium nchloride which, like magnesium chloride, also provides 8 mmol of chloride ion. n60 mmol of sodium ion is provided by 60 x 58.5 mg of sodium chloride which nprovides a further 60 mmol of chloride. The formula becomes— mmol WXM K+ Mg2+ nCa2+ Na+ Cl- Potassium chloride 5 x 74.5 = 0.373 g 5 5 Magnesium chloride 4 x n203 = 0.812 g 4 8 Calcium chloride 4 x 147 = 0.588 g 4 8 Sodium chloride 60 x n58.5 = 3.510 g 60 60 Water for injection, to 1 litre 73 31Although there nappears to be inequality between the anions and cations, the charges are nequallybalanced. ~mlk~ 18Milimole(mmol)
· n19. Example 6.2 Calculate the number of millimoles of n(a) dextrose and (b) sodium ions in 1 litre of Sodium Chloride and Dextrose nInjection containing 4.3 per cent w/v of anhydrous dextrose and 0.18 per cent nw/v of sodium chloride. Use the conversion equation— ⁄ = × ÷ n10000 ⁄ × 10000 = a) For dextrose Since dextrose is nnon-electrolyte, W = M.Wt Hence, 4.3 × 10000 = = 239 180.2 b) For sodium nchloride 0.18 × 10000 = = 31 58.5 Since 1 mmol sodium chloride provides 1 nmmol of sodium ion and 1 mmol of chloride ion, 1 litre of the solution will ncontain 31 mmol of sodium ion (and 31 mmol of chloride ion ). Example 6.3 nCalculate the number of milimoles of calcium and chloride ion in a litre of a n0.029 per cent solution of calcium chloride. 0,029 × 10000 = =2 147 But, neach mole of calcium chloride provides 1 mol of calcium ions and 2 moles of nchloride ions. Therefore, 1 litre of solution contains 2mmol of calcium ion and n4 mmol of chloride ion. ~mlk~ 19Milimole(mmol)
· n20. ppm calculationEXamplePrepare 90ml NaF solution so nthat 5ml of the solution when diluted with water to 1cupfull(240ml), a3ppm nsolution obtained. Solution; 5ml solution → 240ml final volume 90ml nsolution → ? ml final volume (240 )(90 ) = = 4329 5 3g Na → n1000000ml ? ml ←4320ml ( )( ) = = 0.01296 ∴ 0.01296mg needed ≈ 13mg NaFFormula NaF 13mg Distilled water q.s ad. n90ml ~mlk~ 20
· n21. Calculation involving density factorEXampleGiveacetic acid BPC (33%w/w, d=1.04g/ml).What is the concentration of acetic acid nin %w/wSolution; Acetic acid 33%w/w= 33g acetic acid in 100g solution d= n1.04g/ml ? w/v33g acetic acid →100g acid solution33g acetic acid → n?ml solutionFrom density factor information; 104g acid → 100ml acid nsolution 100g acid → ?ml acid solution ( )( ) = = 96.15 ? gram acetic nacid → 100ml solution 33 = × 100 = 34% ⁄ 96.15 ~mlk~
• 1. nCALCULATION FOR SOLUTION ISOTONIC WITH BLOOD AND TEARSSome of the many ways; 1. nFreezing point depression method 2. NaCl equivalent method 3. White Vincent nmethod 4. Sprowl method 5. Molecules concentration method 6. Graphical method non vapor pressure and freezing point determination 1. FREEZING POINT DEPRESSION nMETHOD • The plasma and blood freezing point temperature = -0.52°C • The ndissolved substances in plasma or tear depress the solution freezing point nbelow 0.52°C • Any solution that freeze at T=-0.52°C is isotonic with blood and ntear • weight of substance that need to be adjusted to make it hypotonic giveby the following formula; 0.52 − = Where, W=the weight, in g, of the nadded substance in 100ml of the final solution a =the depression of the nfreezing point of water produced by the medicament already in the solutio(calculated by multiplying the value for b for the medicament by the strength nof the solution express as % w/v) b=the depression of freezing of water nproduced by 1 per cent w/v of the added substance ~mlk~ 1Freezing Point nDepression Method
• 2. nExample 1.1 Rx ∆Tf 1% NaCl 900mg 0.576 Dextrose q.s (20.5g) 0.101 Ft. nisotonic soulution500ml Or; 0.9% ⁄ = Solution; 500 100 . . . ( × . n) = 0.18% ⁄ = = . = 4.12 100 For 500ml; 500 = (4.12 ) × 100 = n20.6Example 1.2 Rx Ephedrine HCl 1g Chlorobutol 0.5g NaCl (1.4g) q.s Distilled nwater q.s ad 200ml Solution; 0.52 − 0.52 − [(0.5 × 0.165) − n(0.25 × 0.138)] = = 0.576 = 0.7 100 200 ; 200 = (0.7 ) × 100 = 1.4 n~mlk~ 2Freezing Point Depression Method
• 3. nEXercise 1.3Prepare 500ml NaHCO3 (∆T.f 1%=0.38) so that when it is dilute nwith the same amount of water, it wouldbe isotonic.Solution; Rx ∆T.f 1% nNaHCO3 Xg(13.7) 0.38 Water qs ad 500mlThe solution is hypertonic When diluted nNaHCO3 Xg Water qs ad 1000mlThis solution is isotonic 0.52 − 0.52 − n0 = = = 1.37 ℎ nℎ 0.38NaHCO3 nrequired to make the solution isotonic upon dilution is;1.37 χ =100 n10001.37 X 10 = 13.7g (to be dissolved to 500ml solution) ~mlk~ 3Freezing nPoint Depression Method
• 4. 2. NaCl nEQUIVALENT METHOD (E) • Based on the factor called the sodium chloride nequivalent which can be used to convert a specified concentration of medicament nto the concentration of medicament to the concentration of sodium chloride that nwill produce the same osmotic effect. • Standard → 0.9 % w/v NaCl at nisotonic • Method to calculate % at isotonic, ∆Tf 1% =0.576 ~ NaCl 0.52 − n0.52 − 0 = = = 0.9 0.9% ⁄ 0.576 Known as normal saline • Sodium nchloride equivalents (E1%) can be calculated from the following formula; ( , . n927) = 0.576 ( ℎ nℎ )Formula & nMethod 1 •Look up in the table the sodium chloride equivalent for the srenght nof solutioearest to the strenght = 0.9% – ([E1%X %w/v] + …….) of medicament nin the preparation. 2 •Multiply this by the strenght of the medicament. 3 n•substract the result from 0.9 per cent; the difference is strenght of sodium nchloride necessary to adjust the solution to iso-osmoticity. ~mlk~ 4 aCl nEquivalent Method
• 5. nExample 2.1 ∆Tf 1% ascorbic acid =0.105°C ∆Tf 1% NaCl = 0.576°CWhat nis the E for 1% ascorbic acid?Solution . ° E 1% ascorbic acid= . ° = n0.18Example 2.2 Rx NaCl 0.2%w/v Dextrose q.s Ft. isotonic solutio500mlSolution; ∆Tf 1% NaCl = 0.576°C ∆Tf 1% dextrose = 0.101 °C . n°For dextrose 1 % (E1%) = . ° = 0.18Conclusion → 1% dextrose has osmotic npressure equals to 0.18% NaClNaCl in the solution = 0.2 %∴Remaining amount of NaCl to be added for isotonic 0.9% n- 0.2% =0.7%1% dextrose equivalent to 0.18% NaCl Convert amount of NaCl nrequiredχ% dextrose equivalent to 0.7% NaCl to amount of dextrose required n(0.7%)(1%)χ= = 3.89% 0.18%For 500ml;3.89 χ =100 500χ = 19.45 g n~mlk~ 5 aCl Equivalent Method
• 6. nExample 2.3 Rx E 1% Ephedrine HC 1g 0.3 Chlorobutol 0.5g 0.24 NaCl q.s nDistilled water 200ml isotonic Solution; For ephedrine HCl × 100 % n× 0.3 = 0.15% . For chlorobutol × 100 % × 0.24 = 0.06% Amount nof NaCl for isotonic; 0.9% – (0.15% + 0.06%) = 0.69% So for 200ml……………….= 1.38 nExample 2.4 Rx E 1% Let adjust isotonicity using KCl instead Ephedrine HCl 1g 0.3 nof NaCl Chlorobutol 0.5g 0.24 KCl q.s 0.4 Distilled water 200ml isotonic nSolution; 1% KCl equivalent to 0.4% NaCl Convert amount of NaCl required to nχ% ← 0.69% NaCl (refer to example 3) amount of KCl required 0.69 nχ= = 1.725% 0.4 So for 200ml …..= 3.45g KCl ~mlk~ 6 aCl Equivalent nMethod
• 7. nExample 2.5 Calculate the percentage of sodium chloride required to render a n0.5 per cent solution of potassium chloride iso-osmotic with blood plasma. nSolution Sodium chloride equivalent of 0.5 per cent potassium chloride = 0.76 ∴ Percentage of sodium chloride for adjustment = 0.9 − n(0.5 Χ 0.76) = 0.9 − 0.38 = 0.52 Example 2.6 Calculate the npercentage of anhydrous dextrose required to render a 1 per cent solution of nephedrine hydrochloride iso-osmotic with body fluid. ∴ Percentage of sodium chloride for adjustment = 0.9 − n(1 Χ 0.3) = 0.6 Equivalent percentage of anhydrous dextrose = 0.6/0.18 = n3.33 Example 2.7 Select a suitable substance for an eye lotion 0.5 per cent of nsilver nitrate and calculate the percentage required to render the lotioiso-osmotic with lachrymal secretion. Sodium chloride is unsuitable because nsilver nitrate is incompatible with chloride. Potassium nitrate will be used nSodium chloride equivalent of 0.5 per cent silver nitrate = 0.33 = 0.9 − n(0.5 Χ 0.33) = 0.9 − 0.165 = 0.753 . Equivalent percentage of npotassium nitrate = = 1.3 . ~mlk~ 7aCl Equivalent Method
• 8. 3. nWHITE VINCENT METHOD Principle: A Isotonic solution + B isotonic solution = C nisotonic solution→This method involves the addition of water to nmedicament to obtain an isotonic solution. Thisfollowed by the addition of nisotonic buffer solution or preservatives isotonic solution to the nrequiredvolume. V = W X E1% X 111.1Where;V= V (ml) of isotonic solution that ncould be obtain in wg of drug in water (the amount of water to addedto form nisotonic solution)W= amount of drug in the formulaE1%=NaCl equivalent of the ndrug111.1= constant that could be find from volume for 1% isotonic NaCl 0.9% nNaCl → 1% NaCl 0.9g NaCl → 100 ml 1 g NaCl → 111.1 ml ~mlk~ n8White Vincent & Sprowl Method
• 9. nExample 3.1 Rx E 1% Ephedrine HCl 1g 0.3 Chlorobutol 0.5g 0.24 NaCl q.s nDistilled water 200ml isotonicSolution; This solution is isotonic (A)For nephedrine HCl, V=1 X 0.3 X 111.1 = 33.33ml (final volume) So the overallFor nchlorobutol, V=0.5 X 0.24 X 111.1 = 13.33ml (final volume) This solution is nisotonic (B) solution is isotonic200ml – (33.33 + 13.33) ml = 153.34ml∴ Amount of NaCl required to adjust 153.34ml to nisotonic; This solution is isotonic (C) 0.9= × 153.34 = 1.38 100 4. SPROWL nMETHOD • Using white Vincent method but w is set to constant, 0.3 V = W X E1% X n111.1 or V = 33.33E1% ~mlk~ 9White Vincent & Sprowl Method
• 10. 5. nMILIEQUIVALENT (mEq) • Definition; the gram equivalent weight of an ion is the ionic nweight (the sum of atomic weights of the element in an ion) in gram divided by nthe valence of that ion. • A mililequivalent is one thousandth part of the gram nequivalent weight, the same figure expressed in milligram. Table 5.1 makes this nclear Gram equivalent weight Ion Ionic weight ℎ Weight of 1mEq (mg) ( ) Sodium Na+ 23 23 23 Potassium nK+ 39.1 39 39.1 Calcium Ca2+ 40 20 20 Chloride Cl− 35.5 35.5 35.5 nBicarbonate HCO3− 61 61 61 Phosphate HPO4− 96 48 48 • ∴ Gram equivalent weight i) For ion; ℎ ℎ n= Example 1 Eq sodium → 23/1 = 23g 1Eq chloride → 35.5/1 =35.5 • nThe weight of a salt containing 1mEq of a particular ion is obtained by ndividing the molecular weight of the salt by the valency of that ion multiply nby the number of such ion in the molecules ℎ , , 1 ℎ n= × ℎ nℎ ~mlk~ n10Miliequivalent (mEq)
• 11. nTable 5.2 makes this clear Weight of salt M. of Valency of Number of containing n1mEq Ion Salt used the salt the ion ion in salt of the ion Na+ Sodium chloride n58.5 1 1 58.5mg Na+ Sodium phosphate 358† 1 2 179.0mg Ca2+ Calcium chloride n147† 2 1 73.5mg Cl- Calcium chloride 147 1 2 73.5mg († N.B. the water of ncrystallization must not be ignored) ii) For salt; ℎ ℎ nℎ = Example n(monovalent) 1 Eq → 58.5/1= 58.5g Example (not monovalent) 1Eq CaCl2.H2O → n147/2 =73.5Miliequivalent (mEq) → 1/1000 Gram Eq weightExample1mEq sodium n→23mg1mEq chloride → 35.5g1mEq NaCl → 58.5 mg1mEq CaCl2.H2O →73.5 nmg ~mlk~ 11Miliequivalent (mEq)
• 12. nConversion Equations I. To convert mEq/litre to mg/litre; × II. To nconvert mEq/litre to g/litre; × 1000 III. To convert mEq/litre to %w/v; n× 10000• The BPC includes table showing the weight s of salt that contai1 mEq of specified ion; Mg of salt containing 1 Ion Miliequivalent (mEq) mg nSalt mEq of specified ion Na+ 23.0 Sodium chloride 58.5 Sodium bicarbonate 84 nK+ 39.1 Potassium chloride 74.5 Ca2+ 20.0 Calcium chloride 73.5 Mg2+ 12.5 nMagnesium sulphate 123 Magnesium chloride 101.5 Cl- 35.5 Sodium chloride 58.5 nHCO3- 61.0 Sodium bicarbonate 84 HPO4- 48 Sodium phosphate 179 ~mlk~ 12Miliequivalent n(mEq)
• 13. nExample 5.1 Calculate the quantities of salt required for the following nelectrolyte solution; Sodium ion 20 mEq Potassium ion 30 mEq Magnesium ion 5 nmEq Phosphate ion (HPO4-) 10 mEq Chloride ion 45 mEq Water for injection, to 1 nlitre From table 5.3:- 1 mEq of both potassium and chloride ion are contained nin 74.5 mg of potassium chloride. 1 mEq of both sodium and phosphate ion are ncontained in 179 mg of sodium phosphate. 1 mEq of both magnesium and chloride nion are contained in 101.5 mg of magnesium chloride. 1 mEq of both sodium and nchloride ion are contained in 58.5 mg of sodium chloride. Therefore; 30 mEq of npotassium ion is provided by 30 X 74.5 mg of potassium chloride which will also nsupply 30 mEq of chloride ion. 10 mEq of phosphate ion is provided by 10 X 179 nmg of sodium phosphate which will also supply 10 mEq of sodium ion. 5 mEq of nmagnesium ion is provided by 5 X 101.5 mg of magnesium chloride which will also nsupply 5 mEq of chloride ion. There remains a deficiency of 10mEq of each nsodium and chloride ions which cn be provided by 10 X 58.5 mg of sodium nchloride. The formula becomes; ~mlk~ 13Miliequivalent (mEq)
• 14. mg g nmEq × × K+ Na+ Mg2+ Cl- HPO4- 1000 Potassium chloride 74.5 X 30 n2.235 30 30 Sodium phosphate 179.0 X 10 1.790 10 10 Magnesium chloride 101.5 X n5 0.508 5 5 Sodium chloride 58.5 X 10 0.585 10 10 Water for injection, to 1 nlitre to 1 litre 30 20 5 45 10 55 55 The fact that cation and anion balance nconfirm that the formula has been worked out correctly Example 5.2 Express the nfollowing formula as percentage w/v. Sodium ion 147 mEq Potassium -ion 4 mEq nCalcium ion 4 mEq Chloride ion 155 mEq Water for Injections, to 1 liter. From nTable 20.6 and the appropriate conversion equation, the required percentages nare— Sodium Chloride 5.85 X 147 ÷ 10 000 = 0.860% W/v Potassium Chloride n74.5 X 4 ÷ 10 000 = 0.030% w/v Calcium Chloride 73.5 X 4 ÷ 10 000 n= 0.029% w/v Water for injections, to 1 liter ~mlk~ 14Miliequivalent (mEq)
• 15. nExample 5.3 Prepare 500 ml of an Intravenous solution containing 70 mEq of nsodium, 2 mEq of potassium, 4 mEq of calcium and 76 mEq of chloride. The number nof milligrams of the various chlorides which contain I mEq of the required ions nis obtained from Table 20.6 and the formula becomes— Sodium Chloride 70 X 58.5 n÷ 1000 = 4.0Q5 g Potassium Chloride 2 X 74.5 ÷ 1000 = 0.149 g nCalcium Chloride 4 X 73.5 ÷ 1000 = 0.294 g Water for Injections, to 500 nml It is not unusual for students to miscalculate the amounts for this type of nformula due to failing to appreciate that mEq is a unit of weight and not aabbreviation for mEq per litre. This leads to an incorrect halving of the final nquantities in the above example. Example 5.4 Express 0.9 per cent sodium nchloride solution to mililequivalent per litre. 0.9 × 10000 = 154 / 58.5 nTo Convert Percentage w/v to mEq The number of grams (C) per 100 milliliters is nconverted to mg/liter by multiplying by 10 000. This, divided by the weight (W) nin mg of salt containing 1mEq, will give the number of mEq/liter. × 10000 n= / ~mlk~ 15Miliequivalent (mEq)
• 16. nAdjustment to Iso-osmoticity with Blood Plasma Based on Miliequivalent The nequation used is:- = 310 − Where; a = number of mililequivalent per liter nof medicament present and b = number of mililequivalent per liter of adjusting nsubstance required Example 5.5 Calculate the amount of sodium chloride required nto adjust a solution containing 40 mEq of each potassium and chloride ion to niso-osmoticity with blood plasma. A solution containing 40 mEq of chloride ioprovide a total of 80 mEq of anion and cation. ∴ = 310 − 80 = 230 230 mEq will be provided by n115 mEq of sodium ions and 115 mEq chloride ions. The formula of the solutiowill be:- Potassium chloride 74.5 mg X 40 = 2.98 g Sodium chloride 58.5 mg X n115 = 6.73 g Water for injection, to 1 litre ~mlk~ 16Miliequivalent (mEq)
• 17. 6. nMILIMOLES In the SI, the unit for chemical quantity is mole and the term nequivalent and mililequivalent becomes absolute. Consequently the method of nexpressing the composition of body and infusion fluid is changing from the nmililequivalent to mole notation. By analogy with atoms and molecules, a mole nof an ion is its ionic weight in grams but the number of moles of each of the nions of a salt in solution depends on the number of each ion in the molecule of nthe salt. It follows that the quantity of salt, in mg, containing 1 mmol of a nparticular ion can be found by dividing the molecular weight of the salt by the nnumber of that ion contained in the salt. For example— Salt Ion Quantity of salt n(In mg) containing 1mmol NaCl Na+ M.Wt /1 = 58.5 CI- M.Wt / 1 = 58.5 CaCl2 Ca2+ nM.Wt / 1 = 147 Cl- M.Wt /2 = 73.5 Na2HPO4 Na+ M.Wt /2 = 179 HP042- M.Wt /1 = n258 NaH2PO4 Na+ M.Wt /1 = 156 H2P04- M.Wt /1 = 156 Conversion Equations To nconvert quantities expressed in mmol per litre into weighable amounts. The nfollowing formulae may be used:- = × = × ÷ 1000 ⁄ = n× ÷ 10000 Where W is the number of mg of salt containing 1 mmol of nthe required ion and M is the number of mmol per litre ~mlk~ 17Milimole(mmol) n
• 18. nExample 6.1 Calculate the quantities of salts required for the following nelectrolyte solution. Sodium 60 mmol Potassium 5 mmol Magnesium 4 mmol Calcium n4mmol Chloride 81 mmol Water for injection, to 1 litre From Table 20.8— 5 mmol nof potassium ion is provided by 5 x 745 mg of potassium chloride which also nprovides 5 mmol of chloride ion. 4 mmol of magnesium ion is provided by 4 x 203 nmg of magnesium chloride which also provides 2 x 4 mmol = 8 mmol of chloride nion, since there are two chloride ions in the molecule. 4 mmol of calcium ioare provided by 4 x 147 mg of calcium chloride which, like magnesium chloride, nalso provides 8 mmol of chloride ion. 60 mmol of sodium ion is provided by 60 x n58.5 mg of sodium chloride which provides a further 60 mmol of chloride. The nformula becomes— mmol WXM K+ Mg2+ Ca2+ Na+ Cl- Potassium chloride 5 x 74.5 = n0.373 g 5 5 Magnesium chloride 4 x 203 = 0.812 g 4 8 Calcium chloride 4 x 147 = n0.588 g 4 8 Sodium chloride 60 x 58.5 = 3.510 g 60 60 Water for injection, to 1 nlitre 73 31Although there appears to be inequality between the anions and ncations, the charges are equallybalanced. ~mlk~ 18Milimole(mmol)
• 19. nExample 6.2 Calculate the number of millimoles of (a) dextrose and (b) sodium nions in 1 litre of Sodium Chloride and Dextrose Injection containing 4.3 per ncent w/v of anhydrous dextrose and 0.18 per cent w/v of sodium chloride. Use nthe conversion equation— ⁄ = × ÷ 10000 ⁄ × 10000 n= a) For dextrose Since dextrose is non-electrolyte, W = M.Wt Hence, 4.3 n× 10000 = = 239 180.2 b) For sodium chloride 0.18 × 10000 = = 31 n58.5 Since 1 mmol sodium chloride provides 1 mmol of sodium ion and 1 mmol of nchloride ion, 1 litre of the solution will contain 31 mmol of sodium ion (and n31 mmol of chloride ion ). Example 6.3 Calculate the number of milimoles of ncalcium and chloride ion in a litre of a 0.029 per cent solution of calcium nchloride. 0,029 × 10000 = =2 147 But, each mole of calcium chloride nprovides 1 mol of calcium ions and 2 moles of chloride ions. Therefore, 1 litre nof solution contains 2mmol of calcium ion and 4 mmol of chloride ion. ~mlk~ n19Milimole(mmol)
• 20. ppm ncalculationEXamplePrepare 90ml NaF solution so that 5ml of the solution whediluted with water to 1cupfull(240ml), a3ppm solution obtained. Solution; 5ml solutio→ 240ml final volume 90ml solution → ? ml final volume (240 )(90 ) n= = 4329 5 3g Na → 1000000ml ? ml ←4320ml ( )( ) = = 0.01296 ∴ 0.01296mg needed ≈ 13mg NaFFormula NaF 13mg nDistilled water q.s ad. 90ml ~mlk~ 20
• 21. nCalculation involving density factorEXampleGiven acetic acid BPC (33%w/w, nd=1.04g/ml).What is the concentration of acetic acid in %w/wSolution; Acetic nacid 33%w/w= 33g acetic acid in 100g solution d= 1.04g/ml ? w/v33g acetic acid →100g nacid solution33g acetic acid → ?ml solutionFrom density factor ninformation; 104g acid → 100ml acid solution 100g acid → ?ml acid nsolution ( )( ) = = 96.15 ? gram acetic acid → 100ml solution 33 = n× 100 = 34% ⁄ 96.15 ~mlk~
