Calculating theparenteral solution
concentration&
Alcohol dilution
Main rule
The concentration of the constituents of the infusion solutions, interms of both ion-dissociating and non-dissociating substances, mustbe expressed in millimoles per liter (mmol/l).
The mmol/l value is the quotient of the amount of substance (a) in1 liter of solution for infusion and finished in mg and the relativemolecular weight (Mw) of the solution.
๐ (๐๐๐๐/๐)=a/Mw*n
n=charge number of the dissociated ion
The base of the calculation: Rault law
Determination of isotonia of parenteral solutions
The decrease in the freezing point of dilute solutions is independent of the
material quality of the solute, but depends only on the number of solute particles
(ions, molecules).
The values obtained for different materials can be added together (additive
property).
Types of calculations
Based on:
โข nomograms
โข freezing point depression
โข mosm/l concentration
โข NaCl equivalent
Nomogram
A nomogram or alignment chart, is a graph for calculation. It is a two-dimensional diagram which gives a computation of a mathematical function.
A nomogram consists of a set of n scales, one for each variable in an equation. Knowing the values of n-1 variables, the value of the unknown variable can be found, or by fixing the values of some variables, the relationship between the unfixed ones can be studied.
The result is got by laying a straightedge across the known values on the scales and reading the unknown value from where it crosses the scale for that variable. The virtual or drawn line created by the straightedge is called an index line or isopleth.
Nomogram
Basic example
The composition of the infusion to be isotonized:
Calcii chloridum hexahydricum 3.00 gGlucosum anhydricum 15.00 gNatrii chloridum qu.s.Aqua ad injectabilia ad 1000.0 ml
Calcium chloride is an inorganic compound, a salt with the chemical formula CaCl2. It is a white colouredcrystalline solid at room temperature, highly soluble in water.Calcium chloride is commonly encountered as a hydrated solid with generic formula CaCl2(H2O)x,where x = 0, 1, 2, 4, and 6.Glucose (C6H12O6) is a monosaccharide containing 6 carbon atoms and aldehyde groups.Sodium chloride is an ionic compound with the chemical formula NaCl, representing a 1:1 ratio of sodium andchloride ions.
Mw
219.08180.15
58.44
%0.301.50
Calculation based on: nomograms (1)
The nomograms in Volume 4 of the Hungarian Pharmacopoeia are used for the calculation!
The nomograms indicate the amount of NaCL required to isotonic the 100 ml solution in g/100 ml.
The process of calculation:
If there is only one dissolved component in the solutionโข the composition is used to calculate the concentration of the soluteโข determine the required amount of NaCL per 100 ml of solution using the nomogramโข calculate the amount of NaCL required to the solution to be prepared
Calculation based on: nomograms (1)
Calcii chloridum hexahydricum 3.00 gGlucosum anhydricum 15.00 gNatrii chloridum qu.s.Aqua ad injectabilia ad 1000.0 ml
0.3 %
0.79 g
Calculation based on: nomograms (2)
If the solution contains more than one dissolved componentโข the composition is used to calculate the concentration of each componentโข use the nomogram to find the required quantities of NaCL (per 100 ml of solution)โข calculate the amount of NaCL required for isotonize 100 ml of solution
๐๐๐ถ๐ ๐, 100 ๐๐ = ๐ โ [ ๐ โ 1 โ 0.9]where
S = is the sum of the individual NaCL amountdetermined for each component
n = number of the dissolved compunds
c (CaCl2) = 0.30 g / 100 ml -> 0.79 g NaCl / 100 mlc (glucose) = 1.50 g / 100 ml -> 0.64 g NaCl / 100 ml
S = 1.43 g NaCl / 100 ml
g NaCl / 100 ml = 1.43 g โ 1*0.90 g = 0.53 g NaCl / 100 ml5.30 g NaCl / 1000 ml
Excercises
Isotonize the following parenteral solutions using nomograms:
(A)Lidocaine hydrochloride 2.40 gIsoniazide 4.00 gGlucose anhydr. 3.20 gAqua dest. pro inj. ad 400 ml
(B)Codeinum phosphoricum 8.00 gCoffeinum natr. benzoic. 4.00 gEphedrinium chlor. 4.00 gAqua dest. pro inj. ad 1000 ml
(C)Lidocaine hydrochloride 2.40 gIsoniazide 4.00 gGlucose anhydr. 3.20 gAqua dest. pro inj. ad 1000 ml
(D)Codeinum phosphoricum 6.00 gCoffeinum natr. benzoic. 3.00 gEphedrinium chlor. 3.00 gAqua dest. pro inj. ad 500 ml
Excercises
Lidocaine hydrochloride 2.40 g => 0.60 g/100 mlIsoniazide 4.00 g => 1.00 g/100 mlGlucose anhydr. 3.20 g => 0.80 g/100 mlAqua dest. pro inj. ad 400 ml
๐๐๐ถ๐ ๐, 100 ๐๐ = ๐ โ [ ๐ โ 1 โ 0.9]
Lidocaine hydrochloride 0.6% => 0.77 gIsoniazide 1.0% => 0.63 gGlucose anhydr. 0.8% => 0.75 g
S => 2.15 g
NaCl (g/100 ml) = 2.15 โ [(3-1) * 0.9] = 0.35 gNaCl (g/400 ml) = 0.35 * 4 = 1.4 g
Isotonize the following parenteral solution using nomograms:
Excercises
Isotonize the following parenteral solution using nomograms:
Codeinum phosphoricum 8.00 g => 0.8 g/100 mlCoffeinum natr. benzoic. 4.00 g => 0.4 g/100 mlEphedrinium chlor. 4.00 g => 0.4 g/100 mlAqua dest. pro inj. ad 1000 ml
Codeinum phosphoricum 0.8 % => 0.79 gCoffeinum natr. benzoic. 0.4 % => 0.80 gEphedrinium chlor. 0.4 % => 0.79 g
S => 2.38 g
๐๐๐ถ๐ ๐, 100 ๐๐ = ๐ โ [ ๐ โ 1 โ 0.9]
NaCl (g/100 ml) = 2.38 โ [(3-1) * 0.9] = 0.58 gNaCl (g/1000 ml) = 0.58 * 10 = 5.8 g
Calculation based on: freezing point depression
Freezing-point depression is the decrease of the freezing point of a solvent on the
addition of a non-volatile solute.
Freezing point is the temperature at which the substance is transferred from a
liquid to a solid state.
For pure materials, this temperature is a constant, material-specific value: freezing
is an isothermal process.
Calculation based on: freezing point depression
โ๐ = ๐ โ โ๐๐๐ โ 1000
๐๐ค โ ๐บwhereDT = freezing point depressioni = Vanโt Hoff coefficient: i = a (N - 1) +1,
wherea = the dissociation grade of the electrolytesN = the number of ions formed(a = 0 or 100)non dissociative substances i = 1dissociative substances i = the number of ions formed
DTm = molar freezing point depression (in case of water: - 1.86 ยฐC)g = the mass of solute in 1000 ml solutionMw = relative molar massG = mass of the solvent g (simplification: 1000)
Calculation based on: freezing point depression
โ๐ = ๐ โ โ๐๐๐ โ 1000
๐๐ค โ ๐บ
The process of calculation:
โข calculate the quantities of solutes in 1000 ml of solutionโข determine the decrease in freezing point caused by each componentโข adding the decrease in freezing point caused by each componentโข subtracting the amount from the freezing point decrease of the isotonic solution (- 0.52 ยฐC) โข calculate the amount of isotonizing excipient (determine the โgโ in the equation)
Example:
Calcii chloridum hexahydricum 3.00 gGlucosum anhydricum 15.00 gNatrii chloridum qu.s.Aqua ad injectabilia ad 1000.0 ml
Mw
219.08180.15
58.44
Calculation based on: freezing point depression
โ๐ = ๐ โ โ๐๐๐ โ 1000
๐๐ก โ ๐บ
The process of calculation:
DT(CaCl2) = [3 * (-1.86) * 3 / 219.08] = - 0.076 ยฐC
DT(Glucose) = [1 * (-1.86) * 15 / 180.15] = - 0.155 ยฐC
DT(NaCl) = Tiso - S DT = - 0.52 ยฐC - (-0.231 ยฐC) = - 0.289 ยฐC
g = (DT(Nacl) * MNaCl - G) / (DTm * 1000 * i) = [(-0.289) * 58.44 * 1000] / [(-1.86) * 1000 * 2]
g = 4.54 g NaCl / 1000 ml
S DT = - 0.231 ยฐC
Excercises
Determine the amount of glucose required to isotonic the solutions of the following formulations by freezing point calculation.
Natrium hydrogencarbonicum 0.65 g (Mw = 84.01)
DiNatrium edeticum 0.10 g (Mw = 372.24)
Acidum lacticum 4.20 g (Mw = 90.08)
Aqua destillata pro injectione ad 1000 ml
Glucose ??? (Mw = 180.15)
Excercises
Natrium hydrogencarbonicum 0.65 g (Mw = 84.01) (2)DiNatrium edeticum 0.10 g (Mw = 372.24) (3)Acidum lacticum 4.20 g (Mw = 90.08) (2)Aqua destillata pro injectione ad 1000 ml
Glucose ??? (Mw = 180.15) (1)
โ๐ = ๐ โ โ๐๐๐ โ 1000
๐๐ก โ ๐บ
DT(NaHCO3) = [2 * (-1.86) * 0.65 / 84.01] = - 0.02878 ยฐCDT(Na2EDTA)= [3 * (-1.86) * 0.10 / 372.24] = - 0.0015 ยฐCDT(Acid.Lact)= [2 * (-1.86) * 4.2 / 90.08] = - 0.17345 ยฐC
DT = Tiso - S DT = - 0.52 ยฐC - (-0.20373 ยฐC) = - 0.31627 ยฐC
g = (DT * MGlucose - G) / (DTm * 1000 * i) = [(-0.31627) * 180.15 * 1000] / [(-1.86) * 1000 * 1]
g = 30.63 g Glucose / 1000 ml
S DT(NaCl) = - 0.20373 ยฐC
Excercises
Kalium chloratum 0.15 g (Mw = 74.55) (2)Calcium chloratum 0.25 g (Mw = 219.08) (3)Natrium chloratum 3.50 g (Mw = 58.44) (2)Magnesium chloratum 0.50 g (Mw = 203.30) (3)Aqua destillata pro injectione ad 1500 ml
Glucose ??? (Mw = 180.15) (1)
โ๐ = ๐ โ โ๐๐๐ โ 1000
๐๐ก โ ๐บ
DT(KCl) = [2 * (-1.86) * 0.100 / 74.55] = - 0.00499 ยฐCDT(CaCl2) = [3 * (-1.86) * 0.167 / 219.08] = - 0.00425 ยฐCDT(NaCl) = [2 * (-1.86) * 2.333 / 58.44] = - 0.14853 ยฐCDT(MgCl2) = [3 * (-1.86) * 0.333 / 203.30] = - 0.00915 ยฐC
DT = Tiso - S DT = - 0.52 ยฐC - (-0.16691 ยฐC) = - 0.35309 ยฐC
g = (DT * MGlucose - G) / (DTm * 1000 * i) = [(-0.35309) * 180.15 * 1500] / [(-1.86) * 1000 * 1]
g = 51.29 g Glucose / 1500 ml
S DT(NaCl) = - 0.16691 ยฐC
Calculation based on: Mosm/l concentration
๐ ๐๐๐ ๐/๐ =๐ (๐๐ ๐๐๐๐๐๐๐๐๐ก ๐๐ 1000 ๐๐ ๐ ๐๐๐ข๐ก๐๐๐)
๐๐ก* n
wherec = concentration (mosm/l)g = mg mass of the solute in 1000 ml solutionn = number of the ions formedMt = relative molecular mass
The process of calculation:
โข calculate the masses of the solutes in 1000 ml of solutionโข determine the concentration of mosm/l of each componentโข summarize the concentrations in mosm/l of each componentโข subtracting the amount from the concentration in mosm/l of the isotonic solution
(isotonic solution (Ciso) = 301.4 mosm/l)โข calculate the amount of isotonizing excipient (g)
Calculation based on: Mosm/l concentration
๐ ๐๐๐ ๐/๐ =๐ (๐๐ ๐๐๐๐๐๐๐๐๐ก ๐๐ 1000 ๐๐ ๐ ๐๐๐ข๐ก๐๐๐)
๐๐ค* n
c1 = (3000 * 3) / 219.08 = 41.07 mosm/l
c2 = (15000 * 1) / 180.15 = 83.26 mosm/l
Dc = Ciso โ S c = 301.4 mosm/l โ 124.33 mosm/l = 177.07 mosm/l
g = c * Mt / n = 177.07 * 58.44 / 2 = 5173.98 mg NaCl / 1000 ml => 5.17 g NaCl / 1000 ml
S c = 124.33 mosm/l
Calcii chloridum hexahydricum 3.00 gGlucosum anhydricum 15.00 gNatrii chloridum qu.s.Aqua ad injectabilia ad 1000.0 ml
Mw
219.08180.15
58.44
Excercises
Determine the amount of NaCl (Mw = 58.44) required to isotonic the solutions of the following formulations by mosm/l calculation.
Atropine sulfate 0.5 g (Mw = 694.85)Morphine hydrochloride 20.0 g (Mw = 375.84)Acidum chloratum 0.1 n 10.0 g (Mw = 36.47)Aqua dest. pro inj. ad 1000 ml
catrop = (500 * 2) / 694.85 = 1.43916 mosm/l
cmorph.HCl = (20000 * 1) / 375.84 = 53.21413 mosm/l
cHCl 0.1n = (3647 * 1) / 36.47 = 100 mosm/l
Dc = Ciso โ S c = 301.4 mosm/l โ 154.65329 mosm/l = 146.74671 mosm/l
g = Dc * Mw / n = 146.74671 * 58.44 / 2 = 4287.938866 mg NaCl / 1000 ml => 4.287 g NaCl / 1000 ml
S c = 154.65329 mosm/l
Excercises
Determine the amount of sorbitol required to isotonic the solutions of the following formulations by mosm/l calculation.
Calcium gluconate 7.50 g (Mw = 448.39)Kalium chloratum 1.50 g (Mw = 74.55)Calcium chloratum hexahydr. 0.90 g (Mw = 219.08)Aqua dest. pro inj. ad 1500 mlSorbitol ??? (Mw = 182.17)
cCa-gluconate= (5000 * 3) / 448.39 = 22. mosm/l
cKCl = (1000 * 2) / 74.55 = 26.8276 mosm/l
cCaCl2 = (600 * 3) / 36.47 = 5.48 mosm/l
Dc = Ciso โ S c = 301.4 mosm/l โ 54.61 mosm/l = 246.79 mosm/l
g = c * Mt / n = 246.79 * 182.17 / 1 = 44958.26 mg Sorbitol / 1000 ml = 67.44 g Sorbitol / 1500 ml
S c = 54.61 mosm/l
Calculation based on: NaCl equivalentENaCl is the amount of substance (NaCl), which produces an osmotic pressure equal to 1 g of the active
substance in the solution.
Example:โข the 0.9% NaCl and the 5% glucose solutions are isotonicโข 0.9 g of NaCl / 100 ml of solution has the same osmotic effect as 5.0 g of glucoseโข 5.0 g glucose corresponds to 0.9 g NaCl -> 1.0 g glucose corresponds 0.18 g NaClโข the glucose ENaCl is 0.18.
๐ธ =๐๐๐๐ถ๐ โ ๐ฟ
๐ฟ๐๐๐ถ๐ โ ๐= 17 โ
๐ฟ
๐where
L = dependent on chemical structureM = relative molecular mass g = E * c
whereg = NaCl amount results in a decrease in freezing point equivalent to the isotonizing effect of the
appropriate active ingredientc = concentration of the API (g/100 ml)
Calculation based on: NaCl equivalent
The substances are grouped according to their chemical structure.
Material L Example
Not electrolyte 1.9 Glucose, sorbit, mannit, carbamide, isoniazide
Light electrolyte 2.0 Basic alcaloide, boric acid, citric acid
Bi-bivalent electrolyte 2.0 MgSO4
Mono-monovalent electrolyte 3.4 KCl, NaCl, ephedrine-hydrochloride, phenobarbitone Na, morphine-hydrochloride, lidocaine-hydrochloride, procaine-hydrochloride
Mono-bivalent electrolyte 4.3 Na2SO4, atropine-sulfate, disodium-edetate
Bi-monovalent electrolyte 4.8 CaCl2, MgCl2, calcium-gluconate
Mono-trivalent electrolyte 5.2 Sodium-citrate
Tri-monovalent electrolyte 6.0 AlCl3, FeCl3
Mono-tetravalent electrolyte 7.6 Sodium-tetraborate
๐ธ =๐๐๐๐ถ๐ โ ๐ฟ
๐ฟ๐๐๐ถ๐ โ ๐= 17 โ
๐ฟ
๐
Calculation based on: NaCl equivalent
The substances are grouped according to their chemical structure. ๐ธ =๐๐๐๐ถ๐ โ ๐ฟ
๐ฟ๐๐๐ถ๐ โ ๐= 17 โ
๐ฟ
๐The process of calculation:
โข calculate the quantities of solutes in 100 ml of solution (g)
โข calculate the equivalent values of the dissolved components
โข multiply the equivalent value by the concentration of the dissolved components to get the amount of
NaCl corresponding to that component
โข the calculated NaCl amounts are added and subtracted from the normal value (0.9 g NaCl / 100 ml)
โข calculate the amount of NaCl required to isotonize the solution to be prepared.
Calculation based on: NaCl equivalent
The substances are grouped according to their chemical structure. ๐ธ =๐๐๐๐ถ๐ โ ๐ฟ
๐ฟ๐๐๐ถ๐ โ ๐= 17 โ
๐ฟ
๐The process of calculation:
Example:
g1 = 17 * L / M * c1 = 17 * (4.8 / 219.08) * 0.3 = 0.112 g NaCl / 100 ml
g2 = 17 * L / M * c2 = 17 * (1.9 / 180.16) * 1.5 = 0.269 g NaCl / 100 ml
x (g NaCl / 100 ml) = 0.9 โ 0.381 = 0.519 g NaCl / 100 ml ==> 5.19 g NaCl / 1000 ml
S g = 0.381 g NaCl / 100 ml
Calcii chloridum hexahydricum 3.00 gGlucosum anhydricum 15.00 gNatrii chloridum qu.s.Aqua ad injectabilia ad 1000.0 ml
Mw
219.08180.15
58.44
%0.301.50
L4.81.9
Excercises
Determine the amount of NaCl required to isotonic the solutions of the following formulations by NaCl equivalent calculation.
(A)Atropine sulfate 0.5 g (Mw = 694.85)Morphine hydrochloride 20.0 g (Mw = 375.84)Acidum chloratum 0.1 n 10.0 g (Mw = 36.47)Aqua dest. pro inj. ad 1000 ml
(B)Codeinum phosphoricum 8.00 g (Mw = 299.364)Ephedrinium chlor. 4.00 g (Mw = 165.23)Aqua dest. pro inj. ad 1000 ml
(C)Lidocaine hydrochloride 2.40 g (Mw = 234.34)Isoniazide 4.00 g (Mw = 137.139)Glucose anhydr. 3.20 g (Mw = 180.156)Aqua dest. pro inj. ad 1000 ml
(D)Codeinum phosphoricum 6.00 g (Mw = 299.364)Ephedrinium chlor. 3.00 g (Mw = 165.23)Aqua dest. pro inj. ad 500 ml
Excercises
Determine the amount of NaCl required to isotonic the solutions of the following formulations by NaCl equivalent calculation.
(A)Atropine sulfate 0.5 g (Mw = 694.85)Morphine hydrochloride 20.0 g (Mw = 375.84)Acidum chloratum 0.1 n 10.0 g (Mw = 36.47)Aqua dest. pro inj. ad 1000 ml
๐ธ =๐๐๐๐ถ๐ โ ๐ฟ
๐ฟ๐๐๐ถ๐ โ ๐= 17 โ
๐ฟ
๐
g1 = 17 * L / M * c1 = 17 * (4.3 / 694.85) * 0.05 = 0.0003 g NaCl / 100 ml
g2 = 17 * L / M * c2 = 17 * (3.4 / 375.84) * 2.0 = 0.0180 g NaCl / 100 ml
g3 = 17 * L / M * c2 = 17 * (2.0 / 36.47) * 3.647 = 0.20 g NaCl / 100 ml
x (g NaCl / 100 ml) = 0.9 โ 0.2183 = 0.6817 g NaCl / 100 ml ==> 6.817 g NaCl / 1000 ml
S g = 0.2183 g NaCl / 100 ml
Thank you for your cooperation!