calculating the concentration of parenteral solutions

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.

𝑐 (𝑚𝑚𝑜𝑙/𝑙)=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:

• nomograms

• freezing point depression

• mosm/l concentration

• 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 solution• the composition is used to calculate the concentration of the solute• determine the required amount of NaCL per 100 ml of solution using the nomogram• 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 component• the composition is used to calculate the concentration of each component• use the nomogram to find the required quantities of NaCL (per 100 ml of solution)• calculate the amount of NaCL required for isotonize 100 ml of solution

𝑁𝑎𝐶𝑙 𝑔, 100 𝑚𝑙 = 𝑆 − [ 𝑛 − 1 ∗ 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 – 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

𝑁𝑎𝐶𝑙 𝑔, 100 𝑚𝑙 = 𝑆 − [ 𝑛 − 1 ∗ 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 – [(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

𝑁𝑎𝐶𝑙 𝑔, 100 𝑚𝑙 = 𝑆 − [ 𝑛 − 1 ∗ 0.9]

NaCl (g/100 ml) = 2.38 – [(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

∆𝑇 = 𝑖 ∗ ∆𝑇𝑚𝑔 ∗ 1000

𝑀𝑤 ∗ 𝐺whereDT = freezing point depressioni = Van’t 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 °C)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

∆𝑇 = 𝑖 ∗ ∆𝑇𝑚𝑔 ∗ 1000

𝑀𝑤 ∗ 𝐺

The process of calculation:

• calculate the quantities of solutes in 1000 ml of solution• determine the decrease in freezing point caused by each component• adding the decrease in freezing point caused by each component• subtracting the amount from the freezing point decrease of the isotonic solution (- 0.52 °C) • calculate the amount of isotonizing excipient (determine the „g” 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

∆𝑇 = 𝑖 ∗ ∆𝑇𝑚𝑔 ∗ 1000

𝑀𝑡 ∗ 𝐺

The process of calculation:

DT(CaCl2) = [3 * (-1.86) * 3 / 219.08] = - 0.076 °C

DT(Glucose) = [1 * (-1.86) * 15 / 180.15] = - 0.155 °C

DT(NaCl) = Tiso - S DT = - 0.52 °C - (-0.231 °C) = - 0.289 °C

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 °C

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)

∆𝑇 = 𝑖 ∗ ∆𝑇𝑚𝑔 ∗ 1000

𝑀𝑡 ∗ 𝐺

DT(NaHCO3) = [2 * (-1.86) * 0.65 / 84.01] = - 0.02878 °CDT(Na2EDTA)= [3 * (-1.86) * 0.10 / 372.24] = - 0.0015 °CDT(Acid.Lact)= [2 * (-1.86) * 4.2 / 90.08] = - 0.17345 °C

DT = Tiso - S DT = - 0.52 °C - (-0.20373 °C) = - 0.31627 °C

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 °C

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)

∆𝑇 = 𝑖 ∗ ∆𝑇𝑚𝑔 ∗ 1000

𝑀𝑡 ∗ 𝐺

DT(KCl) = [2 * (-1.86) * 0.100 / 74.55] = - 0.00499 °CDT(CaCl2) = [3 * (-1.86) * 0.167 / 219.08] = - 0.00425 °CDT(NaCl) = [2 * (-1.86) * 2.333 / 58.44] = - 0.14853 °CDT(MgCl2) = [3 * (-1.86) * 0.333 / 203.30] = - 0.00915 °C

DT = Tiso - S DT = - 0.52 °C - (-0.16691 °C) = - 0.35309 °C

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 °C

Calculation based on: Mosm/l concentration

𝑐 𝑚𝑜𝑠𝑚/𝑙 =𝑔 (𝑚𝑔 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑖𝑛 1000 𝑚𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)

𝑀𝑡* 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:

• calculate the masses of the solutes in 1000 ml of solution• determine the concentration of mosm/l of each component• summarize the concentrations in mosm/l of each component• subtracting the amount from the concentration in mosm/l of the isotonic solution

(isotonic solution (Ciso) = 301.4 mosm/l)• calculate the amount of isotonizing excipient (g)

Calculation based on: Mosm/l concentration

𝑐 𝑚𝑜𝑠𝑚/𝑙 =𝑔 (𝑚𝑔 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑖𝑛 1000 𝑚𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)

𝑀𝑤* n

c1 = (3000 * 3) / 219.08 = 41.07 mosm/l

c2 = (15000 * 1) / 180.15 = 83.26 mosm/l

Dc = Ciso – S c = 301.4 mosm/l – 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 – S c = 301.4 mosm/l – 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 – S c = 301.4 mosm/l – 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:• the 0.9% NaCl and the 5% glucose solutions are isotonic• 0.9 g of NaCl / 100 ml of solution has the same osmotic effect as 5.0 g of glucose• 5.0 g glucose corresponds to 0.9 g NaCl -> 1.0 g glucose corresponds 0.18 g NaCl• the glucose ENaCl is 0.18.

𝐸 =𝑀𝑁𝑎𝐶𝑙 ∗ 𝐿

𝐿𝑁𝑎𝐶𝑙 ∗ 𝑀= 17 ∗

𝐿

𝑀where

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

𝐸 =𝑀𝑁𝑎𝐶𝑙 ∗ 𝐿

𝐿𝑁𝑎𝐶𝑙 ∗ 𝑀= 17 ∗

𝐿

𝑀

Calculation based on: NaCl equivalent

The substances are grouped according to their chemical structure. 𝐸 =𝑀𝑁𝑎𝐶𝑙 ∗ 𝐿

𝐿𝑁𝑎𝐶𝑙 ∗ 𝑀= 17 ∗

𝐿

𝑀The process of calculation:

• calculate the quantities of solutes in 100 ml of solution (g)

• calculate the equivalent values of the dissolved components

• multiply the equivalent value by the concentration of the dissolved components to get the amount of

NaCl corresponding to that component

• the calculated NaCl amounts are added and subtracted from the normal value (0.9 g NaCl / 100 ml)

• 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. 𝐸 =𝑀𝑁𝑎𝐶𝑙 ∗ 𝐿

𝐿𝑁𝑎𝐶𝑙 ∗ 𝑀= 17 ∗

𝐿

𝑀The 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 – 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

𝐸 =𝑀𝑁𝑎𝐶𝑙 ∗ 𝐿

𝐿𝑁𝑎𝐶𝑙 ∗ 𝑀= 17 ∗

𝐿

𝑀

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 – 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!