Pharmaceutical Calculations Module Tutorial 9: Millimoles

Pharmaceutical Calculations Module

Tutorial 9: Millimoles, Milliequivalents and Milliosmoles

About This Tutorial This tutorial was developed by the Victorian College of Pharmacy, Monash University in conjunction with the Australian Pharmacy Examining Council Incorporated (APEC Inc) (now known as the Faculty of Pharmacy and Pharmaceutical Sciences, Monash University and the Australian Pharmacy Examining Committee respectively).It remains the property of Monash University. The tutorial cannot be used or copied without prior authorisation.

This tutorial is intended as an aid for students enrolled in Monash University’s Bachelor of Pharmacy & Pharmacy Internship programs.

Original tutorial prepared by in1996:

Arthur Pappas Lecturer in Pharmacy Practice Dr Louis Roller Head, Department of Pharmacy Practice

with the assistance of : May Admans Senior Assistant-Lecturer in Pharmaceutics Dr Denis Morgan Reader in Pharmaceutics Dr Kay Stewart Lecturer in Pharmacy Practice Prof Peter Stewart Professor of Pharmaceutics

Original Tutorial Manual produced by Arthur Pappas - December 1996

Adaptation & updating of the material for online delivery to students of Monash University was undertaken by Phil Bergen, Assistant Lecturer in Pharmacy Practice -February 2005

© Copyright 2005 This publication is copyright. Apart from any fair dealing for the purpose of private study, research, criticism or review as permitted under the Copyright Act, no part may be reproduced by any process or placed in computer memory without written permission. Any assembled extracts each attract individual copyright and enquiries about reproduction of this material should be addressed to the appropriate publishers. Enquiries about all other material should be made to the Faculty named above. Last amended by EM- November 2008

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Pharmaceutical Calculations 9. Millimoles, Milliequivalents and Milliosmoles

9. Millimoles, Milliequivalents and Milliosmoles

Topic Index

Page 9.1 Introduction 73 9.2 Objectives 73 9.3 Moles and Millimoles 73 9.4 Milliequivalents 82 9.5 Milliosmoles 85

9.5.1 Osmotic Pressure 86 9.5.2 Osmolarity and Osmolality 88

Milliequivalents is included in this tutorial only for completeness and interest. Students will not be expected to be familiar with milliequivalents and will NOT be examined on them. Students are expected to be familiar with millimoles and milliosmoles.


9.1 Introduction

"Milliequivalent" and "millimole" are chemical units which denote quantities of ions. Prescriptions may state the quantities of ions required and it is up to the pharmacist to select the amount of appropriate salt for delivery. A knowledge of chemical equations and valencies of atoms is vital.

9.2 Objectives The aim of this tutorial is to provide an understanding of the concepts of millimole, milliequivalent and milliosmole.

At the end of this tutorial you should be able to:

a) define millimole, milliequivalent and milliosmole b) determine the number of millimoles, milliequivalents and milliosmoles of a

substance given its molecular formula and quantity or concentration.

9.3 Moles and Millimoles

Definition of one mole (mol)

One mole (or gram molecular weight) of a substance is defined as the molecular weight of that substance, expressed in grams.

The formula for determining the number of moles of a substance is:

n = m

----

M

n = number of moles

m = mass (g) M = molecular weight (also abbreviated MW)

The number of moles of specific ions produced by dissociation of a salt can be readily determined by knowing the molecular formula of the salt and the valence of the ions.

Worked Example 9.1

How many moles (mol) result from the dissociation of Sodium Sulphate (Na2SO4)?

Working:

Write the equation first.

Na2S04 2Na+ + SO4 2-

1 mol 2mol 1mol

Answer is: 2+1=3 mol

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Pharmaceutical Calculations 9. Mi Hi moles, Milliequivalents and Milliosmoles

Quiz Question 9.1

How many moles of Magnesium Chloride (MgCI2) yield 1 mol of Mg2+ ?

Answer:


MgCI2 Mg2+ + 2CI-

1 mol 1mol 2 mol

ie: 1 mol of MgCI2 provides 1 mol of Mg +

(answer)

Worked Example 9.2

How many grams of NaCI are equivalent to 1 mol? MW(NaCI) = 58.5

Working: m

Using: n = — M

m = 1 x58.5 = 58.5 g

Quiz Question 9.2

How many moles are equivalent to 69.1g of Potassium Carbonate (K2C03) ? MW(K2C03) = 138.2

Answer: m

Using: n = — M

69.1 g n = --------- = 0.50 mol (answer)

138.2

Quiz Question 9.3 How many grams of Sodium Carbonate Decahydrate (Na2CO3.10H2O) contain 1000mg of sodium ? MW(Na2CO3.10H2O) = 286.0

Answer:


Na2CO3.10H2O 2Na+ + CO32- + 10H2O

1000mg of Na+ expressed as mol: n = m/M = 1/23 = 0.043mol

From the equation:

1 mole of salt 2 mol of Na+

x 0.043 mol

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Pharmaceutical Calculations 9. Millimoles, Milliequivalents and MilHosmoles

Therefore x = 0.0215mol of salt Using n = m/M m = 0.0215 x 286 = 6.15g (answer)

Worked Example 9.3

A formulation of Sodium Fluoride (NaF) has a strength of 0.50% w/v NaF. If a dropper delivers 0.8mL of solution per 40 drops, how many drops will contain 2mg of fluoride (F-) ? MW (NaF) = 42

AW(F-) = 19

Working: Write the equation first.

NaF Na+ + F- 1 mol of NaF 1 mol F-

Determine the number of mole of NaF in 0.8mL of

0.50% w/v.

0.50g : 100mL xg : 0.8mL

x = 0.004g

Using n = m/M = 0.004g/42 = 9.52 x 10-5 mole

Determine the mole of NaF in one drop.

This is 9.52x10-5

------------- = 2.38 x 10 -6

40

Determine what weight of F" this is equal to:

m = n x M = 2.38 x 10 -6 x 19 = 4.52 x 10-5 g = 0.0452mg (per drop)

Determine the number of drops per 2mg.

2mg ------------ = 44.2 (44 to the nearest drop) 0.0452mg/drop

Definition of one millimole (mmol)

One millimole is the molecular weight expressed in milligrams Note: this is a

measure of quantity, not concentration.

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Pharmaceutical Calculations 9. Miiiimoles, Milliequivalents and Milliosmoles

The following Tables contain valuable milliequivalent and milliosmole calculations.

information when dealing with millimole,

Table 9.1 Atomic Weight and Valence of Some Elements

Element Atomic Weight (AW) Valence

Calcium 40.0 +2

Carbon 12.0 n/a Chlorine 35.5 -1 Fluorine 19.0 -1 Hydrogen 1.0 +1 Magnesium 24.3 +2 Oxygen 16.0 -2 Potassium 39.1 +1 Sodium 23.0 +1 Sulphur 32.1 0

Table 9.2 Miiiimoles of commonly used salts.

Ion Salt mg mg of salt containing

per mmol mmol of ion

Na+ 23

Sodium acid phosphate 156 (NaH2PO4.2H2O) Sodium bicarbonate 84 Sodium chloride 58.5 Sodium citrate 98 (C6H5Na3O7.2H2O) Sodium hydroxide 40 Sodium lactate 112 Sodium phosphate 179 (Na2HPO4.12H2O)

K+ 39.1

Potassium acid phosphate 136 (KH2PCM) Potassium bicarbonate 100 Potassium chloride 74.6 Potassium phosphate 87.1 (K2HPO4)

Ca2+ 40.0

Calcium chloride 147 (CaCI2.2H2O) Calcium gluconate 448

Mg2+ 24.3

Magnesium chloride 203.3 (MgCI2.6H2O) Magnesium sulphate 246 (MgSO4.7H2O)

NH4* 18.0

Ammonium chloride 53.5

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Table 9.2 Millimoles of commonly used salts, (cont)

Ion Salt mg mg of salt containing

per mmol mmol of ion

Cl- 35.5

Ammonium chloride 53.5 Calcium chloride 73.5 (CaCI2.2H2O) Magnesium chloride 101.7 (MgCI2.6H2O) Potassium chloride 74.6 Sodium chloride 58.5

HCO3- 61.0

Potassium bicarbonate 100 Sodium bicarbonate 84 Sodium lactate* 112

HPO42- 96.0

Potassium phosphate 174.2 (K2HPO4) Sodium phosphate 358 (Na2HPO4.12H2O)

H2PO4- 97.0

Potassium acid phosphate 136 (KH2PO4) Sodium acid phosphate 156 (NaH2PO4.2H2O)

Lactate 89.1

Sodium iactate 112

* Sodium lactate is sometimes required in terms of bicarbonate because lactate is metabolised to bicarbonate.

(Ref APF15 page 448)

Worked Example 9.4

How many mg of NaCI represent 2 mmol ? MW(NaCI) = 58.5

Working: Equation:

NaCI Na+ + Cl-

1 mmol NaCI = 58.5 mg 2 mmol NaCI: x

x= 117mg

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Pharmaceutical Calculations 9. Millimoles, Milliequivalents and Milliosmotes

Quiz Question 9.4

An intravenous solution requires 40 mmol of KCI. What weight of KCI is required ? MW(KCI) = 74.6

Answer:

By definition:

1 mmol KCI = 74.6mg

40 mmol KCI = x mg

Therefore x = 2984mg = 2.984g

Quiz Question 9.5

How many mmol are there in 4.5g of Magnesium Sulphate (MgSO4.7H2O) ? MW(MgSO4.7H2O) = 246.4

Answer:

By definition:

1 mmol of MgSO4.7H2O = 246.4mg

x mmol of MgS04.7H2O : 4,500mg

Therefore x = 18.26mmol

Worked Example 9.5

How many grams of NaCI is required to prepare 100mL of a solution of Sodium Chloride containing 1mmol Na+ per 5ml_ ? MW(NaCI) = 58.5

Working:

Equation:

NaCI Na + Cl- 1mmol 1mmol 1 mmol

By definition 1 mmol NaCI = 58.5mg NaCI

Firstly, determine the number of mmol Na+ in 100mL of solution (1mmol/5mL) 100mL x 1 mmol/5mL = 20mmol Na+

This is equivalent to 20mmol NaCI

Now: 1 mmol NaCI : 58.5mg NaCI 20mmol NaCI : ymg NaCI

Therefore y = 1,170mg = 1.17g NaCI

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Quiz Question 9.6

How many grams of anhydrous Sodium Sulphate (Na2SO4) are required to prepare 100mL of a solution of Sodium Sulphate containing 1mmol Na+ per 5mL? MW(Na2SO4) = 142.0

Answer: Equation:

Na2SO4 2Na+ + SO42-

1 mmol 2 mmol 1 mmol

By definition:

1mmol Na2SO4 = 142.0mg

Firstly, determine the number of mmol Na* in 100mL of solution (1mmol/5ml_) 100mL x 1 mmol/5mL - 20mmol Na+ This is equivalent to 10mmol Na2SO4

Now: 1 mmol Na2SO4 = 142.0mg Na2SO4

10mmol Na2SO4 = ymg Na2SO4

Therefore y = 1,420mg = 1.42g Na2SO4

Quiz Question 9.7

You are required to make 100m of a CaF2 solution which contains 9 ppm of F-. How much CaF2 do you need ? MW(CaF2) = 78.0 AW(F") = 19.0

Answer: Equation:

CaF2 Ca2+ + 2F-

By definition: 1 ppm = 1 g per 106 mL

Therefore 9 ppm = 9g (F-) per 106 mL To

determine how much F- is in 100mL: 9g in 106mL xg in 100 mL

Therefore x = 9 x 10-4 g F-

To determine the amount of CaF2 required: From the above equation:

number of mole CaF2 number of mole of F- 1 : 2 x : 9x10-4g/19

Therefore x = 2.36 x 10 -5 mole CaF2 mass of CaF2 = 2.36 x 10-5 x 78 = 1.85 x 10 -3 g (answer)

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Pharmaceutical Calculations 9, Millimoles, Milliequivalents and Milliosmoles

Worked Example 9.6

How much Potassium Chloride (KCI, MW = 74.6) would be required to prepare 1 litre of an intravenous solution to be infused into a patient continuously over a 12 hour period at the rate of 0.05 millimoles K+ per minute?

Working:

Firstly, determine the total number of mmol K+ required over 12 hours:

12 hours = 12 x 60 = 720 minutes

0.05 mmol ------------ x 720 minutes = 36mmol

minute

Equation: KCI K+ + Cl-

1 mmol KCI 1 mmol K+ = 74.6mg KCI 36 mmol K+ (just cross multiply) = (36 x 74.6)/1 = 2685.6mg KCI (answer)

Worked Example 9.7

You have 60mL of 4.0% w/v CaCI2.2H2O solution. How many millimoles of CI- are present ? MW(CaCI2.2H2O) = 147.0 AW(CI-) - 35.5

Working: Equation:

CaCI2.2H2O Ca2+ + 2CI- + 2H2O Firstly determine how

much CaCI2.2H2O is present in 60mL

4g : 100mL xg : 60mL Therefore x = 2.4g CaCI2.2H2O

1 mmol of CaCI2.2H2O = 147.0mg y mmol of CaCI2.2H2O = 2,400mg Therefore y = 16.33 mmol CaCI2.2H2O

This quantity of CaCI2.2H2O will yield twice the mmol of CI- ie:

16.33 x 2 = 32.66 mmol CI- (answer)

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Quiz Question 9.8

How many millimoles of sodium ions are in 10 mL of the following mixture ?

Sodium Citrate 1 g (C6H5NaaO72H2O; MW = 294.1) Sodium Bicarbonate 750 mg (NaHCO3; MW = 84.01) Orange Syrup 1 mL Purified Water to 10 mL

Answer:

Equations:

1. C6H5Na3O7.2H2O -> 3 Na+ + C6H5O73- + 2H2O

1 mmol of C6H5Na3O7.2H2O 3 mmol Na+ = 294.1 mg C6H5Na3O7.2H2O

To determine number of mmol of Na+ in 1g (1000mg), cross multiply:

= 10.2 mmol Na+

2. NaHCO3 Na+ + HCO3"

1 mmol NaHCO3 1 mmol Na+ = 84.01 mg NaHCO3

To determine the number of mmol of Na+ in 750mg, cross multiply: = 8.9 mmol Na+

Total number of mmol Na+ = 10.2 + 8.9 = 19.1

Quiz Question 9.9

What volume of a sterile Potassium Chloride Injection containing 2.45 g KCI/5 mL would need to be added to 1 litre of Synthamin® 10% with Electrolytes Solution to give a total of 75 millimoles of potassium? Synthamin® 10% with Electrolytes contains 60 mmol K+/litre. MW(KCI) = 74.6

Answer:

Since Synthamin® 10% with Electrolytes contains 60 mmol K+/litre.

we need to add a further 15 mmol K+ from our KCI injection.

Equation:

KCI K++ CI-

1mmol KCI = 1 mmol K+ 74.6 mg KCI

To determine the amount of KCI needed for 15 mmol K+, cross multiply:

= 1119 mg KCI

To determine the volume of KCI injection (2.45g/5mL) that contains 1119mg KCI.

2450 mg KCI : 5mL 1119mgKCI : xmL

Therefore x = 2.28 mL 81


9.4 Milliequivalents

Milliequivalents is included in this tutorial only for completeness and interest. Students will not be expected to be familiar with milliequivalents and will NOT be examined on them. Students are expected to be familiar with millimoles and milliosmoles.

Another unit that is occasionally used in measuring electrolyte quantities is the equivalent. When a salt compound dissociates, the same number of 'equivalents' of positive and of negative ions are produced. Similarly when a new compound is formed, one equivalent of a positively charged ion combines with one equivalent of a negatively charged ion.

The number of equivalents of an ion is determined by multiplying the number of moles by the absolute value of the valence.

More often used is the milliequivalent.

The number of milliequivalents (mEq) of an ion is determined by multiplying the number of millimoles by the absolute value of the valence.

Note: the absolute value of 2+ or 2- is two.

Worked Example 9.8 How many equivalents of CI- are produced by the dissociation of 1 mol of NaCI ?

Working: Equation:

NaCI Na+ + Cl-

Thus 1 mole of Cl- is produced, equivalent = moles x valence

= 1x1 = 1 equivalent of Cl-

Worked Example 9.9

You have a vial of 1.0 g of potassium fluoride.

How many mEq of F- are present ? MW(KF) = 58.1

Working:

Equation: KF K+ + F -

1 mmol KF 1 mmol F - 58.1 mg

xmmol F- : 1000mg KF

Therefore x = 17.21 mmol

milliequivalent - millimoles x valence = 17.21 x1

= 17.21 mEq F-

Quiz Question 9.10

A formula for an intravenous solution calls for 25 mEq of Cl-.

Given that your source of Cl- will be Calcium Chloride (anhydrous), what quantity is required ? MW(CaCI2.2H2O) = 147.0mg

Answer: Equation:

CaCI2.2H2O Ca2+ + 2CI- + 2H2O82

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1/2 mmol CaCI2.2H2O 1 mmol Cl- = 1/2 x 147.0mg = 73.5.5mg CaCI2.2H2O

1 mEq CI- = mmol x valence = 73.5mg x 1 = 73.5mg CaCI2.2H2O

Now for 25mEq CI-:

73.5mg x 25 = 1837.5mg CaCI2.2H2O (answer)

Worked Example 9.10

You are required to prepare 200mL of a solution of a Calcium Chloride solution with a concentration of 5mEq of Ca + per 5mL How much Calcium Chloride is required ? MW(CaCI2.2H2O) = 147.0 AW(Ca2+) = 40.0

Working: Equation:

CaCI2.2H2O Ca2+ + 2CI- + 2H2O Firstly, determine how

many mEq of Ca2+ are needed in 200mL.

5mEq -------- x 200mL = 200mEq Ca2+ 5mL

Secondly, determine the mass of calcium chloride required:

1 mmol CaCI2.2H2O = 1mmol Ca2+ = 2mEq Ca2+= 147.0mg

Therefore to calculate the mass of 200mEq of Ca2-:

2mEq Ca2+: 147mg CaCI2.2H2O

200mEq Ca2+: x mg CaCI2.2H2O

Therefore x = 14.7g CaCI2.2H2O is needed.

Worked Example 9.11

How much NaCI is needed to prepare one litre of the following IV solution ? MW(NaCI) = 58.45mg

Na+ 154mEq/L

Cl- 154mEq/L

Working:

Equation: NaCI Na+ + Cl-

1mmol NaCI 1mEq Na+ = 58.45mg NaCI To determine what mass of NaCI contains 154 mEq Na+, cross multiply:

= 9,0001.3mg NaCI

Therefore, 9.0 g of NaCI dissolved in a litre of water will produce the desired formulation.

Quiz Question 9.11

How many grams of potassium chloride will contain the same quantity of potassium as 7.5 grams of Potassium Carbonate ? MW(KCI) = 74.6 MW(K2CO3) = 138.2 AW(K+) = 39.1

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Pharmaceutical Calculations 9. Mitlimoles, Milliequivalents andMilliosmoles

Answer: Equations:

1. KCI K+ + CI- 2. K2CO3 2K+ + CO3

2-

Determine the number of moles of K2CO3: n = m/M = 7.5/138.2 = 0.0543mole

Determine the number of moles of K+: this will be twice the number of moles of K2CO3 = 0.1085

Determine the mass of KCI that yields 0,1085g of K+:

m= n x M = 0.1085 x 74.6 = 8.09g KCI

Quiz Question 9.12

An oral mixture formulation requires 40mEq of Mg 2+ What volume of stock solution of 10% MgCI2 (anhydrous) is required ? MW (MgCI2) = 95.3

Answer: Equation:

MgCI2 Mg2+ + 2CI-

1 mmol MgCI2 1 mmol Mg2+ = 2mEq Mg2+ = 95.3mg MgCI2

Determine the amount of MgCI2 required for 40mEq of Mg2+ :

2mEq Mg2+ : 95.3 mg MgCI2

40mEq Mg2+ : x mg MgCI2

x= 1906mg MgCI2

Determine the volume of 10% w/v MgCI2 solution that contains this:

10gMgCI2 : 100mL

1.906g MgCI2 : y mL

y = 19.1 mL (answer)

Quiz Question 9.13 You are required to prepare 500 mL of Potassium Sulphate solution with a concentration of 2.5 mEq of potassium ion per millilitre. What quantity of potassium sulphate is required ? MW(K2SO4) = 174.3

Answer: Equation:

K2SO4 2K+ + SO4 2-

Determine the total quantity of mEq of K+ needed in 500mL: 2.5mEq/1mL x 500mL = 1250mEq of K+

From the equation: 1 mmol K2SO4 2mEq K+ = 174.3mg K2SO4 Determine the amount of K2SO4 needed for 1250mEq of K+

2mEqK+ : 174.3 mg K2SO4 1250mEqK+ : y mg K2SO4 84


Therefore y = 108.9g of K2SO4 (answer)

Worked Example 9.12

Calculate the quantities of salts required to produce 1 litre of an intravenous injection containing the following ionic composition:

Na+ 67 mEq MW (NaCI) = 58.5

K+ 6 mEq MW (KCI) = 74.6

Ca2+ 4 mEq MW (CaCI2.2H2O) = 147

CI- 77 mEq

Working: milliequivalent = millimoles x valence

Equations:

1. NaCI Na + + CI'

1mmol NaCI 1 mmol Na+

= 1 mEq Na+ = 58.5mg NaCI

To determine, the amount of NaCI required for 67mEq Na+, cross multiply:

= 3,920mg (Note: this will also provide 67mEq of CI-)

2. KCI K++ CI-

1 mmol KCI 1 mmol K+ = 1 mEq K+ = 74.6mg KCI To determine, the amount of KCI

required for 6mEq K+ cross multiply: = 448mg (Note: this will also provide 6mEq of CX)

3. CaCI2.2H2O Ca2+ + 2CI- + 2H2O

1 mmol CaCI2.2H2O 1 mmol Ca2+

= 2mEq Ca2+ = 147mg CaCI2.2H2O

To determine, the amount of CaCI2.2H2O required for4mEq Ca2+, cross multiply: = 294mg (Note: this will also provide 4mEq of CI-)

Total mEq of CI- = 67 + 6 + 4 = 77

9.5 Milliosmoles

When a solution of a drug comes into contact with a body fluid (eg: blood, or tears) there is potential for damage to cells if the solution is not isotonic. To achieve isotonicity, there needs to be balanced osmotic pressure. Differences in osmotic pressure provide the force for osmosis which is the movement of water from regions of low to high solute concentration. In various body fluids, the osmotic pressure is generated by the presence of electrolytes, proteins and other normal constituents.

As an example, if an IV solution (of low osmotic pressure) is injected into the blood, then there will be a net flow of water into red blood cells (RBC) to decrease the concentration difference. This will have the effect of making the RBC swell or burst.

On the other hand, if the osmotic pressure of the IV solution is too high, the RBC will lose water and shrink.

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This movement of water across cell membranes will be minimised greatly if the drug in solution has the same osmotic pressure as that of the bodily fluid. These solutions are termed isosmotic with that particular fluid.

Balancing osmotic pressure is a major factor in achieving isotonicity. However there are some drugs, because of their nature, which can damage cells even though they have the correct osmotic pressure. We can calculate the theoretical contribution made by solutes (drugs) to the osmotic pressure of a solution.

9.5.1 Osmotic Pressure

Osmotic pressure depends on the number of particles (molecules or ions) dissolved in a unit volume of solvent.

a) Non-electrolytes When an ideally behaving non-electrolyte is dissolved in water, each molecule produces one particle in solution.

This holds true for real drugs provided that they do not dimerise (two molecules linked together) or polymerise (many molecules linked together). Note: The water of hydration present within a crystal becomes part of the solvent when a drug is dissolved in water. This does not affect the number of osmoles (particles) in solution.

Definition of osmole. One osmole (Osm) is defined as the weight (g) of a solute osmotically equivalent to one gram-molecular weight (1 mol) of an ideally behaving non-electrolyte. A milliosmole (mOsm) is 1/1000 of an osmole.

For an ideal non-electrolyte:

1 mol = 1 osmole

1 mmol = 1 mOsm

1 mEq = 1 mOsm

Worked Example 9.13

How many mOsm in 5g of Sucrose ? MW (Sucrose) = 342.

Working: Sucrose is a non-electrolyte.

1mmol = 342mg 1 mmol = 1mOSm = 342 mg Therefore for 5,000mg of Sucrose, the number of mOsm = 1mOsm/342mg x 5,000mg = 14.62mOsm.

Worked Example 9.14 You are required to make one litre of a solution containing 200mOsm of Dextrose, How much Dextrose do you need ? MW (Dextrose) = 180

Working: Dextrose is a non-electrolyte.

1mmol = 180mg 1 mmol = 1m0sm = 180mg

Therefore, for 200mOsm, the mass of Dextrose needed is: 180mg/1 mOsm x 200 mOsm - 36,000mg = 36g of Dextrose.

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b) Electrolytes

It is possible to calculate the theoretical number of mOsm of electrolytes by taking into account the number of ions formed. Assuming complete dissociation of the electrolyte, the number of osmoles is the sum of the number of moles of each ion.

Worked Example 9.15 Calculate the number of mOsm present in 1g of Sodium Chloride. MW (NaCI) = 58.5

Working: Equation:

NaCI Na+ +CI-

1 mol of NaCI yields 2 mol of ions.

Each mole of ions contributes one osmol.

Therefore: 1 mol NaCI 2 Osm or 1 mmol NaCI = 2 mOsm 58.5 mg = 2 mOsm

To calculate the number of mOsm in 1g NaCI: 58.5 mg = 2 mOsm

1000mg = y mOsm y = 34.2 mOsm

NaCI (answer)

Quiz Question 9.14

You are required to produce a solution of Calcium Chloride of 200mOsm strength. What mass of Calcium Chloride do you need ? MW (CaCI2.2H2O) = 147

Answer: Equation:

CaCI2.2H2O Ca2+ + 2 Cl- + 2H2O

147mg of CaCI2.2H2O = 3mmol of ions = 3 mOsm (Note: the water molecules form part of the solvent and add no further particles)

To determine the mass of CaCI2.2H2O for 200 mOsm:

147 mg of CaCI2.2H2O : 3 mOsm y mg : 200 mOsm

Therefore y = 9800mg = 9.8g (answer)

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9.5.2 Osmolarity and Osmolality

Osmotic pressure is determined by the concentration of osmoles in solution rather than the absolute number, ie: the number of ions or molecules that are dissolved. Osmotic pressure is an example of a colligative property (like freezing point depression and boiling point elevation) since it depends on the number of particles in solution.

Two terms are used for osmole concentration:

1. Osmolarity

1 osmolar solution = 1 Osm/litre of solution

Eg: If 200mOsm are dissolved in a litre , the concentration is 200 milliosmolar.

This can be calculated from measured osmolality (see Note below) or by summing the particle concentration of each constituent.

2. Osmolality

1 osmolal solution = 1 Osmol/kg of water

(for practical purposes: density of water = 1g/mL) The osmolality of a solution can be measured using an osmometer which either measures freezing-point depression or vapour pressure. The instrument is calibrated for a direct reading of osmolality.

Note: For dilute solutions, the difference between osmolar and osmolal concentrations is very small. This is because the solute occupies so little space in the solution that the volume of solvent is virtually the same as the total volume of the solution.

Eg: the osmolality and osmolarity of 0.9% NaCI differ by less than 1%. When dealing with concentrated solutions osmolarity and osmolality are not similar because

the mass of solute particles becomes significant.

Worked Example 9.16

What is the osmolarity of 30 mOsm dissolved in 100mL of water ?

Working:

1 osmolal = 1 osmol per 1000mL of solution.

We have 30mOsm/100mL or300mOsm/1000mL

ie: Concentration is: 300 milliosmolar

Worked Example 9.17

What is the osmolarity of a solution containing 0.35 mOsm/mL of solution ?

Working:

1 osmolal = 1 osmol per 1000mL of solution

We have 0.35mOsm/mL or 350mOsm/1000mL

ie: Concentration is: 350 milliosmolar 88

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Pharmaceutical Calculations 9. Millimoles, Milliequivalents and Milliosmotes

Quiz Question 9.15

What is the osmolarity of the following solution?

45 g of Hydrous Dextrose in 500mL water.

MW (Hydrous Dextrose) = 198

Answer: Dextrose is a non-electrolyte.

1 mol 1 mOsm = 198mg To determine how many mOsm in 45g:

1 mOsm : 198mg

x mOsm : 45,000mg

x = 227.3 mOsm of Dextrose, (that is in 500mL)

To determine the osmolarity: 1 osmolar = 1 osmol/litre

227.3 mOsm : 500mL y mOsm : 1000mL

y = 454.6 mOsm/L (osmolarity)

Worked Example 9.18

Calculate the osmolarity (mOsm/L) of 0.9% w/v NaCI in water. MW (NaCI) = 58.5

Working:

Equation:

NaCI Na++ Cl-

1 mmol NaCI 2 mmol ions = 2 mOsm = 58.5mg

1 osmolar = 1 osmol/L

We have 0.9% w/v = 0.9g/100mL or 9g/1000mL

Determine the number of mOsm in 9g of NaCI.

2 mOsm : 58.5g

x mOsm : 9000mg

x = 307.7 mOsm

ie: osmolarity is: 307.7 mOsm/L

Note: 0.9% NaCI is 'normal saline'. This can be safely instilled into the eye or injected. Any solution that has an osmolarity of within 10% of 308 mOsm/L is termed to be isosmotic with body fluids such as blood serum and tears.

Solutions which are isosmotic with body fluids are only considered to be isotonic if: - membranes in contact with the solution are impermeable to the solute; - the solute does not alter the permeability of membranes to any other substance present; - no chemical reaction leads to a change in the total concentration of dissolved ions or

molecule 89

Pharmaceutical Calculations 9, Millimoles, Miltiequivalents and Milliosmoles

Problems are encountered with solutions of some substances such as boric acid, urea, ethanol and certain monohydric or polyhydric alcohols, as well as some local anaesthetics. These solutions cause haemolysis of red blood cells in isosmotic concentrations and hence are not isotonic with red blood cells.

(Ref APF 15 page 443). See also Tutorial 10: Isotonic Solutions

Quiz Question 9.16

Assuming complete dissociation, calculate the osmoiarity of the following 100mL solution with the following concentration.

Dextrose (20mmol/100mL) KCI(4mmol/100mL) MW (Dextrose) = 180 MW (KCI) = 74.6

Answer; Equations: a) Dextrose is a non-electrolyte

Therefore: 20mmol = 20mOsm

b) KCI -> K* + CI"

1 mmol KCI = 2 mmol ions - 2 mOsm Therefore 4 mmol = 8 mOsm

Total = 20 + 8 = 28 mOsm/100mL = 280 mOsm/L (osmolarity)

Worked Example 9.19

What is the osmolarity of the following solution (Ringer's) 0.86% NaCI (MW = 58.5) 0.03% KCI (MW = 74.6) 0.033% CaCI2 (MW =111.0)

Working:

Equations:

1.NaCI Na+ + CI- 0.86% = 860mg/100mL = 8,600mg/1000mL 1mmol NaCI = 2 mmol ions = 2 mOsm = 58.5mg How many mOsm in 8,600mg ? (cross multiply)

= 294mOsm 2. KCI K+ + CI- 0.03% = 30mg/100mL = 300mg/1000mL 1 mmol KCI = 2 mmol ions = 2 mOsm = 74.6mg How many mOsm in 300mg ? (cross multiply)

= 8.04 mOsm

3. CaCI2 Ca2+ + 2CI-

0.033% = 33mg/100mL = 330mg/1000ml_ 1 mmol CaCI2 = 3 mmol ions = 3 mOsm = 111.0mg How many mOsm in 330mg ? (cross multiply)

= 8.92 mOsm Total = 294 + 8.04 + 8.92 = 311 mOsm/L (This is very close to the osmoiarity concentration for normal saline (308 mOsm/L) ie: they are osmotically equivalent. 90

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Quiz Question 9.17

If the osmolarity of a NaCI solution is 309 mOsm/L, what is its percentage strength ? MW (NaCI) = 58.5

Answer:

Equation: NaCI Na* +CI-

1 mmol NaCI 2 mmol ions = 2 mOsm = 58.5mg To determine the weight of NaCI for 309mOsm, cross multiply.

ie: 309x58.5 / 2 = 9038mg (this is in one litre)

Therefore in 100mL there is 903.8mg = 0.9038g (about 0.9g) ie: expressed as a percentage: 0.9% w/v

Worked Example 9.20

Prepare 1 L of a solution of CaCI2 containing 40 mEq/L of Ca2+. Make isotonic with NaCI. MW(CaCI2.2H2O) = 147

MW (NaCI) = 58.5

Working out: Equation:

CaCI2.2H2O Ca2+ + 2CI- + 2H2O

Firstly, determine mass of CaCI2.2H2O required:

1mmol CaCI2.2H2O contains 2mEq Ca2+ = 147mg CaCI2.2H2O

To determine how much CaCI2.2H2O is needed for 40mEq Ca +, cross multiply: = 2,940mg

Secondly, determine how many mOsm this quantity of CaCI2.2H2O contributes:

1 mmol CaCI2.2H2O 3 mOsm of ions = 147mg

To determine the number of mOsm in 2,940mg, cross multiply: = 60 mOsm

Now, an isotonic solution contains 308 mOsm (+/-10%) So we need 308 - 60 = 248 mOsm (to be added using NaCI)

With NaCI: NaCI Na+ + CI-

1 mmol NaCI -> 2 mOsm ions = 58.5mg NaCI To determine the amount of NaCI needed for 248 mOsm, cross multiply: = 7,254 mg NaCI

Final formulation becomes:

CaCI2.2H2O 2,940 mg NaCI 7,254 mg Water for Injections to 1000 mL 91

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Quiz Question 9.18

What quantities of salts are required to prepare 2 litres of a solution of KCI containing 40

mEq/L of K+ and also made isotonic with NaCI ?

MW (KCI) = 74.6 MW (NaCI) = 58.45

Answer: Equation:

KCI K+ + Cl -

1 mmol KCI 1 mmol K+ = 1mEq K+ = 74.6mg

To determine the quantity of KCI needed for 80mEq K+, cross multiply: = 5,968mg

To determine how many mOsm this contributes:

1 mmol KCI 2 mOsm ions = 74.6mg KCI Therefore, to determine the number of mOsm in 5,968mg, cross multiply:

= 160mOsm

Remember 308 mOsm/L is isotonic. We are making 2L of solution here, thus need 308 x 2 = 616mOsm.

The additional amount required is: 616 -160 = 456 mOsm This will be supplied by NaCI

NaCI Na+ + Cl- 1mmol NaCI 2 mOsm ions = 58.45mg

To determine the amount of NaCI needed for 456 mOsm, cross multiply: = 13,327mg

Final formulation becomes:

KCI 5,968 mg (5.97 g) NaCI 13,327 mg (13.3 g) Water for Injections to 2 litres

Quiz Question 9.19

Calculate the number of milliosmoles (mOsm) in Sorensen's Phosphate Buffer Ref: APF15p450(pH6.8)

50 mL of 0.067 M KH2PO4 (0.908%) MW = 136 50 mL of 0.067 M Na2HPO4.12H2O (2.39%) MW = 358

Answer: Firstly, determine the amount of each salt:

1.KH2PO4

0.908g -------- x 50 mL = 0.454g = 454mg

100 mL

2. Na2HPO4.12H2O

2.39g ------- x50 mL= 1.195g - 1,195mg 100 mL 92

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Pharmaceutical Calculations 9, Millimoles, Milliequivalents and Milliosmoles

Equations:

+ 1. KH2PO4 K+ + H2PO-

4

1 mmol KH2PO4 = 2 mOsm ions = 136mg KH2PO4 To determine the number of mOsm in 454mg, cross multiply:

= 6.68 mOsm 2. Na2HPO4 2 Na+ + HPO4

2-

1 mmol Na2HPO4 = 3 mOsm of ions = 358mg of Na2HPO4

To determine the number of mOsm in 1,195mg, cross multiply: = 10.01mOsm

Total mOsm = 6.68 + 10.01 = 16.69 mOsm

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Pharmaceutical Calculations Module Tutorial 9: Millimoles