isotonic calculation

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Freezing Point Depression Method

CALCULATION FOR SOLUTION ISOTONIC WITH BLOOD AND TEARS

Some of the many ways;

1. Freezing point depression method

2. NaCl equivalent method

3. White Vincent method

4. Sprowl method

5. Molecules concentration method

6. Graphical method on vapor pressure and freezing point determination

1. FREEZING POINT DEPRESSION METHOD

• The plasma and blood freezing point temperature = -0.52°C

• The dissolved substances in plasma or tear depress the solution freezing point below 0.52°C

• Any solution that freeze at T=-0.52°C is isotonic with blood and tear

• weight of substance that need to be adjusted to make it hypotonic given by the following

formula;

� = 0.52 − �

Where,

W=the weight, in g, of the added substance in 100ml of the final solution

a =the depression of the freezing point of water produced by the medicament already in the

solution (calculated by multiplying the value for b for the medicament by the strength of the

solution express as % w/v)

b=the depression of freezing of water produced by 1 per cent w/v of the added substance

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Example 1.1

Solution;

� = .�� = .�� (�.�

� ×.��).��

= 4.12 � �� 100� !" #$�"�

For 500ml;

= (4.12 �) × 500� 100�

= 20.6� &'($)"!' )'*#�)'&

Example 1.2

Rx

Ephedrine HCl 1g

Chlorobutol 0.5g

NaCl (1.4g) q.s

Distilled water q.s ad 200ml

Solution;

� = 0.52 − � = 0.52 − [(0.5 × 0.165) − (0.25 × 0.138)]

0.576

= 0.7 � 0�1 �� 100� !" #$�"�

2") 200� ; = (0.7 �) × 200�

100�

= 1.4� 0�1 )'*#�)'&

Rx ∆Tf 1%

NaCl 900mg 0.576

Dextrose q.s (20.5g) 0.101

Ft. isotonic soulution500ml

0.9% � 5⁄500� = 7

100�

( = 0.18% � 5⁄

Or;

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EXercise 1.3

Prepare 500ml NaHCO3 (∆T.f 1%=0.38) so that when it is dilute with the same amount of water, it would

be isotonic.

Solution;

Rx ∆T.f 1%

NaHCO3 Xg(13.7) 0.38

Water qs ad 500ml

The solution is hypertonic

When diluted

NaHCO3 Xg

Water qs ad 1000ml

This solution is isotonic

� = 0.52 − � = 0.52 − 0

0.38 = 1.37� �ℎ�9ℎ �! �!"$"��9

NaHCO3 required to make the solution isotonic upon dilution is;

1.37�100� = χ

1000�

1.37 X 10 = 13.7g (to be dissolved to 500ml solution)

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�aCl Equivalent Method

1

•Look up in the table the sodium chloride equivalent for the srenght of solution nearest to the strenght

of medicament in the preparation.

2

•Multiply this by the strenght of the medicament.

3

•substract the result from 0.9 per cent; the difference is strenght of sodium chloride necessary

to adjust the solution to iso-osmoticity.

2. NaCl EQUIVALENT METHOD (E)

• Based on the factor called the sodium chloride equivalent which can be used to convert a

specified concentration of medicament to the concentration of medicament to the

concentration of sodium chloride that will produce the same osmotic effect.

• Standard → 0.9 % w/v NaCl at isotonic

• Method to calculate % at isotonic, ∆Tf 1% =0.576 ~ NaCl

� = 0.52 − � = 0.52 − 0

0.576 = 0.9 � ") 0.9% � 5⁄

Known as normal saline

• Sodium chloride equivalents (E1%) can be calculated from the following formula;

= (:)'';�� <"��$ &'<)'!!�"� <)"&#9'& = !" #$�"� >, @A1 <�. 927) 0.576 (2)'';�� <"��$ &'<)'!!�"� <)"&#9'& = 0�1 "2 $ℎ' !��' !$)'��ℎ$ )

Formula & Method

= 0.9% - ([E1%X %w/v] + …….)

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Example 2.1

∆Tf 1% ascorbic acid =0.105°C

∆Tf 1% NaCl = 0.576°C

What is the E for 1% ascorbic acid?

Solution

E 1% ascorbic acid= .��°C.��°C = 0.18

Example 2.2

Rx

NaCl 0.2%w/v

Dextrose q.s

Ft. isotonic solution 500ml

Solution;

∆Tf 1% NaCl = 0.576°C

∆Tf 1% dextrose = 0.101 °C

For dextrose 1 % (E1%) = .��°C.��°C = 0.18

Conclusion → 1% dextrose has osmotic pressure equals to 0.18% NaCl

NaCl in the solution = 0.2 %

∴Remaining amount of NaCl to be added for isotonic

0.9% - 0.2% =0.7%

1% dextrose equivalent to 0.18% NaCl

χ% dextrose equivalent to 0.7% NaCl

χ = (0.7%)(1%)0.18% = 3.89% &'($)"!'

For 500ml;

3.89�100� = χ

500�

χ = 19.45 g

Convert amount of NaCl required

to amount of dextrose required

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Example 2.3

Rx E 1%

Ephedrine HC 1g 0.3

Chlorobutol 0.5g 0.24

NaCl q.s

Distilled water 200ml isotonic

Solution;

For ephedrine HCl D �� × 100E % × 0.3 = 0.15%

For chlorobutol D .�� × 100E % × 0.24 = 0.06%

Amount of NaCl for isotonic;

0.9% - (0.15% + 0.06%) = 0.69%

So for 200ml……………….= 1.38

Example 2.4

Rx E 1%

Ephedrine HCl 1g 0.3


KCl q.s 0.4


Solution;

1% KCl equivalent to 0.4% NaCl

χ% ← 0.69% NaCl (refer to example 3)

χ = 0.690.4 = 1.725% F1

So for 200ml …..= 3.45g KCl

Let adjust isotonicity using KCl instead

of NaCl

Convert amount of NaCl required to

amount of KCl required

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Example 2.5

Calculate the percentage of sodium chloride required to render a 0.5 per cent solution of

potassium chloride iso-osmotic with blood plasma.

Solution

Sodium chloride equivalent of 0.5 per cent potassium chloride = 0.76

∴ Percentage of sodium chloride for adjustment

= 0.9 − (0.5 Χ 0.76) = 0.9 − 0.38 = 0.52

Example 2.6

Calculate the percentage of anhydrous dextrose required to render a 1 per cent solution

of ephedrine hydrochloride iso-osmotic with body fluid.

∴ Percentage of sodium chloride for adjustment

= 0.9 − (1 Χ 0.3) = 0.6

Equivalent percentage of anhydrous dextrose = 0.6/0.18 = 3.33

Example 2.7

Select a suitable substance for an eye lotion 0.5 per cent of silver nitrate and calculate the

percentage required to render the lotion iso-osmotic with lachrymal secretion.

Sodium chloride is unsuitable because silver nitrate is incompatible with chloride.

Potassium nitrate will be used

Sodium chloride equivalent of 0.5 per cent silver nitrate = 0.33

= 0.9 − (0.5 Χ 0.33) = 0.9 − 0.165 = 0.753

Equivalent percentage of potassium nitrate = .�G�.�� = 1.3

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White Vincent & Sprowl Method

3. WHITE VINCENT METHOD

Principle:

A Isotonic solution + B isotonic solution = C isotonic solution

→This method involves the addition of water to medicament to obtain an isotonic solution. This

followed by the addition of isotonic buffer solution or preservatives isotonic solution to the required

volume.

V = W X E1% X 111.1

Where;

V= V (ml) of isotonic solution that could be obtain in wg of drug in water (the amount of water to added

to form isotonic solution)

W= amount of drug in the formula

E1%=NaCl equivalent of the drug

111.1= constant that could be find from volume for 1% isotonic NaCl

0.9% NaCl → 1% NaCl

0.9g NaCl → 100 ml

1 g NaCl → 111.1 ml

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White Vincent & Sprowl Method

Example 3.1

Rx E 1%

Ephedrine HCl 1g 0.3


NaCl q.s


Solution;

For ephedrine HCl, V=1 X 0.3 X 111.1 = 33.33ml (final volume)

For chlorobutol, V=0.5 X 0.24 X 111.1 = 13.33ml (final volume)

200ml – (33.33 + 13.33) ml = 153.34ml

∴ Amount of NaCl required to adjust 153.34ml to isotonic;

= 0.9� 0�1 100� !" #$�"� × 153.34 � !" #$�"� = 1.38 � 0�1

4. SPROWL METHOD

• Using white Vincent method but w is set to constant, 0.3

V = W X E1% X 111.1 or V = 33.33E1%

This solution is isotonic (A)

This solution is isotonic (B)

This solution is isotonic (C)

So the overall

solution is isotonic

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Miliequivalent (mEq)

5. MILIEQUIVALENT (mEq)

• Definition; the gram equivalent weight of an ion is the ionic weight (the sum of atomic

weights of the element in an ion) in gram divided by the valence of that ion.

• A mililequivalent is one thousandth part of the gram equivalent weight, the same figure

expressed in milligram. Table 5.1 makes this clear

Ion

Ionic weight

Gram equivalent weight

( �"��9 �'��ℎ$5� '�9= )

Weight of 1mEq (mg)

Sodium Na+

Potassium K+

Calcium Ca2+

Chloride Cl−

Bicarbonate HCO3−

Phosphate HPO4−

23

39.1

40

35.5

61

96

23

39

20

35.5

61

48

23

39.1

20

35.5

61

48

• ∴ Gram equivalent weight

i) For ion;

H)�� '*#�5� '�$ �'��ℎ$ = �"� �'��ℎ$5� '�9=

Example

1 Eq sodium → 23/1 = 23g

1Eq chloride → 35.5/1 =35.5

• The weight of a salt containing 1mEq of a particular ion is obtained by dividing the

molecular weight of the salt by the valency of that ion multiply by the number of such

ion in the molecules

�'��ℎ$, �� , "2 !� $ 9"�$�� 1�I* "2 �� "� = �" '9# �) �'��ℎ$ "2 !� $

5� '�9= "2 �"� × �#�') "2 !#9ℎ �"� �� $ℎ' �" '9# '

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Table 5.2 makes this clear

Ion

Salt used

M. of

the salt

Valency of

the ion

Number of

ion in salt

Weight of salt

containing 1mEq

of the ion

Na+

Na+

Ca2+

Cl-

Sodium chloride

Sodium phosphate

Calcium chloride

Calcium chloride

58.5

358†

147†

147

1

1

2

1

1

2

1

2

58.5mg

179.0mg

73.5mg

73.5mg

(† N.B. the water of crystallization must not be ignored)

ii) For salt;

H)�� '*#�5� '�$ �'��ℎ$ = �'��ℎ$ "2$ℎ' !� $ �)�'!$ �"��9 5� '�9=

Example (monovalent)

1 Eq → 58.5/1= 58.5g

Example (not monovalent)

1Eq CaCl2.H2O → 147/2 =73.5

Miliequivalent (mEq) → 1/1000 Gram Eq weight

Example

1mEq sodium →23mg

1mEq chloride → 35.5g

1mEq NaCl → 58.5 mg

1mEq CaCl2.H2O →73.5 mg

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Conversion Equations

I. To convert mEq/litre to mg/litre;

J × I

II. To convert mEq/litre to g/litre;

J × I1000

III. To convert mEq/litre to %w/v;

J × I10000

• The BPC includes table showing the weight s of salt that contain 1 mEq of specified ion;

Ion

Miliequivalent (mEq) mg

Salt

Mg of salt containing 1

mEq of specified ion

Na+

K+

Ca2+

Mg2+

Cl-

HCO3-

HPO4-

23.0

39.1

20.0

12.5

35.5

61.0

48

Sodium chloride

Sodium bicarbonate

Potassium chloride

Calcium chloride

Magnesium sulphate

Magnesium chloride

Sodium chloride

Sodium bicarbonate

Sodium phosphate

58.5

84

74.5

73.5

123

101.5

58.5

84

179

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Example 5.1

Calculate the quantities of salt required for the following electrolyte solution;

From table 5.3:-

1 mEq of both potassium and chloride ion are contained in 74.5 mg of potassium

chloride.

1 mEq of both sodium and phosphate ion are contained in 179 mg of sodium phosphate.

1 mEq of both magnesium and chloride ion are contained in 101.5 mg of magnesium

chloride.

1 mEq of both sodium and chloride ion are contained in 58.5 mg of sodium chloride.

Therefore;

30 mEq of potassium ion is provided by 30 X 74.5 mg of potassium chloride which will

also supply 30 mEq of chloride ion.

10 mEq of phosphate ion is provided by 10 X 179 mg of sodium phosphate which will

also supply 10 mEq of sodium ion.

5 mEq of magnesium ion is provided by 5 X 101.5 mg of magnesium chloride which will

also supply 5 mEq of chloride ion.

There remains a deficiency of 10mEq of each sodium and chloride ions which cn be

provided by 10 X 58.5 mg of sodium chloride. The formula becomes;

Sodium ion 20 mEq

Potassium ion 30 mEq

Magnesium ion 5 mEq

Phosphate ion (HPO4-) 10 mEq

Chloride ion 45 mEq

Water for injection, to 1 litre

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mg g mEq

J × I J × I1000

K+

Na+

Mg2+

Cl-

HPO4-

Potassium chloride

Sodium phosphate

Magnesium chloride

Sodium chloride

74.5 X 30

179.0 X 10

101.5 X 5

58.5 X 10

2.235

1.790

0.508

0.585

30

10

10

5

30

5

10

10

Water for injection, to 1 litre to 1 litre 30 20

55

5

45 10

55

The fact that cation and anion balance confirm that the formula has been worked out correctly

Example 5.2

Express the following formula as percentage w/v.

Sodium ion 147 mEq

Potassium -ion 4 mEq

Calcium ion 4 mEq

Chloride ion 155 mEq

Water for Injections, to 1 liter.

From Table 20.6 and the appropriate conversion equation, the required percentages are—

Sodium Chloride 5.85 X 147 ÷ 10 000 = 0.860% W/v

Potassium Chloride 74.5 X 4 ÷ 10 000 = 0.030% w/v

Calcium Chloride 73.5 X 4 ÷ 10 000 = 0.029% w/v

Water for injections, to 1 liter

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Example 5.3

Prepare 500 ml of an Intravenous solution containing 70 mEq of sodium, 2 mEq of

potassium, 4 mEq of calcium and 76 mEq of chloride.

The number of milligrams of the various chlorides which contain I mEq

of the required ions is obtained from Table 20.6 and the formula becomes—

Sodium Chloride 70 X 58.5 ÷ 1000 = 4.0Q5 g

Potassium Chloride 2 X 74.5 ÷ 1000 = 0.149 g

Calcium Chloride 4 X 73.5 ÷ 1000 = 0.294 g

Water for Injections, to 500 ml

It is not unusual for students to miscalculate the amounts for this type of formula due to

failing to appreciate that mEq is a unit of weight and not an abbreviation for mEq per litre. This

leads to an incorrect halving of the final quantities in the above example.

Example 5.4

Express 0.9 per cent sodium chloride solution to mililequivalent per litre.

0.9 × 1000058.5 = 154 �I*/ �$)'

To Convert Percentage w/v to mEq

The number of grams (C) per 100 milliliters is converted to mg/liter by multiplying by 10

000. This, divided by the weight (W) in mg of salt containing 1mEq, will give the number of

mEq/liter.

1 × 10000J = �I*/ �$')

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Adjustment to Iso-osmoticity with Blood Plasma Based on Miliequivalent

The equation used is:-

= 310 − �

Where;

a = number of mililequivalent per liter of medicament present and

b = number of mililequivalent per liter of adjusting substance required

Example 5.5

Calculate the amount of sodium chloride required to adjust a solution containing 40

mEq of each potassium and chloride ion to iso-osmoticity with blood plasma.

A solution containing 40 mEq of chloride ion provide a total of 80 mEq of anion and

cation.

∴ = 310 − 80

= 230

230 mEq will be provided by 115 mEq of sodium ions and 115 mEq chloride ions.

The formula of the solution will be:-

Potassium chloride 74.5 mg X 40 = 2.98 g

Sodium chloride 58.5 mg X 115 = 6.73 g


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Milimole(mmol)

6. MILIMOLES

In the SI, the unit for chemical quantity is mole and the term equivalent and

mililequivalent becomes absolute. Consequently the method of expressing the composition of

body and infusion fluid is changing from the mililequivalent to mole notation.

By analogy with atoms and molecules, a mole of an ion is its ionic weight in grams but

the number of moles of each of the ions of a salt in solution depends on the number of each ion

in the molecule of the salt.

It follows that the quantity of salt, in mg, containing 1 mmol of a particular ion can be

found by dividing the molecular weight of the salt by the number of that ion contained in the

salt. For example—

Salt Ion Quantity of salt (In mg)

containing 1mmol

NaCl Na+ M.Wt /1 = 58.5

CI- M.Wt / 1 = 58.5

CaCl2 Ca2+

M.Wt / 1 = 147

Cl- M.Wt /2 = 73.5

Na2HPO4 Na+ M.Wt /2 = 179

HP042-

M.Wt /1 = 258

NaH2PO4 Na+ M.Wt /1 = 156

H2P04- M.Wt /1 = 156

Conversion Equations

To convert quantities expressed in mmol per litre into weighable amounts. The

following formulae may be used :-

� <') �$)' = J × M

�� <') �$)' = J × M ÷ 1000

<')9'�$ � 5⁄ = J × M ÷ 10000

Where W is the number of mg of salt containing 1 mmol of the required ion and M is the

number of mmol per litre

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Milimole(mmol)

Example 6.1

Calculate the quantities of salts required for the following electrolyte solution.

Sodium 60 mmol

Potassium 5 mmol

Magnesium 4 mmol

Calcium 4mmol

Chloride 81 mmol


From Table 20.8—

5 mmol of potassium ion is provided by 5 x 745 mg of potassium

chloride which also provides 5 mmol of chloride ion.

4 mmol of magnesium ion is provided by 4 x 203 mg of magnesium chloride which also

provides 2 x 4 mmol = 8 mmol of chloride ion, since there are two chloride ions in the molecule.

4 mmol of calcium ion are provided by 4 x 147 mg of calcium chloride which, like

magnesium chloride, also provides 8 mmol of chloride ion.

60 mmol of sodium ion is provided by 60 x 58.5 mg of sodium chloride which provides a

further 60 mmol of chloride.

The formula becomes—

W X M

mmol

K+

Mg2+

Ca2+

Na+

Cl-

Potassium chloride

Magnesium chloride

Calcium chloride

Sodium chloride

Water for injection, to

5 x 74.5 = 0.373 g

4 x 203 = 0.812 g

4 x 147 = 0.588 g

60 x 58.5 = 3.510 g

1 litre

5

4

4

60

5

8

8

60

73 31

Although there appears to be inequality between the anions and cations, the charges are equally

balanced.

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Milimole(mmol)

Example 6.2

Calculate the number of millimoles of (a) dextrose and (b) sodium ions in 1 litre of

Sodium Chloride and Dextrose Injection containing 4.3 per cent w/v of anhydrous dextrose and

0.18 per cent w/v of sodium chloride.

Use the conversion equation—

<')9'�$ � 5⁄ = J × M ÷ 10000

M = <')9'�$ � 5⁄ × 10000J

a) For dextrose

Since dextrose is non-electrolyte, W = M.Wt

Hence, M = 4.3 × 10000

180.2 = 239 ��"

b) For sodium chloride

M = 0.18 × 1000058.5 = 31 ��"

Since 1 mmol sodium chloride provides 1 mmol of sodium ion and 1 mmol of chloride ion, 1 litre

of the solution will contain 31 mmol of sodium ion (and 31 mmol of chloride ion ).

Example 6.3

Calculate the number of milimoles of calcium and chloride ion in a litre of a 0.029 per

cent solution of calcium chloride.

M = 0,029 × 10000

147 = 2 ��"

But, each mole of calcium chloride provides 1 mol of calcium ions and 2 moles of chloride ions.

Therefore, 1 litre of solution contains 2mmol of calcium ion and 4 mmol of chloride ion.

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ppm calculation

EXample

Prepare 90ml NaF solution so that 5ml of the solution when diluted with water to 1cupfull(240ml), a

3ppm solution obtained.

Solution;

5ml solution → 240ml final volume

90ml solution → ? ml final volume

= (240� )(90� )5� = 4329�

3g Na → 1000000ml

? ml ←4320ml

= (OG�PQ)(GPQ)�PQ = 0.01296�

∴ 0.01296mg needed ≈ 13mg NaF

Formula

NaF 13mg

Distilled water q.s ad. 90ml

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Calculation involving density factor

EXample

Given acetic acid BPC (33%w/w, d=1.04g/ml).What is the concentration of acetic acid in %w/w

Solution;

Acetic acid

33%w/w= 33g acetic acid in 100g solution

d= 1.04g/ml

? w/v

33g acetic acid →100g acid solution

33g acetic acid → ?ml solution

From density factor information;

104g acid → 100ml acid solution

100g acid → ?ml acid solution

= (�PQ)(�R)�OR = 96.15� �9�& !" #$�"�

? gram acetic acid → 100ml solution

= 33�96.15� × 100 = 34% � 5⁄

isotonic calculation

Documents

solution isotonic

isotonic method

dextrose equivalent

nacl equivalent method

g nacl

solution freezing point

wv nacl

percentage of sodium