Isotonic calculation

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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Freezing Point Depression Method

Example 1.1

Rx ∆Tf 1%
NaCl 900mg 0.576
Dextrose q.s (20.5g) 0.101

Ft. isotonic soulution500ml Or;

0.9% ⁄
=
Solution; 500 100
.
. . ( × . ) = 0.18% ⁄
= =
.

= 4.12 100

For 500ml;
500
= (4.12 ) ×
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

= 0.7 100

200 ;
200
= (0.7 ) ×
100

= 1.4

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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
= = = 1.37 ℎ ℎ
0.38

NaHCO3 required to make the solution isotonic upon dilution is;

1.37 χ
=
100 1000

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

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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
= = = 0.9 0.9% ⁄
0.576

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

( , . 927)
=
0.576 ( ℎ ℎ )

Formula & Method 1
•Look up in the table the sodium chloride equivalent
for the srenght of solution nearest to the strenght
= 0.9% - ([E1%X %w/v] + …….) 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.

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aCl Equivalent Method

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= . °
= 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%) = . °
= 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 Convert amount of NaCl required
χ% dextrose equivalent to 0.7% NaCl to amount of dextrose required

(0.7%)(1%)
χ= = 3.89%
0.18%

For 500ml;
3.89 χ
=
100 500

χ = 19.45 g

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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 × 100 % × 0.3 = 0.15%

.
For chlorobutol × 100 % × 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% Let adjust isotonicity using KCl instead
Ephedrine HCl 1g 0.3 of NaCl
KCl q.s 0.4

Solution;
1% KCl equivalent to 0.4% NaCl Convert amount of NaCl required to
χ% ← 0.69% NaCl (refer to example 3) amount of KCl required

0.69
χ= = 1.725%
0.4

So for 200ml …..= 3.45g KCl

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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 = = 1.3
.

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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; This solution is isotonic (A)
For ephedrine HCl, V=1 X 0.3 X 111.1 = 33.33ml (final volume)
So the overall
For chlorobutol, V=0.5 X 0.24 X 111.1 = 13.33ml (final volume) This solution is isotonic (B)
solution is isotonic
200ml – (33.33 + 13.33) ml = 153.34ml

∴ Amount of NaCl required to adjust 153.34ml to isotonic; This solution is isotonic (C)
0.9
= × 153.34 = 1.38
100

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%

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White Vincent & Sprowl Method

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

Gram equivalent weight
Ion Ionic weight ℎ Weight of 1mEq (mg)
( )

Sodium Na+ 23 23 23
Potassium K+ 39.1 39 39.1
Calcium Ca2+ 40 20 20
Chloride Cl− 35.5 35.5 35.5
Bicarbonate HCO3− 61 61 61
Phosphate HPO4− 96 48 48

• ∴ Gram equivalent weight
i) For ion;

ℎ
ℎ =

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 , , 1
ℎ
=
× ℎ ℎ

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Miliequivalent (mEq)

Table 5.2 makes this clear

Weight of salt
M. of Valency of Number of containing 1mEq
Ion Salt used the salt the ion ion in salt of the ion

Na+ Sodium chloride 58.5 1 1 58.5mg
Na+ Sodium phosphate 358† 1 2 179.0mg
Ca2+ Calcium chloride 147† 2 1 73.5mg
Cl- Calcium chloride 147 1 2 73.5mg

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

ii) For salt;
ℎ ℎ
ℎ =

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;

×

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

×
1000

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

×
10000

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

Mg of salt containing 1
Ion Miliequivalent (mEq) mg Salt mEq of specified ion

Na+ 23.0 Sodium chloride 58.5
Sodium bicarbonate 84
K+ 39.1 Potassium chloride 74.5
Ca2+ 20.0 Calcium chloride 73.5
Mg2+ 12.5 Magnesium sulphate 123
Magnesium chloride 101.5
Cl- 35.5 Sodium chloride 58.5
HCO3- 61.0 Sodium bicarbonate 84
HPO4- 48 Sodium phosphate 179

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Example 5.1
Calculate the quantities of salt required for the following electrolyte solution;

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

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;

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mg g mEq

×
× K+ Na+ Mg2+ Cl- HPO4-
1000

Potassium chloride 74.5 X 30 2.235 30 30
Sodium phosphate 179.0 X 10 1.790 10 10
Magnesium chloride 101.5 X 5 0.508 5 5
Sodium chloride 58.5 X 10 0.585 10 10

Water for injection, to 1 litre to 1 litre 30 20 5 45 10

55 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 × 10000
= 154 /
58.5

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.

× 10000
= /

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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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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:-

= ×

= × ÷ 1000

⁄ = × ÷ 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—

mmol

WXM K+ Mg2+ Ca2+ Na+ Cl-

Potassium chloride 5 x 74.5 = 0.373 g 5 5
Magnesium chloride 4 x 203 = 0.812 g 4 8
Calcium chloride 4 x 147 = 0.588 g 4 8
Sodium chloride 60 x 58.5 = 3.510 g 60 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—

⁄ = × ÷ 10000

⁄ × 10000
=

a) For dextrose

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

Hence,
4.3 × 10000
= = 239
180.2

b) For sodium chloride
0.18 × 10000
= = 31
58.5
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.

0,029 × 10000
= =2
147

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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Milimole(mmol)

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 )
= = 4329
5

3g Na → 1000000ml

? ml ←4320ml

( )( )
= = 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

( )( )
= = 96.15

? gram acetic acid → 100ml solution

33
= × 100 = 34% ⁄
96.15

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Isotonic calculation

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