By Mario
Bacar (family medical specialist)
Diploma in
Medical Surgical Emergencies.
To
understand and carry out the preparation of solutions in
different percentages, we must first regulate the presentations that exist in
our region. In the case of Mexico we have that our saline serum comes at
0.9% and 0.45%, this means that in every 100 ml of solution
there will be diluted 0.9 grams and 0.45 grams respectively. There is also
an IV ampoule presentation in the emergency room called hyperton which contains
17.7% sodium chloride in 10 ml, this presentation means that there are 17.7
grams of NaCl in each 100 ml of the solution, even though the ampoule comes
from 10 ml.
Now, we want
to prepare a 3% saline solution since, due to service needs, we require a
hypertonic solution and that presentation does not exist. Tell yourself to
correct severe hyponatremia, for resuscitation or any other medical
use. So we must apply science and mathematics to prepare it. Below is
an explanation of how such a solution could be forged and the cooking recipes
for its application in clinical practice.
The 0.9%
saline solution contains, as we have already said, 0.9 grams of NaCl diluted in
100 ml, so it has 9 grams in a liter of water, which is equivalent to 154
milliequivalents (mEq) of Sodium and 154 of chlorine in a liter of
water . , these equivalents are calculated using formulas for
molar weight and ionic particle equivalent, which is not the subject of
now. What we need to know is that a 0.9% saline solution has 154 mEq of
sodium in one liter. Now we want to calculate a 3 % solution , so a simple
rule of 3 will suffice
A .9% saline
solution is equal to = 1000 ml = 154 mEq of Na
Yes 0.9%
------ 154
So 3%
----- how much?
3 x 154 / 0.9
= 513.3 (rounded to 513).
So we have to
achieve a solution that remains with 513 mEq of Na in one liter of solution,
for this we must balance the solutions of low load and very heavy load.
If we take
into account the ampulla hyperton, which has 10 ml of 17.7%
saline solution (17.7 grams of NaCl in each 100 ml), which means 30 mEq of
sodium per 10 ml, or 300 mEq of Na in 100 ml, or 3000 mEq in a liter, simple
exponentiation. Then 12 hyperton ampoules would then be 360 mEq, for our
calculation we would need for the adjustment:
0.9% saline
solution ----------- 154 mEq of Na in one liter
3% saline
solution ------------- 513.3 mEq of Na per liter
ampoule
hyperton ----------- 30 mEq of Na in each 10 ml.
So
12 hyperton
ampoules = 12*30= 360 mEq of Na
In our case,
after looking for many combinations, we found that:
12.6 hyperton
ampoules would be 12.6*30= 378 mEq, which would be in 126 ml
of solution (30 mEq per 10 ml)
If we gauge
those 126 ml in one liter of 0.9% saline solution, then we would have to empty
126 ml of the 0.9% solution and we would have 874 ml of 0.9% saline solution,
which we would complete with the 126 ml of hyperton. We would have like
this:
1000 ml of
sun. 0.9% saline = 154 mEq of Na
874 ml of
sun. 0.9 saline = 134.596 mEq of
Na
+
The 126 ml of
hyperton (378 mEq) =
134.596 mEq
(874 ml of 0.9% saline sol) + 378 mEq (126 ml of hyperton) =
512.5 mEq in a liter of solution, which rounded off is 513 mEq, so we will have
achieved a solution that in a liter of water has 513 mEq of Na or 3% saline
solution, or 3 grams of NaCl in 100 ml.
Then our
indication would be as follows:
Dilute 126 ml
of hyperton in 1000 cc of 0.9% saline solution and pass IV to ----- (whatever
is indicated according to) or
Dilute 12.6
hypertonic ampoules in 1000 0.9% saline solution... or
874 ml 0.9%
saline + 126 ml IV hyperton…..
Any of these
ways will have achieved a 3% saline solution, that is, 513 mEq of Sodium in
1000 cc of solution.
Now it turns
out that 1000 ml is a lot of waste!!! since we will only use a 150
cc bolus; Let's prepare a 3% saline solution but in less quantity.
This time we
will no longer split our heads, we will only split everything in half in the
same proportions.
First we need
to know that 3% saline solution contains 514 mEq in one liter of
solution. Therefore in 100 ml you will have:
1000/ 10 =
100, then 514 / 10 = 51.4 mEq of Na in 100 mL
A 3% saline
solution has the following measurements:
In 100 ml
there are 51.4 mEq of Na or 3 grams of NaCl
In 200 ml
there are 102.8 mEq of Na or 6 grams of NaCl
In 250 ml
there are 128.5mEq of Na or 7.5 grams of NaCl
In 300 ml
there are 154.2 mEq of Na or 9 grams of NaCl
In 500 ml
there are 257 mEq of Na or 15 grams of NaCl
We also have
under the same mathematical algorithm that a 0.9% saline solution has the
following measurements
In 50 ml
there are 7.7 mEq of Na
In 100 ml
there are 14.4 mEq of Na
In 200 ml
there are 28.8 mEq of Na
In 210 ml
there are 32.34
In 220 there
are 33.88
In 230 there
are 35.42
In 240 there
are 36.96
In 250 there
are 38.5 ….And so gradually
Then we can
play with the numbers at our convenience trial trial error: in this case we
will only halve the totals.
If we had to
measure a liter of 0.9% physiological saline solution with 12.6 ampoules of
hyperton to obtain a 3% saline solution, then to obtain 500 cc of 3% solution
we only had to measure half of those ampoules, that is, 6.3 ampoules, then we
would obtain :
6.3 ampoules
of hyperton are 63 ml
If we remove
500 cc of the 0.9% saline solution, those 63 mml would remain 437 ml.
437 mL of
0.9% saline solution has 67.298 mEq of sodium by rule of 3.
And the 6.3
hyperton ampules would be 6.3 * 30 = 189 mEq
67,298+189=
256,298
Rounded
to 257 mEq of Na
And if we
look at the table above
500 cc of 3%
saline solution has 257 mEq of Na or 15 grams of NaCl
So the prompt
would be:
Gauge 6.3
hypertonic ampoules in 500 cc of 0.9% saline solution….. and thus we would
obtain 500 cc of 3% saline solution
The same
procedure would apply when splitting those 500 cc in half and
gauging them.
To obtain 250
cc of 3% saline solution, the indication would be:
Dilute 3.15
(half of 6.3) hypertonic ampoules in 250 cc (half of 500) of 0.9% saline
solution, and thus we would obtain 250 cc of 3% saline solution.
I HOPE IT IS USEFUL FOR UNDERSTANDING
KNOWLEDGE AND ITS USEFULNESS IN THE CLINIC.
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