How to prepare a 3% saline solution? In different proportions.

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.