How to make 50% citric acid solution

I can't seem to find info on how to make a bottle of 50% citric acid solution. Can someone tell me how much citric acid to 3oz of purified water? Thank you very much!

Aloha
Keiko

Hi Keiko,

The best way to do it, if you have a sensitive enough scale, is to do it by weight. Then, you just use equal weights of water and citric acid powder and mix them. This will give you a 50% citric acid solution.

The other way to do it is by using equal volume amounts of both. You would want to pack the citric acid down as much as possible to remove as much air space between the crystals as possible. The problem with this second method is that because of the air spaces between the crystals, you won't have quite a 50% mix, but Jim does say that it will be close enough. I would say, just to be sure to get a little closer to the 50% mix ratio, just add an additional 5% citric acid crystal and that should get you pretty close.

When I did it by weight, I used a digital postal scale that was sensitive to 1/10 of an ounce.

Hope this helps

Scott

Thank you Scott for the information! Really appreciate it. I also was thinking about getting the Sodium Chlorite powder and mix my own. How much of Sodium Chlorite do you use to mix a 4oz bottle? Thank you again!

Aloha
Keiko

Wait a minute...

Does 50% citric acid solution mean, 50% of VOLUME or of WEIGHT is (dry) citric acid?
While for water 50 grams equals 50 milliliters, this may or may not be so for dry citric acid crystals.
If it means by VOLUME, 50 milliliters of acid might very well be only 29 grams...

So, WHICH is it?

See also:
www.ruf.rice.edu/~bioslabs/methods/solutions/formulas.html

Hi tomcat,

It is 50% by weight to be technically correct like Scott said. However, from practical experience, 50% by volume is pretty close. With my own citric acid, it takes 50% by volume plus about another 5-10% by volume to get the weights to be even.

So, if you want to be exact, then do it by weight. If you don't have a scale, then volume is close enough to get working MMS.

Michael

Hi Michael,

Thank you!
You are right, Scott did mention the weight.
Unfortunately, he also did mention "the space between the crystals" and made it sound like, once that problem is solved, one can do it by volume.
However, this is a non-problem, as one can fill in half the volume in water first and then add the crystals.
All of this left me wondering...

Greetings
Thomas

Hi again,

I assume the same is true for a 28% Sodium Clorite solution?
For a total of 100 ml/grams solution I use 28 grams of the dry powder?

Thank you
Thomas

Correcto. And 72 grams water...

With the 80% sodium chlorite powder (approx. 20% salt must be added to make it safe to transport and handle) as 28% of the solution, this makes the solution an actual 22.4% sodium chlorite by weight.

Bruce

please correct me if I'm wrong ... I'm adding this to clarify the above (for me anyway)

1. I want to finish with a 4 oz bottle of std 28% MMS (liquid volume, US oz) --

dissolve 1.6 oz NaClO2 crystals at 80% (weight, avoirdupois oz) into 4 oz pure water (liquid volume, US oz)

thus 1 lb = 16 oz (weight, avoirdupois oz) of NaClO2 (80%) will make ten 4-oz bottles finished MMS.

2. I want to finish with a 4 oz bottle (liquid volume, US oz) of 50% Citric Acid Activator --

dissolve 2.5 oz citric acid crystals at 99% (weight, avoirdupois oz) into 4 oz pure water (liquid volume US oz)

thus 25 oz (weight, avoirdupois oz) of citric acid (99%) will make ten 4-oz bottles of 50% activator solution.

3. thus, for 16 oz of NaClO2 at 80% (weight, avoirdupois oz) you need 25 oz of citric acid at 99% (weight, avoirdupois oz) to produce ten 4-oz bottles each of MMS (28%) and activator (50%)

thanks

Molecule

I believe the number 2 is not going to produce a 50% solution. Is a ratio of 1:1 so, 2.5 ounces of Citric Acid and 2.5 ounces of Water.