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How to make 50% citric acid solution
- molecule
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16 Apr 2014 16:34 #43210
by molecule
Replied by molecule on topic How to make 50% citric acid solution
thanks Gregorio -- I tried to edit 4 oz water to 2.4 oz water, but edit option had expired. I hope anyone new continues to this page.
I think I finally figured out the numbers -- I was trying to calculate how much Citric Acid to buy
====
1. MMS - 28% solution
1 lb = 16.00 oz (avoirdupois wt) NaClO2 (80% pure) = 453.6 g
453.6 g is 28% of 1620.0 g total (= 453.6 / 0.28)
1620.0 g total – 453.6 g NaClO2 = 1166.4 g H2O (pure)
1166.4 g H2O = 1166.4 cc H2O
1166.4 cc H2O = 39.44 oz (fluid vol US) H2O
from volume 1--
13 standard 4-oz bottles = 52 oz (fluid vol US) requires 1627.2 g H2O (pure)
therefore 1166.4 g H2O produces 37.24 oz (fluid vol US) MMS 28%
(i.e., 1166.4 g / 1627.2 g x 52 oz (fluid vol US) = 37.24 oz (fluid vol US) MMS 28%)
therefore 16.00 oz (avdp wt) NaClO2 produces 37.24 oz (fluid vol US) MMS 28%
2. Citric Acid - 50% activator solution
(1 drop CA-50 per drop MMS, plus 20 second wait, plus 4 oz (fluid vol US) water per 3 drops MMS, ...)
23.00 oz (avdp wt) Citric Acid (99%) = 652.0 g CA (99%)
652.0 g H2O (pure) = 652.0 cc H2O (equal weights of CA and H2O)
652.0 cc H2O = 22.0 oz (fluid vol US) H2O
from this site (and volume 2?)--
16.00 oz (fluid vol US) Citric Acid 50% activator solution requires 280 g H2O (pure)
therefore 652.0 g H2O produces 37.24 oz (fluid vol US) CA 50% activator solution
(i.e., 652.0 g / 280 g x 16.00 oz (fluid vol US) = 37.25 oz (fluid vol US) CA 50% activator)
therefore 23.00 oz (avdp wt) CA produces 37.24 oz (fluid vol US) CA 50% activator solution
we now have equal volumes of MMS and CA-50 activator
summary
16.0 oz (avdp, wt) of NaClO2 (80%) plus
40.0 oz (fluid, vol, US) of H2O (pure)
produces 37.24 oz (fluid vol US) of MMS 28%
23.0 oz (avdp, wt) of CA (99%) plus
22.0 oz (fluid, vol, US) of H2O (pure)
produces 37.24 oz (fluid vol US) of CA 50% activator
each batch will almost fill 10 standard 4-oz bottles
each 4-oz bottle will be just shy of 3-3/4 oz (fluid vol US) of respective solution
(20 std 4-oz bottles total for both solutions)
to order chemicals
16.0 oz (avdp wt) NaClO2 (dry, 80% pure)
23.0 oz (avdp wt) Citric Acid (dry, 99% pure)
for every 1.00 oz (avdp, wt) of NaClO2 (80%), I need 1.44 oz (avdp, wt) of Citric Acid (99%).
(i.e. 23.0 oz CA / 16.0 oz NaClO2 = 1.44 dry weight ratio)
====
I think I finally figured out the numbers -- I was trying to calculate how much Citric Acid to buy
====
1. MMS - 28% solution
1 lb = 16.00 oz (avoirdupois wt) NaClO2 (80% pure) = 453.6 g
453.6 g is 28% of 1620.0 g total (= 453.6 / 0.28)
1620.0 g total – 453.6 g NaClO2 = 1166.4 g H2O (pure)
1166.4 g H2O = 1166.4 cc H2O
1166.4 cc H2O = 39.44 oz (fluid vol US) H2O
from volume 1--
13 standard 4-oz bottles = 52 oz (fluid vol US) requires 1627.2 g H2O (pure)
therefore 1166.4 g H2O produces 37.24 oz (fluid vol US) MMS 28%
(i.e., 1166.4 g / 1627.2 g x 52 oz (fluid vol US) = 37.24 oz (fluid vol US) MMS 28%)
therefore 16.00 oz (avdp wt) NaClO2 produces 37.24 oz (fluid vol US) MMS 28%
2. Citric Acid - 50% activator solution
(1 drop CA-50 per drop MMS, plus 20 second wait, plus 4 oz (fluid vol US) water per 3 drops MMS, ...)
23.00 oz (avdp wt) Citric Acid (99%) = 652.0 g CA (99%)
652.0 g H2O (pure) = 652.0 cc H2O (equal weights of CA and H2O)
652.0 cc H2O = 22.0 oz (fluid vol US) H2O
from this site (and volume 2?)--
16.00 oz (fluid vol US) Citric Acid 50% activator solution requires 280 g H2O (pure)
therefore 652.0 g H2O produces 37.24 oz (fluid vol US) CA 50% activator solution
(i.e., 652.0 g / 280 g x 16.00 oz (fluid vol US) = 37.25 oz (fluid vol US) CA 50% activator)
therefore 23.00 oz (avdp wt) CA produces 37.24 oz (fluid vol US) CA 50% activator solution
we now have equal volumes of MMS and CA-50 activator
summary
16.0 oz (avdp, wt) of NaClO2 (80%) plus
40.0 oz (fluid, vol, US) of H2O (pure)
produces 37.24 oz (fluid vol US) of MMS 28%
23.0 oz (avdp, wt) of CA (99%) plus
22.0 oz (fluid, vol, US) of H2O (pure)
produces 37.24 oz (fluid vol US) of CA 50% activator
each batch will almost fill 10 standard 4-oz bottles
each 4-oz bottle will be just shy of 3-3/4 oz (fluid vol US) of respective solution
(20 std 4-oz bottles total for both solutions)
to order chemicals
16.0 oz (avdp wt) NaClO2 (dry, 80% pure)
23.0 oz (avdp wt) Citric Acid (dry, 99% pure)
for every 1.00 oz (avdp, wt) of NaClO2 (80%), I need 1.44 oz (avdp, wt) of Citric Acid (99%).
(i.e. 23.0 oz CA / 16.0 oz NaClO2 = 1.44 dry weight ratio)
====
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- gjplaceres
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- The truth will make you free!!!
16 Apr 2014 17:01 #43212
by gjplaceres
Replied by gjplaceres on topic How to make 50% citric acid solution
Molecule
You don't need to do all that calculation. MMS distributor have already that math done for you. They sold the reactive Sodium Chlorite and Citric Acid in packages or Kit that you can purchase to make yourself the solution. Check the link of Buy MMS and there one supplier wet4... that have that already setup.
You don't need to do all that calculation. MMS distributor have already that math done for you. They sold the reactive Sodium Chlorite and Citric Acid in packages or Kit that you can purchase to make yourself the solution. Check the link of Buy MMS and there one supplier wet4... that have that already setup.
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