2 part imbalance and adjusting

LFS_worker

'ignoramus maximus'
After doing some calculations it seems that after making some of the 2 part stuff that there is an excess of Cl ions, after a full gallon of each component there is 69.552g of free Cl. I am wondering if it would be ok to keep things in balance using NaOH appx. 45.09g dissolved in about 1 gallon of water and added in an amount equal to the amount of each component?

Thanks
Brian
 

Randy Holmes-Farley

Reef Chemist
Premium Member

bertoni

Premium Member
I don't think that's a good idea. Sodium hydroxide will add a lot of alkalinity. Water changes should deal with any ionic imbalance.
 

LFS_worker

'ignoramus maximus'
I think I did something wrong then:
500g CaCl2 = 110.986g/mol
Ca - 40.08 g/mol 36.1126% = 180g Ca
Cl2 - 70.906 g/mol 63.8874% = 319.437

594g 2(NaHCO3)
374g Na2CO3 after baking = 105.99g/mol
Na2 = 45.978 = 43.3795% = 162g Na
CO3 = 60.008 = 56.6166% = 211.746g CO3

1g NaCl weighs : .3933g Na and .6067g Cl
NaCl ratio is 1 : 1.5425

Na Cl
162 X
so 162 x 1.5425 = 249.885g Cl for balance

CaCl2 adds 319.437g Cl leaving excess of 69.552g Cl
 

bertoni

Premium Member
The ratio of ions is what's important, and that doesn't change.

CaCl<sub>2</sub> + Na<sub>2</sub>O<sub>3</sub> -> CaCO<sub>3</sub> + 2 NaCl
 

bertoni

Premium Member
Well, that's not very clear. There will be buildup of NaCl over time, but the Na:Cl ratio remains okay. Did I understand your question?
 

LFS_worker

'ignoramus maximus'
<a href=showthread.php?s=&postid=11670586#post11670586 target=_blank>Originally posted</a> by bertoni
The ratio of ions is what's important, and that doesn't change.

CaCl<sub>2</sub> + Na<sub>2</sub>O<sub>3</sub> -> CaCO<sub>3</sub> + 2 NaCl

CaCl<sub>2</sub> + Na<sub>2</sub>CO<sub>3</sub> -> CaCO<sub>3</sub> + 2 NaCl

I see thanks ...

Is there any reason that it doesn't work when dealing with weight?
Brian
 

bertoni

Premium Member
It should work with the weights, as well. I don't understand the last multiply, but the other numbers are correct, as far as I can see.
 

LFS_worker

'ignoramus maximus'
the alk component adds 162g of Na so to get the ratio correct of 1 : 1.5425

1/162 1.5425/x 162x1.5425 = 249.885

Brian
 

LFS_worker

'ignoramus maximus'
if the math error is due to purity on CaCl being appx 80%

this may also be why 20% less is used with the anhydrogenous (sp)

B
 

bertoni

Premium Member
No, the purity is not the issue with the math. I'll go through the numbers tonight.

Anhydrous CaCl<sub>2</sub> does require 20% less when mixing, though.
 

LFS_worker

'ignoramus maximus'
It is purity I used the numbers that randy got :

500g * .785 (avg%purity) * .36 (%Ca by weight) = 141.3g Ca

I got

500g * 1 *.361126 = 180g Ca

and for the Cl ion
Randy (that genius)
500g * .785 *.64 = 251.2 Cl

And from the "DE DE DE"
500g * 1 *.638874 = 319.437Cl

leaving a margin for error of 2g per Gallon of total mixture.

Thanks for the patience guys!
Brian
 
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