Second New DIY Two Part Recipe with Higher pH Boost

jeffp1

Active Member
View Badges
Joined
Feb 28, 2015
Messages
229
Reaction score
155
Location
Walnut Hill, Fl
Rating - 0%
0   0   0

Phillyd1990

Well-Known Member
View Badges
Joined
Apr 22, 2017
Messages
512
Reaction score
235
Rating - 0%
0   0   0
This might be a dumb question but can i use just the alk solution to help with ph and continue to use my current cal and mag (esv)
 
OP
OP
Randy Holmes-Farley

Randy Holmes-Farley

Reef Chemist
View Badges
Joined
Sep 5, 2014
Messages
72,100
Reaction score
69,741
Location
Massachusetts, United States
Rating - 0%
0   0   0
This might be a dumb question but can i use just the alk solution to help with ph and continue to use my current cal and mag (esv)

In general, yes. What brand of calcium?
 
OP
OP
Randy Holmes-Farley

Randy Holmes-Farley

Reef Chemist
View Badges
Joined
Sep 5, 2014
Messages
72,100
Reaction score
69,741
Location
Massachusetts, United States
Rating - 0%
0   0   0
Ok so best to just use the esv calcium chloride recipe with the alk recipe.

Not sure what you mean, but expect to dose less of the ESV calcium than the DIY recipe alk part. It would no longer be 1:1.
 

nanomania

Valuable Member
View Badges
Joined
Apr 12, 2016
Messages
1,873
Reaction score
365
Rating - 0%
0   0   0
If i want to add strontium chloride in the cal/mag part, howmuch do i add? I know u dont support strontium, but i i want to add as iv been using seachem reef adv cal.
 
OP
OP
Randy Holmes-Farley

Randy Holmes-Farley

Reef Chemist
View Badges
Joined
Sep 5, 2014
Messages
72,100
Reaction score
69,741
Location
Massachusetts, United States
Rating - 0%
0   0   0

Tmmste

Active Member
View Badges
Joined
Aug 15, 2016
Messages
285
Reaction score
90
Rating - 0%
0   0   0
If i want to add strontium chloride in the cal/mag part, howmuch do i add? I know u dont support strontium, but i i want to add as iv been using seachem reef adv cal.
If you are aiming for 1:50 mole ratio; if you use calcium chloride dehydrate 55,5gr would be 20 gr calcium / 40 gr ( = 1 mole calcium) which is 0.5 mole. 50% of this is 0,02 x 0,5 mole = x mole strontium. 1 mole of strontium is 87,62 gram. x mole x 87,62 = x gram strontium. x gram strontium / mass % (32,863%) = 2,6 gram of strontium chloride hexahydrate... this does not include the amount lost which the amount of water removed to compensate for the rise of salinity. Maybe this overview may help you out;

0d3e0366e27e307d9db445dd506c888a.jpg
 
Last edited:

nanomania

Valuable Member
View Badges
Joined
Apr 12, 2016
Messages
1,873
Reaction score
365
Rating - 0%
0   0   0
If you are aiming for 1:50 mole ratio; if you use calcium chloride dehydrate 55,5gr would be 20 gr calcium / 40 gr ( = 1 mole calcium) which is 0.5 mole. 50% of this is 0,02 x 0,5 mole = x mole strontium. 1 mole of strontium is 87,62 gram. x mole x 87,62 = x gram strontium. x gram strontium / mass % (32,863%) = 2,6 gram of strontium chloride hexahydrate... this does not include the amount lost which the amount of water removed to compensate for the rise of salinity.
500 g of calcium chloride dihydrate plus 261.2 g of magnesium chloride hexahydrate and 8.3grams of strontium chloride in a total of 1 gallon..

THIS IS WHAT RANDY APPROVED IN THE OLDER 2part post.
 

Tmmste

Active Member
View Badges
Joined
Aug 15, 2016
Messages
285
Reaction score
90
Rating - 0%
0   0   0
500 g of calcium chloride dihydrate plus 261.2 g of magnesium chloride hexahydrate and 8.3grams of strontium chloride in a total of 1 gallon..

THIS IS WHAT RANDY APPROVED IN THE OLDER 2part post.
55.5 anhydrate = 73,5 dihydrate for 2.6gr strontium. 500 / 73.5 = a factor of 6.8 x 2.6 = 17.7 gr. This is identical to Balling & DSR Easy ratio for strontium. The ratio of randy is not 1:50 but more towards 1:100.
 

nanomania

Valuable Member
View Badges
Joined
Apr 12, 2016
Messages
1,873
Reaction score
365
Rating - 0%
0   0   0
55.5 anhydrate = 73,5 dihydrate for 2.6gr strontium. 500 / 73.5 = a factor of 6.8 x 2.6 = 17.7 gr. This is identical to Balling & DSR Easy ratio for strontium. The ratio of randy is not 1:50 but more towards 1:100.
Im confused, so do u add 2.6g or 17.7g? Instead of 8.3g
 
OP
OP
Randy Holmes-Farley

Randy Holmes-Farley

Reef Chemist
View Badges
Joined
Sep 5, 2014
Messages
72,100
Reaction score
69,741
Location
Massachusetts, United States
Rating - 0%
0   0   0

Tmmste

Active Member
View Badges
Joined
Aug 15, 2016
Messages
285
Reaction score
90
Rating - 0%
0   0   0
Im confused, so do u add 2.6g or 17.7g? Instead of 8.3g
Wel.. I will add a bit more than 2.6gr per 55.5gr (for dihydrate 2.6 per 73,5gr) as I will also compensate for the salinity correction. I am not going to exactly follow the new recipe as I am missing a lot of sulfate in the recipe to compensate for the amounts removed with the salinity correction.. (randy is this looking at this I think). I will probably do this (also including trace elements)
7272f77e3570716d9072ac5e09be4c7f.jpg
 
OP
OP
Randy Holmes-Farley

Randy Holmes-Farley

Reef Chemist
View Badges
Joined
Sep 5, 2014
Messages
72,100
Reaction score
69,741
Location
Massachusetts, United States
Rating - 0%
0   0   0
. I am not going to exactly follow the new recipe as I am missing a lot of sulfate in the recipe to compensate for the amounts removed with the salinity correction..

Can you clarify what that sentence means?

You add less sulfate to offset sulfate that is being lost in the salinity correction? That doesn't seem right, does it?
 

Tmmste

Active Member
View Badges
Joined
Aug 15, 2016
Messages
285
Reaction score
90
Rating - 0%
0   0   0
Can you clarify what that sentence means?

You add less sulfate to offset sulfate that is being lost in the salinity correction? That doesn't seem right, does it?

I will add much more sulfate. 37,52 gr per 40gr sodium hydroxide to be exact. This 37,52gr contains 8,47 gr of sulfate. which is 14% of the 60,54gr Chloride which is added by the stock solution. Around 3.1 L of aquariumwater is removed by the end of dosing this 1 liter of this stock (depends on the concentration but the principle remains the same). Apart from 60,54 chloride 35.15gr of sodium is added.. in total 95,7 gr of Sodium + chloride... 3.1l aquariumwater contains 30,5 x 3.1 = 95,7 gr of sodium & chloride.

There will be 3.1 l removed per 1l stock to keep salinity stable (again just to illustrate the calculation).. 3.1l contains;
3.1 x 2700 sulfate = 8,47 gr / 0,22574 = 37,52 gr of sodium sulfate. I believe this is accurate:


079b1cebc5bbd3f453601ec5414692a7.jpg
 
OP
OP
Randy Holmes-Farley

Randy Holmes-Farley

Reef Chemist
View Badges
Joined
Sep 5, 2014
Messages
72,100
Reaction score
69,741
Location
Massachusetts, United States
Rating - 0%
0   0   0
This will get very confusing since the sulfate discussion is now appended to a recipe thread that does not include sodium sulfate. I'll continue that discussion on the other thread.
 
Back
Top