Randy Recommended DIY Two Part Recipes

[Randy Holmes-Farley]

Will calcium sulfate redissolve in the aquarium?

Yes, assuming it is in adequate flow.

 

[Randy Holmes-Farley]

I'm kinda lost too. Let's redo the math.
Since your recommendation is doubling Balling Part C, then:

• 594 grams of baking soda = 163 grams of sodium = 19.5 grams of magnesium needed to offset sodium chloride buildup (12% of 163 grams).
• Balling Part C has 3350 mg/L of magnesium. So that’s 3350 x 3.78 x 2 (2 gallons added for every 1 gallon of Recipe #1) = 25.3 grams of magnesium. That would be 15.5% of 163 grams, which exceeds the 12%.

To calculate the additional magnesium needed for coral consumption, we actually need to subtract the excess from Balling Part C (25.3 - 19.5) / 47 g/L = 123 ml of Part C from your recipe needs to be added on top of Balling Part C.

Correct me if I’m wrong,
Thanks!

Where did you get the Part C magnesium concentration? Perhaps they calculated the balanced amount in a different way or maybe they or I made a mistake.

 

[VLJ]

Where did you get the Part C magnesium concentration? Perhaps they calculated the balanced amount in a different way or maybe they or I made a mistake.

Concentration of each Part can be found here:
Tropic Marin® Original Balling Components

 

[VLJ]

I was curious enough to calculate at which ratio Balling Part C magnesium is balanced to their Parts A and B.

Given that the concentration of Balling Part B is 2800 dKH/L and is a mix of carbonate and bicarbonate with a ratio of about 20%:80% (calculated based on the concentration), we can easily calculate how magnesium is balanced.

Alkalinity part:
The mixture of 286 grams of 20% sodium carbonate and 80% sodium bicarbonate contains approximately 87 grams of sodium in one gallon.
0.12 x 87 = 10.4 grams

Calcium part:
The 288 grams of calcium chloride dihydrate (CaCl₂·2H₂O) contains approximately 78 grams of calcium in one gallon.
0.065 x 78 = 6 grams

Totaling 16.4 grams of magnesium deposition for every gallon dosed, based on Randy’s calculations in the previous post.

Now, since Balling Part C is concentrated at 3350 mg/L of magnesium, 3350 x 3.78 (liters in a gallon) = 12.6 grams.

That's a little shy of 16.4 grams.

Would be grateful if @Hans-Werner could chime in on how magnesium is balanced in the original Balling method.

Thanks!