Math help

[Lazys Coral House]

Hoping somebody smarter than me can help me out. I'm trying to determine how many ml I need to dose to raise strontium level 1 ppm in 300g.

I dissolved the entire 300 gram jar into 1 gallon of Ro/di. Trying to determine how many ml of the solution will raise strontium level by 1ppm in 300 gallon system.

Here is instructions on container. Thank you for your help

Instructions and Guidelines
Basic : Dissolve 1 gram (~ teaspoon) of product per 40 US-gallons of aquarium water in 8-fl. oz. of fresh water; add every other day or as needed to maintain the strontium concentration within a range of 8 - 12 ppm. When used in this fashion, 1,200 g treats up to 48,000 US-gallons (181,700 L).
Advanced: Create a stock solution by dissolving 2 grams (~1 teaspoon) of Strontion-P in 8-fl. oz. of fresh water (preferably purified); each ml of the solution will increase the strontium concentration ("[Sr2+]") in 1 US-gallon (3.785 L) of water by 0.75 ppm. [Reference: Each g of Strontion-P will increase the [Sr2+] in 1 US-gallon (3.785 L) of water by 143 ppm.] If initial [Sr2+] in aquarium is below 412 ppm, add stock sol'n at maximum rate of 10 ml per 20 US-gallons daily until desired [Sr2+] is attained, then dose daily or weekly as needed (see below). Maintain [Sr2+] within a range of +/-2 ppm. Once desired [Sr2+] has been acquired, measure aquarium's [Sr2+] at the same time each day over a one- to two-week period to determine the daily rate of strontium uptake (i.e. the decrease in strontium). To determine daily dosing rate (preferable to weekly dosing): estimate volume of water in entire aquarium system (US-gal.); divide the daily decrease in [Sr2+] by 0.75; multiply this number by volume of water in system to obtain daily ml of stock sol'n required to maintain stable [Sr2+].

 

[abecker]

128 oz/gal / 8oz/solution = 16
300grams/2grams/16 = 9.375 times potency
So if 1 ml of their solution will raise 1 gallon by 0.75ppm, 1.33ml will raise 1 gallon 1ppm.
1.33/9.375 = 0.142ml of your solution to raise 1 gallon 1ppm.
0.142 * 300 = 42.6ml to raise 1 ppm.

I believe my math is correct on this, but I would wait for a second answer to confirm.

 

[Marcus E. Mommer]

Sorry if I am incorrect here, but I am getting a different answer, this may be a result of different solution concentrations, but my resulting answer is 400ml of stock solution.
You are mixing 2grams of powder in with 8fl oz of RO water; 8fl oz translating to 236.56ml. The instructions say that each one of those ml should increase the water by 0.75ppm, then the 236.56 ml will only increase the water of aprom. 177gallons of 1ppm per gallon. (1.33ml per 1ppm per gallon=> 236.56ml / 1.33 ml resulting in only treating 177gallons)

So 300g water 1ppm Sr2+: 300gallons * 1.33 ml per 1ppm should result in 400ml
(400 ml of the solution you create by adding 2 grams of powder to 8fl oz)

Another way to think of it is that each gram of powder raises the 1g of water by 143ppm. So 1g of powder regardless of how much water you mix it with would raise the Sr2+ in your 300g tank by 0.47ppm. With that math you would need 2.09 grams of powder mixed with however much RO water you want to add in a single dose to raise your tank 1ppm.

Though i notice that based off their statement of 8fl oz of solution changing the Sr2+ by 0.75ppm you would need 400ml of solution to change the Sr2+, an amount that at given instructions requires 3.381grams of powder vs. the instructions in grams resulting in you only needing 2.09grams. There may be a inconsistency in the instructions. They also make it more complicated by using both English and metric units in a single instruction.

 

[abecker]

Sorry if I am incorrect here, but I am getting a different answer, this may be a result of different solution concentrations, but my resulting answer is 400ml of stock solution.
You are mixing 2grams of powder in with 8fl oz of RO water; 8fl oz translating to 236.56ml. The instructions say that each one of those ml should increase the water by 0.75ppm, then the 236.56 ml will only increase the water of aprom. 177gallons of 1ppm per gallon. (1.33ml per 1ppm per gallon=> 236.56ml / 1.33 ml resulting in only treating 177gallons)

So 300g water 1ppm Sr2+: 300gallons * 1.33 ml per 1ppm should result in 400ml
(400 ml of the solution you create by adding 2 grams of powder to 8fl oz)

Another way to think of it is that each gram of powder raises the 1g of water by 143ppm. So 1g of powder regardless of how much water you mix it with would raise the Sr2+ in your 300g tank by 0.47ppm. With that math you would need 2.09 grams of powder mixed with however much RO water you want to add in a single dose to raise your tank 1ppm.

Though i notice that based off their statement of 8fl oz of solution changing the Sr2+ by 0.75ppm you would need 400ml of solution to change the Sr2+, an amount that at given instructions requires 3.381grams of powder vs. the instructions in grams resulting in you only needing 2.09grams. There may be a inconsistency in the instructions. They also make it more complicated by using both English and metric units in a single instruction.

I read his post differently that OP didn't make a stock solution per the instructions but dissolved 300 grams into 1 gallon of water. If OP made the stock solution it would have only dissolved 32 grams into the gallon of water.

 

[Marcus E. Mommer]

Indeed he did, missed that part, my bad. Read the premiss then went strait to the instructions. result is the same though. Guess it just comes up to if your just gonna put the whole container of powder into solution at once or make it as you go. So as far as math goes mine confirms your answer to within +- .025ml

 

[JimWelsh]

The instructions say that the "Reference" for the stock solution is that 1g of the product in 1 US Gallon of water will raise the Sr content by 143 PPM. (One can infer from this, by the way, that the product is anyhdrous SrCl2, with a MW of 158.53, such that 1g in 3.7854 L of water gives a 1/158.53/3.7854 = 0.00167 M solution, and since the AW of Sr is 87.62, then 0.00167 * 87.62 = 0.146, or 146 PPM -- the discrepancy is almost certainly due to the purity of the substance being 98%.)

So, 300 g of the product dissolved in 1 US Gallon of water contains 143 * 300 = 42,900 PPM of Sr. To raise 300 gallons of tank water by 1 PPM, you will need (300 * 3.7854 /42900) = 0.02647 L = 26.47 mL of of the 42,900 PPM solution.

Another way of working this out is by molarity: 300g of SrCl2 = 300 g / 158.53 MW = 1.89 moles. 1.89 moles in 3.7854 L of water = a 0.5 M solution of SrCl2. In order to add 1 PPM of Sr to 300 gallons of water, you need 0.001 g / 87.62 AW * 300 gal. * 3.7854 = 0.01296 moles of Sr. Since your solution is 0.5 M, then 0.01296 moles is in 0.01296 / 0.5 * 1000 = 25.92 mL of solution, which when adjusted for the 98% purity factor is 25.92 / 0.98 = 26.45 mL.

So, the answer to your question is you need 26.5 mL of your 300 g in 1 gallon solution to raise your 300 gallon tank by 1 PPM Sr.

 

[abecker]

I would probably go with Jim on this one.
But, on a bright note @Marcus E. Mommer , we may not have been accurate, but we were precise!

 

[Lazys Coral House]

Thank you all very much, that was helpful. My Stronium is pretty low at 2ppm so I will try adding 50ml for starters and test in 24 hours to see where that puts it. If it raises it appropriately I will dose the rest to bring it to correct level. Thank you again, there are some smart folks here

 

[JimWelsh]

How are you testing for strontium? I see you posted Triton results back in April. Have you done more Triton testing since then, or are you using a kit to test for strontium yourself?

 

[Lazys Coral House]

It is either Red Sea or Saifert. I took test on the day i pulled the sample to baseline it against the Triton test so I could do any error correction. I did another triton test in Aug. Similar results with low Strontium, Iodine and Potassium; high Aluminum and Tin

 

[JimWelsh]

There may be a inconsistency in the instructions.

I agree.

If I use the "2 grams in 8-fl oz. water" formula, while assuming that the product is anhydrous SrCl2, then instead of 1 mL of the solution raising the Sr of 1 gal. of water by 0.75 PPM I get a number closer to 1.25 PPM. If, on the other hand, I assume that the product is SrCl2*6H2O (strontium chloride hexahydrate) with its greater MW, then I get 0.73 PPM.

So, the two different statements about the effect of adding the product are inconsistent, but one is consistent with the product being the anhydrous salt, and the other is consistent with the product being the hexahydrate.

The prudent approach is to assume the product is the more concentrated anhydrous salt, and use my earlier dose of 26.5 mL, but that may not increase the Sr as much as expected. If it is the less concentrated salt, then the correct dose would be 26.5 * 266.62 / 158.53 = 44.57 mL.

 

[mitch91175]

I am in the same boat as the OP here. I am starting the path of making stock solutions to dose a 240G aquarium. It would be nice to have a stock solution of 1000ml already made to increase my Strontium levels in my system without using 1/3rd of the premixed solution. I would like to dose no more than 50ml to increase my 240G Strontium level by say 1ppm. That way I do not have to mix the solution so frequently.

Trying to determine the amount from the instructions from Brightwell has my head hurting, lol.

Anyone out there can help me determine how much Brightwell Strontion-P to add to 1000ml of water that will allow me to dose 50ml to increase the Strontium by 1ppm using 1000ml water?

I will be using a Salifert Strontium test kit to measure the consumption.

 

[JimWelsh]

In order to raise the Sr concentration of 240 gallons of water by 1 mg/L (PPM), you will need 1 (mg/L) * 240 (gallons) * 3.785 (liters per gallon) = 908 mg of Sr. Your requirement is that 50 mL of your stock solution will contain this 908 mg of Sr. The Brightwell Strontion-P product page describes the product as being anhydrous SrCl2, which agrees with the claim of at least 54% Sr (Atomic weight of Sr = 87.62, while the molar mass of anhydrous SrCl2 = 158.53). So, in order to have 908 mg of Sr per 50 mL of solution, you will need 0.908 * 158.53 / 87.62 = 1.643 grams of SrCl2 in each 50 mL of solution, which implies that you will need 20 times that much per liter, or 32.857 grams of Strontion-P per liter.

 

[mitch91175]

In order to raise the Sr concentration of 240 gallons of water by 1 mg/L (PPM), you will need 1 (mg/L) * 240 (gallons) * 3.785 (liters per gallon) = 908 mg of Sr. Your requirement is that 50 mL of your stock solution will contain this 908 mg of Sr. The Brightwell Strontion-P product page describes the product as being anhydrous SrCl2, which agrees with the claim of at least 54% Sr (Atomic weight of Sr = 87.62, while the molar mass of anhydrous SrCl2 = 158.53). So, in order to have 908 mg of Sr per 50 mL of solution, you will need 0.908 * 158.53 / 87.62 = 1.643 grams of SrCl2 in each 50 mL of solution, which implies that you will need 20 times that much per liter, or 32.857 grams of Strontion-P per liter.

Thanks for the quick response bro. Will make up the solution tomorrow

 

[mitch91175]

In order to raise the Sr concentration of 240 gallons of water by 1 mg/L (PPM), you will need 1 (mg/L) * 240 (gallons) * 3.785 (liters per gallon) = 908 mg of Sr. Your requirement is that 50 mL of your stock solution will contain this 908 mg of Sr. The Brightwell Strontion-P product page describes the product as being anhydrous SrCl2, which agrees with the claim of at least 54% Sr (Atomic weight of Sr = 87.62, while the molar mass of anhydrous SrCl2 = 158.53). So, in order to have 908 mg of Sr per 50 mL of solution, you will need 0.908 * 158.53 / 87.62 = 1.643 grams of SrCl2 in each 50 mL of solution, which implies that you will need 20 times that much per liter, or 32.857 grams of Strontion-P per liter.

Any particular reason you’d suggest against making the stock solution so potent? I’ll manually dose this and not have it on a doser to ensure not dosing too much.

 

[JimWelsh]

Any particular reason you’d suggest against making the stock solution so potent? I’ll manually dose this and not have it on a doser to ensure not dosing too much.

Not really. It's quite soluble, and should keep indefinitely.

 

[mitch91175]

In order to raise the Sr concentration of 240 gallons of water by 1 mg/L (PPM), you will need 1 (mg/L) * 240 (gallons) * 3.785 (liters per gallon) = 908 mg of Sr. Your requirement is that 50 mL of your stock solution will contain this 908 mg of Sr. The Brightwell Strontion-P product page describes the product as being anhydrous SrCl2, which agrees with the claim of at least 54% Sr (Atomic weight of Sr = 87.62, while the molar mass of anhydrous SrCl2 = 158.53). So, in order to have 908 mg of Sr per 50 mL of solution, you will need 0.908 * 158.53 / 87.62 = 1.643 grams of SrCl2 in each 50 mL of solution, which implies that you will need 20 times that much per liter, or 32.857 grams of Strontion-P per liter.

@JimWelsh going to have to adjust my pre-mix, lol. Tested Strontium today and it was 0mg/L again. Taking too much of pre-made to bring it up. Didn't think that part through enough lol.

So essentially I can take the 32.857 grams * 10 which would be 328.57 grams of Strontium-P per liter. Essentially making the solution 10x stronger. Just want to make sure it's that simple on the math before I make the next batch this weekend. I am pretty sure it'll be depleted by that time.

 

[JimWelsh]

Yes, it is that simple, although at some point you run into issues with solubility, but Wikipedia says that anyhdrous SrCl2 is soluble up to 538 g per liter, so you should be OK.

I would be suspicious of that 0 mg/L result, though. Are you saying that you've dosed almost a whole liter of the original concentration we discussed, and have seen no change in your Sr level? If so, it sounds like it could be testing error to me.

 

[mitch91175]

Yes, it is that simple, although at some point you run into issues with solubility, but Wikipedia says that anyhdrous SrCl2 is soluble up to 538 g per liter, so you should be OK.

I would be suspicious of that 0 mg/L result, though. Are you saying that you've dosed almost a whole liter of the original concentration we discussed, and have seen no change in your Sr level? If so, it sounds like it could be testing error to me.

Hadn't dosed any of the liter that I did make. It was from the original Strontium dosage that I made when I started dosing Strontium on 3/20. First time dosing some of the mix that was made this morning. Dosed 400ml of it.