Barr gh booster

[Soilwork]

Why the 3:3:1 ratio in the gh booster?

Does anyone know how to calculate how much of this will raise a said amout of water by x degrees dGH without using a calculator?

Thanks.

 

[dw1305]

Hi all,

Why the 3:3:1 ratio in the gh booster?

Does anyone know how to calculate how much of this will raise a said amout of water by x degrees dGH without using a calculator?

Thanks.

I don't know why he decided on a 3:3:1 ratio of K2SO4:CaSO4.2H2O:MgSO4.7H2O. There is discussion at <"EI light:..">. We have a thread which <"specifically looks at magnesium">.

The name is a bit of a misnomer as well, because potassium sulphate doesn't add any dGH (K+ is a monovalent ion).

Calcium sulphate (Gypsum) is <"pretty insoluble">. The other two salts are soluble.

cheers Darrel

 

[Manuel Arias]

Barr GH booster is just a slight variant of any normal GH booster.

The numbers 3:3:1 means that final products is made of 7 parts, with 3 of the parts being potassium sulfate K2SO4, 3 parts calcium sulfate CaSO4, 1 part magnessium sulfate MgSO4. In general terms GH in water corresponds to a 75% Ca and 25% Mg, reason why normal GH boosters are just 4 parts, 3 of CaSO4 and one of MgSO4. The addition of potassium is just to reach some levels of potassium in the water to start with, which is totally optional but, under my experience, it can be always a good addition, as it is part of the NPK relationship. Calling it Barr GH booster is just a bit of hype, as K2SO4 is used regularly for most people using dry ferts to adjust potassium levels, anyway.
The reason why I say this is because potassium is not taken into account into the measurement of GH, so adding it in a GH booster is just unnecessary:

https://www.theaquariumwiki.com/General_Hardness

Knowing that K2SO4 will not change our hardness, is only the other two elements that have any effect. GH is measured in dGH or dH, and one degree corresponds to 10 mg/l of CaO. As calcium oxide has a molar weight of 56,0774 gr/l, the amount of mol of CaO into 10 mg is:

1dGH = 10 (mg/l) / (1000 (mg/gr) * 56.0774 (gr/mol) = 0.0001783 M,

where M = mol/l.

Now, GH is the measure of both Ca and Mg but as if every bit of them was accounted as CaO. However, as we are "counting" only the Ca and Mg, in terms of mols it is true that:

GH (mol) = CaO(mol) = Ca (mol)

This happens because there is a single atom of Ca inside CaO, so relationship is one to one.

Now, in the "Barr" GH booster, we have a 3:1 in terms of Ca and Mg. Both CaSO4 and MgSO4 have also a single atom of Ca/Mg, so the addition of a mix of Ca/Mg SO4 will also raise GH depending on the number of mols you add of it. In other words: To raise GH into 1 degree, you need to increase the concentration of Ca+Mg in water into 0.0001783 mol/l. As the formula includes 3 parts of Ca and 1 part of Mg, this means:

Amount of CaSO4 (in mol) = 0.0001783 M * (3/4) = 0.00013374 M
Amount of MgSO4 (in mol) = 0.0001783 M * (1/4) = 0.00004458 M

Now, using the molecular weights of these molecules, we can transform these values into gr/l:

CaSO4 (gr/mol) = 136.14
MgSO4 (gr/mol) = 120.366

Amount of CaSO4 (in gr/l) = 0.00013374 (mol/l) * 136.14 (gr/mol) = 0.0182 gr/l
Amount of MgSO4 (in gr/l) = 0.00004458 (mol/l) * 120.366 (gr/mol) = 0.00537 gr/l

So, adding such concentrations of both products allows you to increase 1dGH in 1 litre of water. The way to determine the total amounts you need to increase it X dGH into an aquarium with a volume of water of V then is obtained by multiplying these number by the wished increase in GH and by the number of litres of water:

Total amount of CaSO4 (gr) = 0.0182 (gr/l) * X dGH * V (l)
Total amount of MgSO4 (gr) = 0.00537 (gr/l) * X dGH * V (l)

For instance, in an example: I have an aquarium of 100L and I use RO water, which has 0 dGH to start with. I wish to reach 4dGH. How much of each substance I have to add?

Total amount of CaSO4 (gr) = 0.0182 (gr/l) * 4 dGH * 100 (l) = 7.28 gr
Total amount of MgSO4 (gr) = 0.00537 (gr/l) * 4 dGH * 100 (l) = 2.15 gr

If you have a pre-made mix of Mg/CaSO4 in a proportion 3:1 (in terms of volume), then the total amount you need of the mix to raise 1 one dGH one litre of water is just the addition of terms:

Total amount of GH booster (gr): 0.02357 (gr/l) * X dGH * V (l)

And in our example, this would mean:

Total amount of GH booster (gr): 0.02357 (gr/l) * 4 dGH * 100 (l) = 9.43 gr.

Now, we consider the "Barr" GH booster. Then we need to add to the mix the amount of K2SO4 included into it. As GH can be only raised by added Ca and Mg, the amounts of above must be preserved. However, we add now also K2SO4, which due to the relationship 3:3:1 means that we add the same amount, in mols, than we add CaSO4. This means:

Amount of K2SO4 (in mols) = 0.00013374 M

Knowing the molecular weight of K2SO4:

K2SO4 (gr/mol) = 174.259

Amount of CaSO4 (in gr/l) = 0.00013374 (mol/l) * 174.259 (gr/mol) = 0.0233 gr/l

Following the same idea as above, regarding the amounts needed to reach GH in water in one degree:

Total amount of K2SO4 (gr) = 0.0233 (gr/l) * X dGH * V (l)
Total amount of CaSO4 (gr) = 0.0182 (gr/l) * X dGH * V (l)
Total amount of MgSO4 (gr) = 0.00537 (gr/l) * X dGH * V (l)

And in our example:

Total amount of K2SO4 (gr) = 0.0233 (gr/l) * 4 dGH * 100 (l) = 9.32 gr
Total amount of CaSO4 (gr) = 0.0182 (gr/l) * 4 dGH * 100 (l) = 7.28 gr
Total amount of MgSO4 (gr) = 0.00537 (gr/l) * 4 dGH * 100 (l) = 2.15 gr

As said before, if you have the right mix of 3:3:1, then the amount of mix you need is

Total amount of "Barr" GH booster (gr): 0.04687 (gr/l) * X dGH * V (l)

And in our example, this would mean:

Total amount of "Barr" GH booster (gr): 0.04687 (gr/l) * 4 dGH * 100 (l) = 18.75 gr.

We can reverse the computations to ensure we did it right. Let's assume then that we have a solution of water with a volume of 100 l and in which we have added 18.75 gr of a mix of K2SO4, CaSO4 and MgSO4 in molar proportion of 3:3:1. What is the expected GH?

To find this, we need to determine the total amount of Mg and Ca added to water and then translate this to equivalent in CaO. To do so, we need to discover how many mols of Ca and Mg we have added, which then needs some calculations. We now the equivalence in mols between the three substances (3:3:1), which means that, in terms of mols, 3 parts are from K2SO4, 3 parts from CaSO4 and 1 part is MgSO4. We also know the molecular weights of these components, so:

(3 / 7) * mK2SO4 + (3 / 7) * mCaSO4 + (1 / 7) * mMgSO4 = 150.221

So this is the apparent molecular weight of the mix. Knowing that, we can compute the apparent mols of the added mix. Essentially:

Mix of GH booster (mols) = 18.75 gr / (150.221 (gr / mol) = 0.124815 mol

And as per the proportions, we know that:

K2SO4(mol) = 0.124815 * (3/7) = 0.05349 mol
CaSO4(mol) = 0.124815 * (3/7) = 0.05349 mol
MgSO4(mol) = 0.124815 * (1/7) = 0.01783 mol

Now, as GH is only related to Ca and Mg, and both ions are in just one unit in the corresponding substances, the total added Ca and Mg is:

Ca+Mg = CaSO4 (mol) + MgSO4(mol) = 0.071321 mol

As GH is measured as mg/l of CaO, and CaO counts also with only one atom of Ca:

CaO (mol) = (Ca+Mg) (mol) = 0.071321 mol

Converting that into mass:

CaO (gr/mol) = 56.0774

CaO(gr) = .071321 (mol) * 56.0774 (gr/mol) = 3.9994 = ~ 4gr.

Now, as 1 dGH is equal to 10 (mg/l) of CaO, we need to find the concentration in the water. As the tank has 100 litres, and we have equivalently dissolved 4 gr of CaO, then concentration in mg/l is:

CaO(mg/l) = 4 gr * 1000 (mg / gr) / 100 (l) = 40 mg/l.

Whis is then equivalent to:

GH = CaO (mg/l) / 10 (mg/l) = 40 / 10 = 4.

Which perfectly matches the value we were expecting. Hence, calculus is correct.

--Notes--

Now, be aware that this is only valid if you are using salts that are not hydrated. The results will be different if you do so. However, the correction is easy to do, and is just a matter of following the same calculus but just updating the associated molecular weights of the salts. As pointed out by Darrel, most of occasions is much better use hydrated salts because they are much more soluble.
For instance, if you use the ones he mentioned: CaSO4.2H2O and MgSO4.7H2O, then the corresponding molecular weights are

CaSO4.2H2O (gr/mol) = 172.1712
MgSO4.7H2O (gr/mol) = 246.4746

For K2SO4 there is no this problem, unless you use KHSO4, which is also quite common (and usually easier to find). If this is the case then:

KHSO4 (gr/mol) = 136.1688

As said, just replace the corresponding molecular weights in the forms of the above to get the right weights based in any new molecular weight, which depends on the chemical form of the components you use. But taking that into account the results are easy.
Considering that mix is done as per Darrel's proposition: K2SO4:CaSO4.2H2O:MgSO4.7H2O, then

Corrected Total amount of "Barr" GH booster (gr)= 0.04687 (gr/l) * X dGH * V (l) * [(mK2SO4+mCaSO4.2H2O +mMgSO4.7H2O ) / (mK2SO4+mCaSO4+mMgSO4) ] = 0.04687 * [ (174.259 + 172.1712 + 246.4746) / (174.259+136.14+120.366) ] * X dGH * V(l)

Corrected Total amount of "Barr" GH booster (gr) = 0.06451 (gr/l) * X dGH * V(l)

In our example of 4 dGH and 100 litres:

Corrected Total amount of "Barr" GH booster (gr) = 0.06451 (gr/l) * 4 dGH * 100 (l) = 25.80 gr.

Hope this help.

Cheers,
Manuel

 

[Soilwork]

Thanks Darrel. Is it true the values needed are derived from the total amount of dry salts trust make up the gh booster.

From what I have seen. To purchase this stuff you get a 1lb bag 450g. A 3:3:1 ratio in this respect would be 192g of potassium sulfphate 192g of calcium sulphate and the rest magnesium sulphate. This would mean an amount in tsp/Tbs of this mixture in X amount of water would rise dGH by Y amount? Is that what the ratio means here?

 

[Manuel Arias]

From what I have seen. To purchase this stuff you get a 1lb bag 450g. A 3:3:1 ratio in this respect would be 192g of potassium sulfphate 192g of calcium sulphate and the rest magnesium sulphate. This would mean an amount in tsp/Tbs of this mixture in X amount of water would rise dGH by Y amount? Is that what the ratio means here?

I am afraid that is not that simple. Amounts you need to add are also related to volume of the tank, and the proportion is in terms of number of molecules, which is not the same than weight. In fact, by doing the approach in weight, you can be ending adding too low Mg if you are using, for example, MgSO4.7H2O. No that 246.48 gr of MgSO4.7H2O have the same number of molecules than 172.17 gr of CaSO4.2H2O. This means that proportion in weight of 3:1 will give you 3 atoms of Ca for each 0.7 of Mg, so a 30% less of Mg than the target (75% Ca, 25% Mg). And that only considering these two substances. Error will be larger if you do the approach including to K2SO4.

The way to compute this is as I explained above, but using the mixture given by Darrel:

"Barr" GH booster (gr) = 0.06451 (gr/l) * X dGH * V(l)

where X is the number of dGH you wish to add, and V the volume of the tank. Transposing that with accuracy into tsp/Tbs can be quite tricky. However, considering that GH is not required with that precision, you can approach it. The resulting amount in grams can be converted in Tbs by dividing roughly by 15 (1 Tbs is about 15 gr). And in tsp by dividing by 5 (1 tsp is about 5 gr).

Hope this help.

Cheers,
Manuel

 

[Soilwork]

Thanks for taking the time to respond Manuel. This is really good stuff. I'm kinda following it as I don't have any background in chemistry it is taking time but I understand what is going on here.

Sorry I wasn't clear before. I wasn't expecting it to be that simple. I knew there is much more involved to get to the answer when considering our water volume and target dGH. I was just trying to clarify the parts/ratio aspect.

If you look here http://www.aquariumfertilizer.com/index.asp?Option1=inven&EditU=2&Regit=7

You can buy the 3:3:1 mixture premade in a bag that looks to weigh 1lb or 450g

Therefore it is true that of our 7 parts the ratio in this case would be 192:192:64g from this we can then use the molecular weights and formulas you discussed to calculate the dGH for a given volume of water in litres.

There is maths in that link also which helps you on your way. I was wondering how this calculation was achieved.

 

[Manuel Arias]

hanks for taking the time to respond Manuel. This is really good stuff. I'm kinda following it as I don't have any background in chemistry it is taking time but I understand what is going on here.

You are welcome, no problem at all. Most difficult part use to be the idea of "mol". A mol is just a way of counting things. The good thing in Chemistry is that is very useful to determine how components/atoms relate each other. Having that in mind, the rest is just to describe how weight relates to this counting so you add the amounts you need, in terms of number of molecules, rather in weight. And rest is just simple algebra.

You can buy the 3:3:1 mixture premade in a bag that looks to weigh 1lb or 450g

Therefore it is true that of our 7 parts the ratio in this case would be 192:192:64g from this we can then use the molecular weights and formulas you discussed to calculate the dGH for a given volume of water in litres.

Yes, they come up with 16 gr to increase 3 dGH in 20 gallons of water. This is exactly the same dosage than Seachem Equilibrium, which also includes potassium. However, as said before, the exact weight will be related to the exact molecular form. Unfortunately, the link you provide is not mentioning this nor specifying the exact salts (surely due to commercial purposes). But considering our numbers, it is quite probable is related to hydrated forms provided by Darrel. Note that MgSO4 can come in different hydratation levels, which also complicates a bit this. That is why I gave a general approach that will work in any case by just adjusting the molecular weights. W can compute their coefficient, though:

20 gallons = ~ 80 litres.

Coefficient = 16 gr / (3dGH*80 l) = 0.0666 (gr/l) of the mixture.

So, in their case:

"Barr" GH booster (gr) = 0.0666 (gr/l) * X dGH * V(l)

Note that value is practically the same than I computed. Differences will come only from rounding in the computations and it is about 3% error, which in terms of GH is nothing.

Cheers,
Manuel

 

[dw1305]

Hi all,

As pointed out by Darrel, most of occasions is much better use hydrated salts because they are much more soluble.

It isn't just a solubility issue, they will be always be hydrated salts when you use them, because they pick up atmospheric moisture. You can get MgSO4 etc, but you would need to heat MgSO4.7H2O to drive the "water of crystallization" off, and then keep it in a desiccator until you want to use it.

I would always assume that "MgSO4" really refers to MGSO4.7H2O ("Epsom Salts") unless it specifically mentions heating the salt, and that you only have ~10% Mg.

To purchase this stuff you get a 1lb bag 450g. A 3:3:1 ratio in this respect would be 192g of potassium sulfphate 192g of calcium sulphate and the rest magnesium sulphate.

That would be what Tom Barr wrote in the original recipe (454/7 ~ 65g and 3 x 65g ~195g).

As Manuel's, very informative, posts says to get from the salt to how much dGH it adds requires a bit of maths, due to the strange derivation of <"dGH">.

cheers Darrel

 

[Soilwork]

Thanks Darrel. For me I am not too interested in the maths. This is for the benefit of someone else who was asking in another forum. Hopefully, with their chemical background this very informative and much appreciated thread will be if more use to them than to me.

I don't typically worry too much about gh and I use tapwater. I have never seen any issues relating to calcium or magnesium deficiencies whilst doing this. The ratio thing was interesting though as Barr is adamant that plants do not care about ratios just non limiting numbers.

I understand though that in this case we are just trying to achieve a higher dGH which will hopefully address any potential deficiencies in these minerals.

Thanks again to both.

 

[Manuel Arias]

That would be what Tom Barr wrote in the original recipe (454/7 ~ 65g and 3 x 65g ~195g).

This interesting and would explain too part of the 3% of difference I find, which will be related to do the mix in terms of mass and not in mol. Following the idea:

195 gr CaSO4.2H2O = 1.1326 mol
195 gr K2SO4 = 1.1190 mol
64 gr MhSO4.7H2O = 0.2597 mol

Now, proportion Mg/Ca should be 0.25. In this case: 0.23. So error is: 0.25-0.23 = 0.02 => 2% down in terms of Mg. Not important so the approach works well.

understand though that in this case we are just trying to achieve a higher dGH which will hopefully address any potential deficiencies in these minerals.

Yes, you are right here. The important point is to ensure the elements are not in deficiency. The exact ratios are not that critical (in Nature these ratios change a lot).

Cheers,
Manuel

 

[Soilwork]

You are welcome, no problem at all. Most difficult part use to be the idea of "mol". A mol is just a way of counting things. The good thing in Chemistry is that is very useful to determine how components/atoms relate each other. Having that in mind, the rest is just to describe how weight relates to this counting so you add the amounts you need, in terms of number of molecules, rather in weight. And rest is just simple algebra.



Yes, they come up with 16 gr to increase 3 dGH in 20 gallons of water. This is exactly the same dosage than Seachem Equilibrium, which also includes potassium. However, as said before, the exact weight will be related to the exact molecular form. Unfortunately, the link you provide is not mentioning this nor specifying the exact salts (surely due to commercial purposes). But considering our numbers, it is quite probable is related to hydrated forms provided by Darrel. Note that MgSO4 can come in different hydratation levels, which also complicates a bit this. That is why I gave a general approach that will work in any case by just adjusting the molecular weights. W can compute their coefficient, though:

20 gallons = ~ 80 litres.

Coefficient = 16 gr / (3dGH*80 l) = 0.0666 (gr/l) of the mixture.

So, in their case:

"Barr" GH booster (gr) = 0.0666 (gr/l) * X dGH * V(l)

Note that value is practically the same than I computed. Differences will come only from rounding in the computations and it is about 3% error, which in terms of GH is nothing.

Cheers,
Manuel

I didn't see this post. Thank you again.