# Boron and Zinc dosing levels



## Zapins (Jul 28, 2004)

Well I was browsing the internet and found this interesting quote. It basically says that in an experiment with wheat adding boron in a 1:2 ratio with boron dosed at 25 ppm and zinc dosed at 50 ppm the plant grew best. Also it seems like boron might regulate the uptake of Fe, Mn, Cu, K and Ca.

Perhaps it would be useful for those interested in squeezing every bit of growth out of a planted tank to raise the boron/zinc levels a bit in order to increase the uptake of other nutrients.



> A pot experiment with wheat plants (Triticum aestivum L. cv. Giza 167) grown on 2.0% or 10.0% CaCO3 in the soil was conducted to study the effect of foliar fertilization of boron, zinc and their combinations on zinc/boron ratio, elemental status and shoot growth. Growth of plants grown on low lime levels was not affected. *The best treatments led to highest growth of shoots grown on soil of high lime content were 25 ppm B or 25 ppm B + 50 ppm* Zn in the spray solution. Boron concentrations were increased in the shoot tissues of plants grown on both CaCO3 levels as boron or zinc was applied. Meanwhile, zinc concentrations were slightly increased when only boron was sprayed; however it decreased when combinations of boron and zinc were applied. Boron or zinc uptake by the shoots grown on low lime level was less affected compared to those grown on high lime level.* Best uptake of boron and zinc was obtained by the low dose of the two elements combination (25 ppm B +50 ppm Zn).* Uptake increases over control obtained by this treatment reached about 230% for boron and 650% for zinc in the plant shoots grown on high lime level. Uptake percentage of other elements by the shoots grown under lime stress conditions was higher than that of plant shoots grown under low lime level in the soil. *Highest uptake of Fe, Mn, Cu, K and Ca were obtained by boron treatments alone or low dose of boron and zinc combinations, which suggested that boron stimulates metabolic functions facilitate absorption and translocation of these nutrients,* especially in plants grown under lime stress conditions


From: http://74.125.113.132/search?q=cach.../+dosing+zinc+plants&cd=8&hl=en&ct=clnk&gl=us


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## Avi (Apr 7, 2004)

Zapins...that's a bit technical for me. If I'd want to try elevating zinc and boron levels, how would I know how much to add to the tank and what form would the additives take?


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## Zapins (Jul 28, 2004)

I think you would add boric acid, and zinc sulfate (not 100% sure about the zinc chemical though maybe someone can confirm?). I've heard its an exceptionally small amount, smaller then a pinky fingernail of the dry compound for a 55g tank.

Could someone provide the dosing levels for both? Kekon?


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## kekon (Aug 1, 2005)

Hi 
When it comes to zinc i added about 0.02 ppm weekly with no negative results. I noticed that my Aromatica really likes higher zinc doses. Also, i noticed that zinc is much less toxic than any other micronutrient. This amount of zinc is also added when using Flourish Trace. I based my micro fert on this product. But boron... is very mysterious element for me. I was always told the range between toxicity and deficiency is very narrow. Most fertilizers usually contains B:Zn in a ratio od 2:1. 
I use RO water and added typically 0.008..0.02 pmm of B weekly. I saw negative results when boron dose was higher than 0.015 ppm weekly. However, my Macradra began to grow exceptionally well when such dose was added to the tank (0.015 ppm weekly). But other plants seemed to grow slower.
I don't know how much boron is removed by my RO filter. I was told that reverse osmosis mebrane removes only 50% boron from the tap water. I estimated that in my tank the best result are abtained when I add 0.008..0.015 ppm of boron (weekly) and 0.006..0.02 ppm of zinc (weekly). Of course this work in my tank when only RO water is used; in other ones those numbers may be different.
I used EDTA zinc chelate (Dissolvine Zn-E-15) and also not chelated ZnSO4 (zinc sulfate) - both compounds worked well.


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## Zapins (Jul 28, 2004)

Hmm... Good info! Thanks Kekon!

The fertilator doesn't have boric acid and zinc sulfate listed. How many grams per gallon of water gives 0.001 ppm boron and 0.001 Zn?

Also, isn't it supposed to be 1:2 of B:Zn?


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## deicide (Sep 1, 2009)

.1g of ZNSO4 in 920ml = .001ppm per 10gal

.4g of H3BO3 in 920ml= .0019ppm per 10gal

.5g of H3BO3 in 920ml= .0023ppm per 10gal


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## Zapins (Jul 28, 2004)

Thanks a ton!! That is exactly the info I was after! I'll go ahead and order the chemicals now.

Kekon, come and chat with us  I'd love to discuss plants with you.


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## kekon (Aug 1, 2005)

> Kekon, come and chat with us I'd love to discuss plants with you.


Thanks ! With pleasure ! 

As far as ZnSO4 etc. are concerned we must take into account that some of them exist in hydrated form:

ZNSO4 * 7H2O = 22.7% Zn 
CuSO4 * 5H2O = 25.45% Cu 
FeSO4 * 7H2O = 20% Fe
MnSO4 * H2O = 32.5% Mn


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## Avi (Apr 7, 2004)

Hmmm...the right (or is it left....or both, for that matter,) side of my brain doesn't work with numbers like these very well. Could you tell me where to order the additives that you've mentioned? THANKS.


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## Zapins (Jul 28, 2004)

Ebay from gumby96 (seller) his shop is here http://shop.ebay.com/gumby96/m.html?_nkw=&_sacat=0&_trksid=p3911.m270.l1313&_odkw=sulfate&_osacat=0

Search for zinc sulfate and boric acid. He also sells urea and some other things that might be useful.


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## ray-the-pilot (May 14, 2008)

deicide said:


> .1g of ZNSO4 in 920ml = .001ppm per 10gal
> 
> .4g of H3BO3 in 920ml= .0019ppm per 10gal
> 
> .5g of H3BO3 in 920ml= .0023ppm per 10gal


I'm a little unclear about this?

ppm or parts per million is the same thing as mg/liter so: 
.1g = 100 mg 
and 920 ml = .92 L

Therefore 100/.92 = 108 ppm I realize that this does not account for the molar amount of zinc in the compound but this seems to be off by a factor of almost 100,000!

I also do not understand what .oo1 ppm per 10 gal means?
The unit is mg. / L /gal ???

Maybe I'm missing something?


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## Zapins (Jul 28, 2004)

>.< good catch.


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## JeffyFunk (Apr 6, 2006)

deicide said:


> .1g of ZNSO4 in 920ml = .001ppm per 10gal
> 
> .4g of H3BO3 in 920ml= .0019ppm per 10gal
> 
> .5g of H3BO3 in 920ml= .0023ppm per 10gal


Okay, so after reading Ray's response, i thought i'd take a look at the math here and indeed the original poster's math is correct (at least for B; I'm not sure which ZnSO4 was used (the anhydrous, monohydrate or heptahydrate form). Here's my math to check:

Molecular Weight [ B ] = 10.81 g/mol
Molecular Weight [H3BO3] = 61.83 g/mol
% B in H3BO3 = [10.81]/[61.83] = 17.48%

ppm B = [0.4 g H3BO3 / 920 mL]*[1000 mg / g]*[0.1748] = 0.078 ppm B

Using M1*V1 = M2*V2 to convert this into a (final) dosed B concentration of 10 gallons...

M1*V1 = M2*V2
[0.078 ppm B]*[0.92 L] = [X ppm B]*[10 gallons]*[1 L / 0.2642 gallons]

Rearranging for X, we find...

X ppm B = [0.078]*[0.92]*[0.2642] / 10 
X ppm B = 0.00189

I think a better way to state this would be through following the calculations below:

0.001 ppm B = [0.001 mg B / 1 L]*[1 L / 0.2642 gallons]*[10 gallons] = 0.03785 mg B per 10 gallons

If we are using H3BO3 as the B Source, we get the following:

X mg H3BO3 = [0.03785 mg B]/[0.1748] = 0.2165 g H3BO3

Using an identical equation but for Zn [Using ZnSO4*7H2O as the Zn Source], we get the following:

Molecular Weight [Zn] = 65.38 g/mol
Molecular Weight [ZnSO4*7H2O] = 287.53 g/mol
% Zn in ZnSO4*7H2O = [65.38]/[287.53] = 22.74%

X mg ZnSO4*7H2O = [0.03785 mg Zn]/[0.2274] = 0.1664 g ZnSO4*7H2O

Finally, a word of warning from the the EPA regarding the preparation of B standard solutions (taken from Method SW-846 6010C, Analysis of metals by ICP-OES Analysis): "Transfer [the B Solution] immediately after mixing into a clean polytetrafluoroethylene (PTFE) bottle to minimize any leaching of boron from the glass container. The use of a non-glass volumetric flask is recommended to avoid boron contamination from glassware".

Gotta love borosilicate glassware.


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## deicide (Sep 1, 2009)

> I'm a little unclear about this?
> 
> ppm or parts per million is the same thing as mg/liter so:
> .1g = 100 mg
> ...


I did this in a rush and I failed to include the % of the compounds used to get the results, so to clarify:

a) Calculation was based on *35.5%* ZNSO4 & *16%* H3BO3
b) See dose rates updates below

.1g of 35.5% ZNSO4 in 920*ml* = for every 1ml dose = .001ppm Zn per 10gal

.4g of 16% H3BO3 in 920*ml* = for every 1ml dose = .0019ppm B per 10gal

.5g of 16% H3BO3 in 920*ml* = for every 1ml dose = .0023ppm B per 10gal

.................................................................................................................

c) Calculation based on *22.74%* ZNSO4 & *17.48%* H3BO3
d) See dose rates updates below

.2g of *22.74%* ZNSO4 in 1000*ml* = for every 1ml dose = .0012ppm Zn per 10gal

.5g of *17.48%* H3BO3 in 1000*ml* = for every 1ml dose = .0023ppm B per 10gal


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## ray-the-pilot (May 14, 2008)

JeffyFunk said:


> Here's my math to check:
> 
> Molecular Weight [ B ] = 10.81 g/mol
> Molecular Weight [H3BO3] = 61.83 g/mol
> ...


I'm sorry, I couldn't get past this step? The units for ppm are the same as mg./L. Your units are mg. / ml. It seems that you could be off by a factor of 1000?


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## ray-the-pilot (May 14, 2008)

deicide said:


> a) Calculation was based on *35.5%* ZNSO4 & *16%* H3BO3
> b) See dose rates updates below
> 
> .1g of 35.5% ZNSO4 in 920*ml* = for every 1ml dose = .001ppm Zn per 10gal


Now this I can believe if you mean dissolve .1 g ZnSO4 in .92 L of water then dose 1 ml for each 10 gal!
.1g / .92L x .355 x 1000mg/g x 1ml /(3750ml x10) = 0.001 mg/L = .001ppm


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## JeffyFunk (Apr 6, 2006)

ray-the-pilot said:


> I'm sorry, I couldn't get past this step? The units for ppm are the same as mg./L. Your units are mg. / ml. It seems that you could be off by a factor of 1000?


Opps - You're right (which is why I'm glad I asked for people to check my math); the answer is off by a factor of 1000 so the final ppm B should be 78 ppm, not 0.078 ppm. (How I managed to properly convert mg to g but not mL to L is pretty amazing...)

The final equations I checked again and they too are incorrect; again off by a factor of 1000. (Grrrr...) The answers should be in mg, not g.


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## Zapins (Jul 28, 2004)

Hehe, sooo for ease of dosing could someone put the fully correct amounts to be dosed in this format? Thanks!!!



> .1g of 35.5% ZNSO4 in 920ml = for every 1ml dose = .001ppm Zn per 10gal


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## ray-the-pilot (May 14, 2008)

Zapins said:


> Hehe, sooo for ease of dosing could someone put the fully correct amounts to be dosed in this format? Thanks!!!


Using ZnSO4.7H2O, dissolve .1 g ZnSO4.7H2O in .61 L of water then dose 1 ml for each 10 gal.

.1g / .61L x .2274 x 1000mg/g x 1ml /(3750ml x10) = 0.001 mg/L = .001ppm

Using anhydrous Boric acid dissolve .4 g H3BO4 in .93 L of water then dose 1 ml for each 10 gal.
.4g /.93L x .1748 x1000mg/g x 1 ml /(3750ml x 10) = .002ppm

Note: .2274 and .1748 are the relative molecular amounts of zinc and boron found in the listed compounds.


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