Vanadium pentoxide- need help from the big brains

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Randy Holmes-Farley

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Randy can you help me with DIY dose of sodium molybdate?
Very appreciate for help

Sure.

Do you have a link to a product you want to use?
 

dani1970

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Sure.

Do you have a link to a product you want to use?
It's problem I tell you why I buy from chemical store that not have website I can only order from them when I get the product only when I can take pic and show you the details of product let me try talk with store that they send me details of chemical on email maybe that works
 
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dani1970

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Randy Holmes-Farley

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As long as you buy the same type of product, it doesn't matter from who. I wanted to see the chemical form, degree of hydration, etc.

Sodium molybdate has a molecular weight of 206 grams per mole.

Molybdenum has a molecular weight of 96 g/mole.

Thus, sodium molybdate is 96/206 x 100 = 46.6% molybdenum by weight

If you dissolve 1 gram (1,000 mg) in 1 L (1000 mL), that has a concentration of 1 mg/mL sodium molybdate or 0.5 mg/mL molybdenum

Natural levels of molybdenum are close to 10 ug/L.

If you want to boost 100 liters of aquarium water by 10 ug/L, you need 100 L x 10 ug/L = 1,000 ug or 1 mg.

Thus, 2 mL of your stock solution (that is 0.5 mg/mL) will add that 1 mg to the 100 liters of aquarium water.
 

dani1970

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As long as you buy the same type of product, it doesn't matter from who. I wanted to see the chemical form, degree of hydration, etc.

Sodium molybdate has a molecular weight of 206 grams per mole.

Molybdenum has a molecular weight of 96 g/mole.

Thus, sodium molybdate is 96/206 x 100 = 46.6% molybdenum by weight

If you dissolve 1 gram (1,000 mg) in 1 L (1000 mL), that has a concentration of 1 mg/mL sodium molybdate or 0.5 mg/mL molybdenum

Natural levels of molybdenum are close to 10 ug/L.

If you want to boost 100 liters of aquarium water by 10 ug/L, you need 100 L x 10 ug/L = 1,000 ug or 1 mg.

Thus, 2 mL of your stock solution (that is 0.5 mg/mL) will add that 1 mg to the 100 liters of aquarium water.
Oh thx for explaining I try ask the store chemical form and degree of hydration
 
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dani1970

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Top Shelf Aquatics

Randy Holmes-Farley

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That material looks fine.

It is 46% molybdenum by weight.

If you dissolve one gram in one liter of water, that gives a concentration of 460 mg/L or 0.46 mg/mL.

If you want to add 10 ug/L to a 100 L aquarium, you need 10 ug/L x 100 L = 1,000 ug or 1 mg.

Thus, 2 mL added to a 100 L aquarium will boost molybdenum by about 10 ug/L, which is close to the NSW level.
 

AlbertGF

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Success! I ended up adding 1.8 grams of sodium hydroxide. 0.9 grams didn’t dissolve it completely.
Now I have vanadium supplement for life for $40. I’ll be giving about 80% of the vanadium pentoxide and sodium hydroxide powder away to my local fellow reefers.

953E0358-0B55-4E92-82A9-951E7B12FF70.jpeg
Hi!
At the end which was you exact formula? I tried to dissolve 2.68gr. of V2O5 to 1.5 liters of RODI water and it is not dissolving. Following the instructions above I added 1.5gr. NaOH and still nothing.
I read at Wikipedia that it dissolves 8 grams per liter but in many other places they say that it dissolve at 0.8 grams per liter, so I don't know what to do...
Do you confirm that 4 grams dissolved well in 1 liter of water (after adding 1.8gr. NaOH)?

Thanks!
 
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