Maths question. Please make it easy.

[Soilwork]

Please keep it step by step and repeatable.

My example it manganese sulphate.

I want to put 2.5g in to 500ml of water and dose 0.1ml how do I calculate the PPM of Mn in that dose?

Ta

 

[dw1305]

Hi all,
You need to know which manganese (Mn) compound you have, and it is almost certainly MnSO4·H2O.

Then you need to add together the RAM of each element (Mn = 54.3, H = 1, O = 16 etc.) to get the RMM of the compound, in this case 169. The percentage Mn in MnSO4·H2O is then 54.3/169 = 32%.

When you add 2.5g MnSO4·H2O to 500cm3 (5000 mg) it is equivalent to 5g in a litre, which means you have added 5 x 0.32 = 1.60g Mn per litre.

Milli-gram per litre and ppm are equivalents, so your 2.5g in 500ml solution of MnSO4·H2O is 1600 ppm Mn.

The final ppm you add depends upon the volume of the tank.

If you add 0.1ml (of the 1600ppm solution) to 1 litre with RO water you have diluted by 10,000 and 1600/10,000 = 0.16ppm.

cheers Darrel

 

[Soilwork]

Thanks Darrel

So using the same method for boric acid at 0.1ml from 2.5g in 500ml of water in to a volume of 76.5 litres = 0.011ppm?

If this is correct I think I understand.

 

[Soilwork]

As it happens mnso4 and a couple of others are actually on rotala butterfly calculator but boric acid isn't there.

 

[dw1305]

Hi all,

So using the same method for boric acid at 0.1ml from 2.5g in 500ml of water in to a volume of 76.5 litres = 0.011ppm?

Yes, but I think you are a power on 10 out, and it is 0.0011ppm.

The boric acid (H3BO3) stock solution (17.5% B) is 870 ppm boron, and the dilution factor from 0.1ml to 1000 ml (1 litre) is 10000, so you have 0.087 ppm in 1 litre of tank water. then 0.087/76.5 (volume of the tank in litres) gives you 0.0011ppm B.

cheers Darrel