randyloveschemistry
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why am i getting precipitation?!?!
i am trying to mix copper sulfate pentahydrate, manganese sulfate monohydrate, zinc sulfate monohydrate and boric acid together.
The problem is i keep getting precipitation and i cannot understand why? I am mixing them into 2 cups of distilled water.
The amounts are .5 grams boric acid, .6 grams copper sulfate, .8 grams manganese sulfate and .3 grams zinc sulfate.
solubility shouldn't be a problem. I add each ingredient one at a time. thoroughly mix until dissolved and then add the next one. i have tried all
24 ways of the order in which they could be added. all turn out the same.
the final pH comes out between 4.6 and 4.7. When i add citric acid the precipitation goes away.
so what is going on here? it seems something (or everything) is combining with the boron but how could that be if it is in the boric acid form?
please someone help me understand what i am doing wrong.
randy
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fusso
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Whats the purpose of adding all these together?
Can you try mixing only 1 metal salt with boric acid at a time to see what happen?
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AJKOER
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In the presence of air/O2 and H+, MnSO4 can form a basic salt, the underlying electrochemical reaction is:
4 Mn(ll) (aq) + O2 + 2 H+ → 4 Mn(lll) + 2 OH-
Related reactions occur with other transition metals (including Fe, Cu, Co,..).
With the addition of acid, the basic salt dissolves.
The source of H+ is more likely from the hydrolysis of an aqua complex than boric acid. For example:
Cu[(H2O)6]2+ (aq) + H2O (l) = [Cu(H2O)5(OH)]+ (aq) + H3O+ (aq)
[Edited on 8-7-2019 by AJKOER]
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Abromination
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Why are you mixing random metal salts with boric acid in a mixed form? Do your experiment separately, recrystallize your reagents and use a
stoichiometric amount of each chemical. If you don't understand that, leave the acid in excess.
List of materials made by ScienceMadness.org users:
https://docs.google.com/spreadsheets/d/1nmJ8uq-h4IkXPxD5svnT...
--------------------------------
Elements Collected: H, Li, B, C, N, O, Mg, Al, Si, P, S, Fe, Ni, Cu, Zn, Ag, I, Au, Pb, Bi, Am
Last Acquired: B
Next: Na
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Bedlasky
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Two cups are how much water? If they are small cups maybe some copper sulfate precipitate (you have in solution other sulfates which decrease
solubility of it). Citrate is complexing agent which solubility increases.
[Edited on 8-7-2019 by Bedlasky]
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Abromination
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Quote: Originally posted by Bedlasky  | Two cups are how much water? If they are small cups maybe some copper sulfate precipitate (you have in solution other sulfates which decrease
solubility of it). Citrate is complexing agent which solubility increases.
[Edited on 8-7-2019 by Bedlasky] |
Copper sulfate is really soluble in water.
It was probably boric acid falling out of solution, it really isn’t soluble at room temp.
List of materials made by ScienceMadness.org users:
https://docs.google.com/spreadsheets/d/1nmJ8uq-h4IkXPxD5svnT...
--------------------------------
Elements Collected: H, Li, B, C, N, O, Mg, Al, Si, P, S, Fe, Ni, Cu, Zn, Ag, I, Au, Pb, Bi, Am
Last Acquired: B
Next: Na
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UC235
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If citrate solubilizes the precipitate, that would mean the solid is a poorly soluble borate salt and the citrate, acting as a chelating agent instead
forms a soluble complex of that metal.
If I had to guess, it's a micronutrient blend for hydroponics or something. Boron, copper, manganese, and zinc are all needed in quite low levels by
plants.
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randyloveschemistry
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fusso: i am trying to make a personalized micronutrient fertilizer for my garden. and trying to learn some more chemistry as i go
AJ Joker - so i tried just mixing the cu, mn, and zn sulfates together. in every possible order. copper sulfate always lowered the pH. the other
two always slightly raised it.
bedlasky - it is 473 mL
abromination -when i would add boric acid to the water to it would just sit on top until i stirred it in. i guess i am confused because in aqueous
solution i thought boric acid was the predominant form of boron below a ph of 7. and since it is uncharged that it would not combine with anything.
and i thought it was soluble enough
UC235 - so if i am forming poorly soluble boron salts is the only solution to lower the pH? i am going to try and use borax and disodium octaborate
tetrahydrate and see how that turns out
thank you for the replies everyone
Randy
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Sulaiman
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You can usually estimate the existing micro-nutrient levels by observing which plants grow well, and which do not.
e.g. Attachment: az1106.pdf (125kB)
This file has been downloaded 350 times
Unless your plants show signs of malnutrition I would be wary of adding heavy metals to your soil.
CAUTION : Hobby Chemist, not Professional or even Amateur
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Abromination
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| Quote: Originally posted by randyloveschemistry | fusso: i am trying to make a personalized micronutrient fertilizer for my garden. and trying to learn some more chemistry as i go
AJ Joker - so i tried just mixing the cu, mn, and zn sulfates together. in every possible order. copper sulfate always lowered the pH. the other
two always slightly raised it.
bedlasky - it is 473 mL
abromination -when i would add boric acid to the water to it would just sit on top until i stirred it in. i guess i am confused because in aqueous
solution i thought boric acid was the predominant form of boron below a ph of 7. and since it is uncharged that it would not combine with anything.
and i thought it was soluble enough
UC235 - so if i am forming poorly soluble boron salts is the only solution to lower the pH? i am going to try and use borax and disodium octaborate
tetrahydrate and see how that turns out
thank you for the replies everyone
Randy
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My bad, I saw 5 and not .5 grams. It should dissolve no problem in that amount.
UC235 is probably right.
List of materials made by ScienceMadness.org users:
https://docs.google.com/spreadsheets/d/1nmJ8uq-h4IkXPxD5svnT...
--------------------------------
Elements Collected: H, Li, B, C, N, O, Mg, Al, Si, P, S, Fe, Ni, Cu, Zn, Ag, I, Au, Pb, Bi, Am
Last Acquired: B
Next: Na
-------------- |
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AJKOER
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Yes, "copper sulfate always lowered the pH." per the cited reaction I gave right out of the literature.
Cu[(H2O)6]2+ (aq) + H2O (l) = [Cu(H2O)5(OH)]+ (aq) + H3O+ (aq)
Also the comment: "It should dissolve no problem in that amount. UC235 is probably right."
I know for sure that basic salt do dissolve in added acid.
Here is a reference to the actual action of oxygen creating a basic salt, see tribasic copper chloride formation reaction in Wikipedia at https://en.wikipedia.org/wiki/Dicopper_chloride_trihydroxide and in particular Equation (7), detailing a commercial preparation path via oxygen.
Some references on the web, do NOT cite the reaction as I prefer, which in the case of copper would be:
4 Cu(l) (complexed) + O2 + 2 H+ → 4 Cu(ll) + 2 OH-
but, by adding 2 H+ to each side, one arrives at:
4 Cu(l) (complexed) + O2 + 4 H+ → 4 Cu(ll) + 2 H2O
which, in my opinion, belies the tendency of this reaction to proceed through the formation of a basic salt. One explanation, as to why it is often
presented in the less informative manner, is likely because of the frequently cited half cell reaction in electrochemistry:
2 H2O (l) = 4 H+ (aq) + O2 (g) + 4 e- (Source, see for example, p. 21 at https://christou.chem.ufl.edu/wp-content/uploads/sites/43/20... and written in the reverse direction at bottom of page 7 at https://miamisprings-shs.enschool.org/ourpages/auto/2017/6/9... )
[Edited on 11-7-2019 by AJKOER]
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presserffg
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| Quote: Originally posted by AJKOER | Yes, "copper sulfate always lowered the pH." per the cited reaction I gave right out of the literature.
Cu[(H2O)6]2+ (aq) + H2O (l) = [Cu(H2O)5(OH)]+ (aq) + H3O+ (aq)
Also the comment: "It should dissolve no problem in that amount. UC235 is probably right."
I know for sure that basic salt do dissolve in added acid.
Here is a reference to the actual action of oxygen creating a basic salt, see tribasic copper chloride formation reaction in Wikipedia at https://en.wikipedia.org/wiki/Dicopper_chloride_trihydroxide and in particular Equation (7), detailing a commercial preparation path via oxygen.
Some references on the web, do NOT cite the reaction as I prefer, which in the case of copper would be:
4 Cu(l) (complexed) + O2 + 2 H+ → 4 Cu(ll) + 2 OH-
but, by adding 2 H+ to each side, one arrives at:
4 Cu(l) (complexed) + O2 + 4 H+ → 4 Cu(ll) + 2 H2O
which, in my opinion, belies the tendency of this reaction to proceed through the formation of a basic salt. One explanation, as to why it is often
presented in the less informative manner, is likely because of the frequently cited half cell reaction in electrochemistry:
2 H2O (l) = 4 H+ (aq) + O2 (g) + 4 e- (Source, see for example, p. 21 at https://christou.chem.ufl.edu/wp-content/uploads/sites/43/20... and written in the reverse direction at bottom of page 7 at https://miamisprings-shs.enschool.org/ourpages/auto/2017/6/9... )
[Edited on 11-7-2019 by AJKOER] |
Thanks for the link.
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