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Topic: polyatomic ion bonding in solutions???  (Read 1788 times)

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Offline long3465

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polyatomic ion bonding in solutions???
« on: October 23, 2013, 03:21:56 AM »
Ok so I am trying to synthesis NH4NO3 by following the instructions from nurd rage. http://www.youtube.com/watch?v=w2IG49Nw1Fo
In this video we make ammonium nitrate from various off-the-shelf chemicals.
Note: This is not cheaper than buying it directly, once again our objective is to explore the chemistry.
Warning: The chemicals are corrosive and ammonia has a horrible smell, work outside or in a fumehood and wear gloves.
First get 138 grams of sodium bisulfate and add just enough water to dissolve it, usually 300mL.
Then get one mole equivalent of a pure nitrate salt. Some common salts include:
Sodium nitrate: 85 grams.
Potassium nitrate: 101 grams.
Calcium nitrate: 118 grams (if using tetrahydrate)
Dissolve the nitrate in a minimum of water, then mix the two solutions together.
Neutralize the mixture with ammonia (using a pH meter to determine the endpoint).
Filter the mixture to get rid of any insoluble materials if needed, then boil until sodium sulfate begins to precipitate. Then cool the mixture to 0 Celsius and filter. Dry the filtrate to obtain ammonium nitrate mixed with some leftover metal sulfates.
For higher purity the solids can be mixed with 500mL of methanol which selectively dissolves ammonium nitrate. Filter again and evaporate the filtrate to obtain pure ammonium nitrate.

If KNO3 was used in place of NaNO3, how would the equation go? I have been working to balance this out and I am just not sure how this would “play out”. The balanced equation using NaNO3 is NaHSO4 + NaNO3 NH3 = Na2SO4 + NH4NO3 , now if KNO3 is used in place of NaNO3 what is on the other side??? Based on EN values and reactivity it would seem that all the K+ would get used up first, in the form of K2SO4, then we are left with NaOH in solution Or vice versa, Na2SO4 is formed and we are left with KOH. I have found papers discussing the properties of NaKSO4 but it seems to be fairly uncommon. Or is it just that half a mole of each Na2SO4, K2SO4 is formed? Again back to the original question, how do I separate out the NH4NO3 from the compounds containing K? I am thinking to just dry the filtrate out and refine with the methanol, since it appears that K2SO4 is not soluble in methanol.
So I guess my basic question is this, how do I  determine which ions will attract each other in solution. With elements I know it is based on electronegativity, but what about polyatomic ions?

*to the moderators, feel free to edit for content or length, or to move to a different forum if it is needed. thanks.

Offline Dan

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Re: polyatomic ion bonding in solutions???
« Reply #1 on: October 23, 2013, 12:54:32 PM »
Please read the Forum Rules. No explosives.
My research: Google Scholar and Researchgate

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