Sciencemadness Discussion Board

Calculating equivalents of NaH2PO4 (2H20) question

maxpayne - 14-4-2012 at 23:46

Hello!

In one procedure I have NaH2PO4 stated as equivalents rather than grams or moles, other substances are in moles and excess of moles and I have no problem with that.

However, I tried to calculate how much it would be 2 equivalents of NaH2PO4 and my result is simply 2 moles. However if this is correct, I must say that procedure is run in H2O as solvent and my NaH2PO4 is dihydrate or NaH2PO4.2H2O

If I want to be precise should I calculate the water in my NaH2PO4.2H2O and take it into account when calculating the solvent (H2O).

Am I correct?

Magpie - 15-4-2012 at 16:01

Quote: Originally posted by maxpayne  

However, I tried to calculate how much it would be 2 equivalents of NaH2PO4 and my result is simply 2 moles.


Edit: No, 1 mole. Each mole has 2 equivalents.
Quote: Originally posted by maxpayne  

However if this is correct, I must say that procedure is run in H2O as solvent and my NaH2PO4 is dihydrate or NaH2PO4.2H2O

If I want to be precise should I calculate the water in my NaH2PO4.2H2O and take it into account when calculating the solvent (H2O).


Yes, if you need to calculate the solvent. It may be a very minor effect and could possibly be ignored. Only a calculation (or estimate) will tell.

[Edited on 16-4-2012 by Magpie]

maxpayne - 15-4-2012 at 18:32

Thank you very much Magpie for your answer.

But If you could be so kind and explain the math here, because I want to understand how did you came up with your result.

I agree with second answer, it's not that hard to check and see if water in dihydrate would make a big difference.

Magpie - 15-4-2012 at 19:25

Yes, I'll explain assuming that you don't yet know the chemical meaning of the term "equivalent:"

An equivalent weight is that weight of a chemical that would accept or donate 1 mole of electrons. This is a little abstract so I will use your compound as an example.

NaH2PO4 + 2NaOH ----> Na3PO4 + 2HOH

Since this compound will donate 2 moles of H+, or pickup 2 moles of OH-, it has 2 equivalents.

Other examples:

Na2HPO4 would have 1 equivalent per mole.

H3PO4 would have 3 equivalents per mole.

Ca(OH)2 would have 2 equivalents/mole, and so on.

Extra credit:

Understanding the term "equivalent" allows one to understand the term "normality." Eg, a solution 1M in H3PO4 is 3N.


maxpayne - 16-4-2012 at 10:01

OK, thanks.

I thought of equivalents completely wrong.

It is hard for me to explain how I complicated it in my mind. But lets just say that i thought that I need to know what will react with what, and then see how many molecules of a compound would react with other, and then multiply it with given equivalent to make reaction balance (excess of a given compound for various reasons).

I'm ashamed, but now enlightened which is very valuable.

Thanks again.




[Edited on 16-4-2012 by maxpayne]

Nicodem - 16-4-2012 at 10:45

Magpie, it is not possible to answer maxpayne's question by adapting the concept of equivalence in using an arbitrary reaction like you did. That is misleading. Obviously the original question remains unanswerable because maxpayne did not provide the procedure he talks about, or its reference. Without it, nobody can know what the reaction is and thus what an equivalent of NaH2PO4.2H2O is.
The term equivalent means "equal value" and in this case refers to amounts. Reaction amounts are reaction specific (stoichiometry!). In fact, even assuming (guessing) that the NaH2PO4.2H2O has the role of an acid in the procedure, the huge difference between the pKa2 and pKa3 makes it less likely to be subject of a double deprotonation.
"Equivalence" in most reactions simply means the equal amount as the substrate or the refereed reactant, but it is stoichiometry dependent, so in some reactions it can be a multiple or fraction of that amount.
Quote: Originally posted by maxpayne  
If I want to be precise should I calculate the water in my NaH2PO4.2H2O and take it into account when calculating the solvent (H2O).

You don't have to calculate the amount of water in NaH2PO4.2H2O. All you need to do, is to use its molar mass in the calculation of the required weight:

m(NaH2PO4.2H2O) = M(NaH2PO4.2H2O) * n(NaH2PO4.2H2O)

where

m : mass [g]
M : molar mass [g/mol]
n : amount [mol]

[Edited on 16/4/2012 by Nicodem]

Magpie - 16-4-2012 at 13:43

Nicodem:

From Wiki for: Equivalent (Chemical):

"The equivalent (symbol: eq or Eq), sometimes termed the molar equivalent, is a unit of amount of substance used in chemistry and the biological sciences.

The equivalent is formally defined as the amount of a substance which will either:

react with or supply one mole of hydrogen ions (H+) in an acid–base reaction; or
react with or supply one mole of electrons in a redox reaction.[1][2]

The mass of one equivalent of a substance is called its equivalent weight.

A historical definition, used especially for the chemical elements, describes an equivalent as the amount of a substance that will react with one gram of hydrogen, or with eight grams of oxygen, or with 35.5 grams (1.25 oz) of chlorine, or displaces any of the three.[3]"

see http://en.wikipedia.org/wiki/Equivalent_%28chemistry%29

This is my understanding of the term and the way in which I use it in practice.

The definition I gave in the post above is formally for the redox reactions, but then I showed an example of an acid base reaction. So I admit that my definition wasn't complete, and that I used the wrong example to illustrate my definition.

But I do see your point, and that the context of the problem at hand is the determining factor as other meanings are possible.

It would be best if maxpayne would present the procedure to us in its entirety. This way we could know the context and thereby assure giving our best response.

[Edited on 16-4-2012 by Magpie]

[Edited on 16-4-2012 by Magpie]

turd - 16-4-2012 at 14:28

The point is that HPO42- is not acidic at all. So unless he's working in strongly alkaline medium, it will not provide that last proton. Indeed Na3PO4/K3PO4 are quite basic. Many people don't realize that even the first proton of H3PO4 is less acidic than HSO4-! The second is already a pretty weak acid and the third, for all intents and purposes, is not an acid. ;)

If a procedure calls for n equivalents of NaH2PO4, the authors definitely want you to add n equivalents and not n/2! Unless it's a very obscure paper. Like from analytic/industrial chemists who write NaAc for sodium acetate and MeOH for a monovalent metal hydroxide. I kid you not!

maxpayne - 17-4-2012 at 18:50

I must apologize for not posting the complete procedure. I did not think it was relevant for the measurement, however it seems that it is. Also every answer here is correct and thank you all for your time.

So here is the procedure (fixed link):

Reductive methylation

In this same procedure this "additive" can be Acetic acid, depending of the di or mono methylation however it is in one example marked as 8 equivalents. Does this mean 2 moles of AcOH since there is 4 hydrogens per molecule of AcOH? I'm almost certain.

And if you look at the 18th and 19th row in the table you see how Glycine is easily attacked and my conclusion is that Magpie gave me the right answer. Making Glycine monomethylated takes half mole of NaH2PO4. I pray that my logic is correct.

Now, where my edge begins to blunt, and why I started this thread is that I can't figure reaction mechanism since I don't know how many molecules of this additive is in the solution. Giving me everything on the table makes me more prone to solving the puzzle.

P.S. Please note that I'm just a beginner, not total because I have some background in chemistry and other science where logic is the key. Also, my everyday work and traveling this week is the reason for updating this thread once a day.

[Edited on 18-4-2012 by maxpayne]

Magpie - 18-4-2012 at 08:17

maxpayne, that link gets me nothing but gibberish. Please fix it.

maxpayne - 18-4-2012 at 10:30

Fixed now, hope it'll work for everyone.

For those with access here is another link:

Reductive methylation


[Edited on 18-4-2012 by maxpayne]

turd - 18-4-2012 at 12:57

Quote: Originally posted by maxpayne  
And if you look at the 18th and 19th row in the table you see how Glycine is easily attacked and my conclusion is that Magpie gave me the right answer. Making Glycine monomethylated takes half mole of NaH2PO4. I pray that my logic is correct.

Magpie is wrong. You're supposed to add 1 mol of NaH2PO4 (or corresponding hydrate) per mol of glycine for monomethylation.

Your "logical" train of thought makes no sense whatsoever. I'm a little bit scared about that "other science where logic is the key". :o

Magpie - 18-4-2012 at 18:01

Quote: Originally posted by turd  
You're supposed to add 1 mol of NaH2PO4 (or corresponding hydrate) per mol of glycine for monomethylation.


For this particular procedure I agree.

maxpayne - 21-4-2012 at 01:56

Quote: Originally posted by turd  
Your "logical" train of thought makes no sense whatsoever. I'm a little bit scared about that "other science where logic is the key". :o


Never mind. It is just training. Don't be "scared".
Quote: Originally posted by turd  
You're supposed to add 1 mol of NaH2PO4 (or corresponding hydrate) per mol of glycine for monomethylation.


Why not explain to me why 1 mole, showing me the reaction mechanism (bonding).

turd - 21-4-2012 at 04:26

I'm just telling you what was determined experimentally according to this paper. One equivalent of NaH2PO4 is apparently just right for monomethylation. I don't see why I should show you the reaction mechanism. You will find imine formation in any beginner organic chemistry textbook and you will likewise find tons of discussion on reduction with Zn, from the antiquated idea of nascent hydrogen to more modern interpretations involving heterogenous catalysis.

maxpayne - 21-4-2012 at 23:42

Ok, thanks.

Maybe I should try to figure simpler reaction in my spare time, this one seems too complicated for me, but because there is a table with many compounds I thought it would be nice for analogy thinking.

One more thing if you want to answer, please:

What is the function besides acidity of NaH2PO4? If it donates two hydrogens per molecule, what is happening to Na and PO4? Does sodium and phosphate reacts somehow with amine?

Nicodem - 22-4-2012 at 05:44

Quote: Originally posted by maxpayne  
What is the function besides acidity of NaH2PO4? If it donates two hydrogens per molecule, what is happening to Na and PO4? Does sodium and phosphate reacts somehow with amine?

You can find good pedagogical explanations on how buffers work on places like wikipedia and elsewhere:
http://en.wikipedia.org/wiki/Buffering_agent
http://en.wikipedia.org/wiki/Buffer_solution
http://www.chemguide.co.uk/physical/acidbaseeqia/buffers.htm...
http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/B...
... and hundreds other sites. That Tetrahedron letters article probably also addresses the issue at least superficially.

Once you get acquainted with the topic, you can return back to this issue with more specific questions worth writing down new and more specific explanations.

Waffles SS - 15-10-2012 at 22:23

I think most challenged point on this article is calculating isoelectric point of N-methylated amino acids(for separation step) not equivalents of NaH2PO4 ;)
Quote:

Typical procedure: A mixture of amine (1 mmol), additive, 37% aqueous formaldehyde (50% excess), zinc dust or Granules(100% excess) and the indicated solvent was stirred at 30 C. After completion of the reaction aqueous ammonia was added and yields were determined directly in the aqueous solution (amino acids) or in the HCCl3-extract of the bases in relation to an internal standard. Analytical samples were obtained as picrates or oxalates in the case of amines; amino acids were isolated at the isoelectrical point by traditional methods.


Nicodem - 16-10-2012 at 07:10

Quote: Originally posted by Waffles SS  
I think most challenged point on this article is calculating isoelectric point of N-methylated amino acids(for separation step) not equivalents of NaH2PO4 ;)

And just how you intent to calculate that?
Last time I checked this property was measured, not calculated.
You just check the literature. No need to calculate anything.

Waffles SS - 17-10-2012 at 06:10

I didnt find isoelectric point of mono N-methyl-L-alanine and di N-methyl-L-alanine.( i know L-alanine isoelectric point is 6.0)
Can you help me?

[Edited on 17-10-2012 by Waffles SS]

maxpayne - 18-10-2012 at 07:53
Quote:

I think most challenged point on this article is calculating isoelectric point of N-methylated amino acids(for separation step) not equivalents of NaH2PO4 ;)


Equivalents are calculated by moles, so for example; 2 equiv of one substance means 2 moles of additive in this particular reaction.

And yes, isoelectric points are also one of the many secrets not revealed by the Tettr.Letts. If what Nicodem said, that it is impossible to calculate isoelectric points, and there is no data about some of the N-methyl-amino acids, then how should anyone proceed forward and follow the Letter??