Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Archdevil155 on May 01, 2013, 07:49:26 AM
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Hey everyone, just created the account to ask something..
Why do we need to add NH3 to CaCl2 on the way to make Ca3(PO4)2?
1st year on University of Chemistry in Greece here and I gotta write a report on an experiment we did the other day.. Based on the directions from a book they had given us, we needed to make Ca3(PO4)2. The steps were these:
1.Get 2,50 g of pure, dry CaCl2 in a 100ml beaker.
2.Add 25ml H2O so they dissolve.
3.Add SOME dense NH3.(Sorry if the word dense isn't correct I'm from, and study in, Greece.. Anyway I mean high concentration.)
4.Filter if neccesary.
5.In a different beaker dissolve 5,00 g Na2HPO4*12H2O in H2O.
6.Heat both solutions and then combine them while stirring.
7.Keep the solution warm for a while so that the sediment becomes granular and drops.
8.Filter using the Buchner filtration system.
9.Wash with warm H2O (twice with 5ml each time).
10.Dry in a drying oven.
So on step 3 why do we need to add the NH3?? It doesn't say and the teacher wouldn't explain.. We need to calculate how much we need to add but we do not know why add it in the 1st place.. So please If anyone can help me I'd be quite grateful..
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Think about pH.
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Umm can you be more specific..??
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What anion do you need to precipitate Ca3(PO4)2? How does its concentration depend on pH?
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(PO4)3- I think.. So we probably get the Ca2+ from the dissolution of CaCl2 right..? So doesn't NH3 react with the Cl- to give NH4Cl? A former chemistry teacher of mine told me that the NH3 reacts witht he Na2HPO4, not the CaCl.. He told me that the Na2HPO4 dissolves and gives Na+ and HPO42- and that then the HPO42- reacts with NH3 and gives (NH4)2PO4 if I remember correctly.. Then that dissolves in the water giving NH4+ and PO42-.. Is that correct??
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It is all happening in water, so there is no reaction followed by dissolution.
You are right about PO43- being the anion that is necessary. HPO42- is a weak acid and NH3 is a (weak) base. What happens when you mix base and acid?
Note that some of your formulae are off, so even if your thinking is correct, it is hard to tell what is a typo and what is an error.
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Oh yea.. I haven't really gotten used to thinkin' like that yet.. In high school that's how we were taught but in University they told us that actually the salt isn't formed, just the ions are around... That's what happens right? Umm they react and neutralize each other, forming a salt. If one of the terms I use (or more) is wrong, please correct me. The terms in Greek and English are probably a bit different.. Oh, I didn't write any typos, just errors then. I have to hurry a bit so I can't check for all of them in case I recognize them myself.. I think that (NH4)2PO4 should be (NH4)3PO4 but I can't find others right now.. :( Please correct my errors if you can and I'll check again as soon as possible..
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not the CaCl..
CaCl2
PO42-
You mean PO43-.
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Wow, I can't believe the 1st one.. THAT was because of me hurrying and not paying enough attention.. :-\ But the 2nd one was because I had written (NH4)2PO4, so THAT would give PO42- if it was correct.. I should have corrected the PO43- along with the (NH4)3PO4 before.. Soo I guess you helped me just as much as I needed :) Thanks a lot and I have just two more small questions.. On step 6 of the experiment I described we heat the two solutions so that they react more easily and quickly I guess right?? And my 2nd question is on step 7, why is it the fact that we keep the mixture warm, that makes the sediment granular and lets it drop??
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Ok.. I'm having trouble putting everything together and I've gotten kind of frustrated. Could you tell me the entire course of the experiment?? What happens in every step each time you add something, along with the reactions..? I'd be veeery grateful!
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I had an impression you mostly got it. Ammonia makes pH high, shifting dissociation of HPO4 to the right and increasing concentration of PO43- - that in turn shift the precipitation equilibrium to the right. Warm solution helps to age the precipitate (see http://en.wikipedia.org/wiki/Precipitation_(chemistry)#Digestion).
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Yea I thought so too.. So CaCl2 doesn't react with NH3?? By the way if instead of the word "react" I should use a different word tell me.. By react I mean that NH3 with CaCl2 form together NH4Cl and CaO. I found this reaction equation:
CaCl2 + 2NH3 + H2O :rarrow: 2NH4Cl + CaO.
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Oh and also, why does high pH shift the dissociation of HPO4 towards PO43-?
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Oh and also, why does high pH shift the dissociation of HPO4 towards PO43-?
Because it is this way. I mean - it is obvious when you know a little bit about acid-base equilibria, but it is not something that can be easily explained in a few words.
Basically it is not different from neutralizing an acid (HPO42-) with a base (ammonia).
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CaCl2 + 2NH3 + H2O :rarrow: 2NH4Cl + CaO
Perhaps it may happen if you react solid with vapors, but for sure not in a solution.
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Yea I met with my former Chem teacher that I mentioned before and he explained it quite well..:) And now I get what you mean about NH3, thanks for all the help man.. He said that the way it's said in the Lab guide book (some dense NH3) it means that it doesn't react because it's too little compared to the CaCl2 and it's just there to adjust the pH and neutralize HPO42-, just as you said.. :) Thanks again and now I'm writing and completing the report :)
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Sorry to ask again on this topic, but can you type the dissociation equation of Na2HPO4 * 12H2O in water please??
It's in solid state in the form of small crystals-powder, if that's any help..
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Sorry to ask again on this topic, but can you type the dissociation equation of Na2HPO4 * 12H2O in water please??
What ions do you expect to be able to get, given the formula? Take a random stab, see what you get, we can help afterward.
It's in solid state in the form of small crystals-powder, if that's any help..
None of that has any bearing on this question. I hope you can see that. Note: the water of hydration likewise has no effect on this problem, I hope you can see that.
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Yea I thought so just making sure.. My problem wasn't about Na2HPO4 but the 12H2O. Whether they do something or not.. I didn't really think so but I wanted to know for sure. Na2HPO4 :rarrow: 2Na+ + HPO42-. So since the 12H2O don't participate in the chemical equation I'm covered.. :)
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So what would happen to 12 molecules of H2O locked into the solid's crystal lattice when the solid dissolves in water? Can you determine how many molecules of water there are in 1 ml? That would help you understand.
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sorry to bother or spam the conversation but I'd like to ask you if you study in Patra because I'm in the exact place with this experiment as you are .