PH of methylammonium bromide

[mnnob07]

What is the pH of .15 M methylammonium bromide, CH3NH3Br (Kb of CH3NH2 = 4.4x10^-4)

I actually asked this to someone else and got this:

CH3NH3 (aq) + H2O (l) <=> CH3NH2 (aq) + H3O+ (aq)
Kw = KaKb
Ka = (10-14)/(4.4 x 10-4) = 2.27 x 10-11

ICE chart:
CH3NH3 (aq) + H2O (l) <=> CH3NH2 (aq) + H3O+ (aq)
I .15 M 0 0
C -x +x +x
E .15-x x x

Ka = (x^2)/(.15 - x) = 2.27 * 10^-11
x = 1.85 x 10-6 M = [H+]

pH = -log([H+]) = -log(1.85 x 10-6) = 5.73

It just doesn't seem right. To use the information given for CH3NH2 we need to first have CH3NH3 or CH3NH2 in the solution but we start with CH3NH3Br. Where did the bromine go? first off here is CH3NH3:

http://img59.imageshack.us/img59/571/chemistryfecee368c664afax2.jpg

I'm not very far in chemistry and maybe that's why I'm confused, but CH3NH3 simply could not have had bromine tacked on somewhere on the above model ready to fall off.

There's simply no room; H, C, and N have their orbitals filled in CH3NH3

It would seem (and I can't find a reference for this so I'm likely wrong) that CH3NH3Br looks like this:

http://img57.imageshack.us/img57/5500/chemistrye6393df93de99fmq2.jpg

It would dissociate as follows:

CH3NH3Br -> CH3NH2 + HBr

now, that CH3NH2 would follow the reaction originally stated with the associated equilibrium constant that was given, but the OH- created from that will pale in comparison to the H30+ generated by the strong acid HBR:

HBr + H20 -> H30+ + Br-

since it is a strong acid, it will dissociate completely and we will be left with [H30+] = .15M

pH = -log[H30+]

pH = -log(.15)

pH = .83

I would appreciate any help, I just want this whole thing to chemically make sense to me. BTW, we haven't covered buffers yet so that shouldn't be a huge part of the solution

OK I'm not sure how to post images. just click on the images below for the basic structures (sorry I didn't explicitly label the lone pairs around Bromine but there should be 3 I think)

 

[Borek]

When dissolved methylammonium bromide dissociates to CH₃NH₂⁺ and Br^-.

Such a compound is commonly called hydrobromide - what it means is that you have an ionic salt consisting of protonated amine and Br^-. Thus to calculate pH you have to assume you really have a solution of CH₃NH₂⁺ - which is your acid.


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[mnnob07]

am I able to use the same equilibrium constant for CH3NH2+ as I do with CH3NH2? The question doesn't give any information about CH3NH2+ and I don't see it in the book's appendix

Also, if CH3NH3Br -> CH3NH2+ and Br-

what happened to the other H? and is that dissociation complete or at equilibrium?

 

[Borek]

am I able to use the same equilibrium constant for CH3NH2+ as I do with CH3NH2? The question doesn't give any information about CH3NH2+ and I don't see it in the book's appendix

These are pair of conjugated acid and its base, pKa + pKb = pKw.

Also, if CH3NH3Br -> CH3NH2+ and Br-

what happened to the other H? and is that dissociation complete or at equilibrium?

Complete dissociation. You have lost H - no idea why. It is not CH₃NH₂⁺ but CH₃NH₃⁺.

Edit: sorry, my typo in previous post. Should be CH₃NH₃⁺.


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[mnnob07]

ok so we have:

CH3NH3Br -> CH3NH3+ and Br-

Br- is a spectator ion and is no longer needed

CH3NH3+ and H20 <-> CH3NH2 and H30+

Ka = Kw/Kb(of reverse which was given)

Ka = 1.0*10^-14 / 4.4*10^-4

and so on. That's exactly what I had in my first post. I later stated what I don't understand about it and my question but I stand with no better understanding than before. It's possible you just didn't read my whole post and I can't be mad at you for trying to help, in fact I am very pleased you tried to help and confident your method is correct but just knowing CH3NH3Br dissociates that way does little to help me in the long run.

My question was, how does that work? like, chemically and structurally. WHERE DOES THE BROMINE COME FROM? It is possible I do not have as complete understanding with ionic bonding and maybe some other ways the atoms bond as I should, but if CH3NH3 looks like this:

http://www.meta-synthesis.com/webbook/40_polyatomics/sp3_sp3.jpg

CH3NH3+ must have a missing electron and somewhere that bromine could attach to (I'm sorry if this is a crude way of explaining this but I am trying) the problem is that there is NO WHERE for the bromine to have been bonded to; it certainly was not attached to hydrogen and there's no room on Nitrogen and Carbon so it couldn't have been attached there (those two don't even have d orbitals (expanded octet) although maybe that's only a consideration in covalent bonding?)

 

[Borek]

something-NH₂ behaves just like NH₃ - it gets protonated and that's where the additional charge comes from. H⁺ goes into the lone electron pair. As you have correctly stated, bromine is just a spectator. Technically ammonia or amine protonation is very similar to water protonation, just ammonia is central atom plus three H plus one lone pair, while water is a central atom plus two H plus two lone pairs.

 

[mnnob07]

So something-NH2 gets protonated to something-NH3+

how does the bromine attach to something-NH3+?

 

[Borek]

It is not bromine, but Br^-, I told you it is an ionic compound. Just like NaBr, or - more closely - NH₄Br.

 

[mnnob07]

alright, I saw in a couple places that CH3NH3Br and CH3NH3Br were salts of CH3NH2 and HBr/HCl but they were probably just wrong. the ionic bonding makes sense, thanks.

 

[Borek]

But they are salts of these. Just nitrogen gets protonated, that's where the cation comes from. HBr dissociates (it is strong acid), proton protonates nitrogen, Br^- is left as counterion

 

[mnnob07]

step-by-step:

CH3NH3Br is a salt and dissociates as:
CH3NH3Br -> CH3NH2 + HBr

CH3NH2 protonates:
CH3NH2 and H20 <-> CH3NH3+ and OH-

but there's also:
HBr -> H+ and Br-

now, while CH3NH2 protonates to give us the CH3NH3+, the Kb given for that seems to suggest it's not going to be much of a factor with as much as the strong acid, HBr from the salt dissociates. That is the step that leads me to believe the pH is actually much lower than simply looking at CH3NH3+ because there is a step between CH3NH3Br dissociating until we get to the CH3NH3+. We haven't gotten to buffers yet, though...

 

[Borek]

No...

You start with CH₃NH₂. Nothing dissociated yet (although it is a weak base, so it already reacts to some extent with water producing OH^-, we will leave it for now).

You have separate solution of HBr - fully dissociated into H⁺ and Br^-.

You mix both solutions.

CH₃NH₂ reacts with H⁺:

CH₃NH₂ + H⁺ = CH₃NH₃⁺

Now you have solution containing CH₃NH₃⁺ cation and Br^- anion. When dried, it will give ionic solid. When dissolved, it will give the same solution of two ions.

Sorry, I have no idea how to explain it further or more clear, if you still don't understand, perhaps someone else will be able to help.

 

[symbolipoint]

Start over a bit. This is mostly simple acid-base chemistry.
You may be a bit familiar with ammonia, NH3. A strong acid can
neutralize this to give the ammonium cation, NH4+.
NH3 + HCl ------------> NH4+ + Cl-

Instead of ammonia, a nitrogen-base can be an amine such as
methylamine, CH3NH2. A strong acid can neutralize this to give
the methylammonium cation, CH3NH3+. (All hydrogen halides are
strong acids, except for HF).
CH3NH2 + HBr ----------> CH3NH3+ + Br-

When you consider each salt by itself, without any added strong acid,
there is NH4Cl for the first situation, and there is CH3NH3Br for the
second situation.
Chlorides and bromides are completely ionic, and generally very soluble
in water.
EACH OF THE COMPOUNDS IS A WEAK ACID.

Just as ammonium dissociates to NH3 and H+,
methylammonium dissociates to CH3NH2 and H+

 

[mnnob07]

This is the third time I'm trying to post and physicsforums keeps saying that some security token is missing and I lose the post.

I think I get it

CH3NH3Br -> CH3NH2 and H30+ and Br-

now, I'm going to just reverse the next step I had in my last post and it makes this much more clear:
CH3NH3+ and H20 <-> CH3NH2 + H30+

because there is SO MUCH H30+ and CH3NH2 from the CH3NH3Br this equilibrium is VERY lopsided. by Le Chatlier, the reaction will shift heavily to the left, effectively eating up most of those H30+ from the HBr. the rest of the problem follows as normal. That might be all I was missing.

 

[symbolipoint]

mnnob07, You seem now to understand most of the quality and reaction. What is not too clear is your description of "lopsided...". Anyway, you have apparently made important progress.

 

[mnnob07]

right; first I was confused why I kept on being told that CH3NH3Br went right to CH3NH3+ and Br- now I see how it gets there.

I meant lopsided by the fact that in the second step, (that is, between the newly formed CH3NH2 and H30+ and the resulting CH3NH3+) there were 0.0M CH3NH3+ and .15M CH3NH2 and H30+... lopsided may not have been the best word choice... oh well

In any case, that is how CH3NH2 and H30+ went to CH3NH3+ so strongly which I think is really what Borek was trying to say while I was still thinking there was a lot of H30+ leftover from the HBr. Now, I think I got this down let me know if there's anything I'm still missing...

mathematically:

CH3NH3+ and H20 <-> CH3NH2 and H30+

Ka = Kw/Kb(of CH3NH2) = (1.0*10^-14)/(4.4*10^-4) = 2.3*10^-11

Ka = ([CH3NH2][H30+])/[CH3NH3+]

___CH3NH3+___CH3NH2___H30+
I_____0_________.15______.15
C____+x_________-x_______-x
E_____x________.15-x____.15-x(2.3*10^-11) = ([.15 - x][.15 - x])/[x]

x = .149998142594 ~ .15 = [CH3NH3+]

Thus, Borek correctly assumed that above reaction virtually went to completion so he went right to CH3NH3+ and calculated H30+. Now, there are several correct ways of doing this and the above is the same equilibrium and if you temporarily ignore sig figs you get the same answer -log[H30+] = -log(.15 - x) = -log(.15 - .149998142594) = 5.73. Of course, with respect to sig figs, Borek's method would be more sound and succinct probably from a chemist's point of view.

Thanks Borek and symbolipoint!

 

[Borek]

I have not presented any method yet, I was referring to qualitative description so far. You are right, protonation reaction is shifted (almost) completely to the right.


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