Complete Dissociation?

  • Thread starter PEZenfuego
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  • #1
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This is probably an easy question, but I can't get it straight and searching has not helped. So when salts go into solution they completely dissociate if they contain anions of strong acids or cations of strong bases, right? I'm still shaky on this concept though. Take Sodium carbonate for example. Na2CO3 completely dissociates because of the Na+, right? NaOH completely dissociates because of the Na+ also. H2CO3 does not completely dissociate. Here's where I am confused...

My understanding is that once Na+ dissociates it is done reacting. It just sticks around as a spectator. So why when I combine H2CO3 and NaOH do we get sodium bicarbonate? Don't the sodium ions want to dissociate from the hydroxide and call it a day?

How can we tell when these ions want to react? I am a little fuzzy. Thanks in advance.
 

Answers and Replies

  • #2
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Salts always completely dissociate in a polar solvent (like water for instance). Think NaCl for example. You drop it in water and if you could magnify enough you would not find NaCl anymore but Na+ in a solvent shell completely separate from Cl- in a solvent shell. You can think of Na+ as being the counter ion to NaOH which is a strong base and thus will not want to associate with any other charges in solution (like an OH- for example). You can think of Cl- as the conjugate base to a strong acid so when it dissociates from Na+ it will not associate with any other counter ions (like H+ for example). Thus what you drop NaCl into solution whatever concentration of H+/OH- will not change.

Think of how weak acids (and their salts) behave in solution.

If I drop acetic acid into water, does it completely dissociate?

Now what will happen if I have the acetate anion with a counter ion (Na+) and I dissolve that salt, in water? You have half the story correct, the Na+ will not want to associate with any other charges in the solution, its 'happy' being solvated by water.

What will CH3COO- do?

What will happen if there is some H+ floating around?

What do we call a solution with a lot of H+?

What is the pH of such a solution?

What will happen to the pH of this solution when we dissolve sodium acetate in it?

The case of Carbonate is a bit more complicated in that Carbonic acid (H2CO3) is a polyprotic acid and is always in equilibrium with HCO3-, CO3(2-), CO2 + H2O. For now, tackle the simpler things and then you can learn about the exceptions and more complicated things.
 
  • #3
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Now what will happen if I have the acetate anion with a counter ion (Na+) and I dissolve that salt, in water? You have half the story correct, the Na+ will not want to associate with any other charges in the solution, its 'happy' being solvated by water.

What will CH3COO- do?


Acetic Acid does not completely dissociate which means that it dables with the proton and reaches equilibrium

What will happen if there is some H+ floating around?

This will make the solution acidic as it dables with the CH3COO-

What do we call a solution with a lot of H+?

Acidic

What is the pH of such a solution?

<7

What will happen to the pH of this solution when we dissolve sodium acetate in it?

The sodium dissociates from the acetate which is a fantastic way to get the CH3COO- into solution and the Na+ becomes a spectator. As the acetate ions increase this shifts the equation to the left. As the equation shifts, the [H+] decreases which raises the pH.

I get all of this (or at least I feel like I do), but I'm still confused about baking soda.
 
  • #4
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Acetic Acid does not completely dissociate which means that it dables with the proton and reaches equilibrium
There is no acetic acid in solution. All you have is acetate anion (and of course sodium cation, water, and H+). Also please clarify what you mean by dabble.

This will make the solution acidic as it dables with the CH3COO-
Go over what acidic solution is and what basic solution is (and what an acid/base is).

Is CH3COO- an acid or a base?

I don't want to give away answers as it is a lot more helpful for you to figure it out on your own but you are wrong here.

What do we call a solution with a lot of H+?

Acidic
Correct

What is the pH of such a solution?

<7
correct

What will happen to the pH of this solution when we dissolve sodium acetate in it?

The sodium dissociates from the acetate which is a fantastic way to get the CH3COO- into solution and the Na+ becomes a spectator. As the acetate ions increase this shifts the equation to the left. As the equation shifts, the [H+] decreases which raises the pH.
Can you clarify which chemical equation you are referring to?

You got the question correct but it may be due to the wrong reasons which is why I'm asking for clarification.

I get all of this (or at least I feel like I do), but I'm still confused about baking soda.
You seem to be on the right track but I think you should go back and re-read your text on acids/bases/equilibria etc. I may be misinterpreting what you are saying but it seems like you have things mixed up somewhere.

The story with baking soda gets messy which is why you're better off getting the basics down first.
 
  • #5
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Okay so back to the acetic acid with clarification.

Here's the equation when added to water

CH3COOH + H20 ⇌ H3O+ + CH3COO-

but from my understanding we can simplify it to


CH3COOH ⇌ H+ + CH3COO-

So there are some CH3COOH still in solution see. The donation of the proton makes this solution acidic.

CH3COO- is a base because when it reacts with water we get this:

CH3COO- + H20 ⇌ CH3COOH + OH-

Since the ion reacts by accepting a proton it is a base, or we could look at it as having hydroxide as a product also making it a base.

Of course the only way to get CH3COO- into solution is by adding a salt with a strong base cation, like NaCH3COO.

Now if we have CH3COOH in solution we have the equation:

CH3COOH ⇌ H+ + CH3COO-

To this we are going to add NaCH3COO. Since the Na+ doesn't react with anything and dissociates completely, CH3COO- is what needs to be focused on.

Since we are adding the CH3COO- ion, the concentration of CH3COO- will increase which will cause reaction to shift left and increase the [CH3COOH]. Since that proton is being shoved over to the left, there will be less proton just sitting in solution (which makes it more basic) To clarify that completely:

CH3COOH ⇌ H+ + CH3COO-
.....up.......down.......up....

Therefore the [H+] decreases which causes a raise in pH.

These are the basics as I understand them, for better or worse.
 
  • #6
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Hi
Brønsted–Lowry acid–base theory may solve the Problematic
 
  • #7
397
21
Okay seems like you have it down. I had my doubts because one of your first answers were that acetate will make the solution more acidic but that may have just been a typo or hasty reply.

So now onto sodium bicarbonate.

Because CO3 has a (-2) charge it can accept 2 protons and become H2CO3. This is what is known as a polyprotic acid. Poly because it has more than one acidic proton. Other examples are sulfuric acid and phosphoric acid, which have 2 and 3 acidic protons respectively.

Now what NaHCO3 can do in solution is this. When you dissolve the salt you will get Na+ and HCO3-. You are correct in that Na+ just sorta hangs out and does nothing but you then have a substance which is amphoteric (meaning can act as both a base and acid). Depending on the pH of the solution the bicarbonate anion can let go of its last proton to form CO3(-2) or it can become protonated to form carbonic acid (H2CO3). These species are all in equilibrium and the equilibrium will shift depending on pH of solution (IOW how much H+ or OH- is floating around). Another confounding factor to all of this is the fact that carbonic acid is not stable and readily decomposes into carbon dioxide and water.

So it somewhat looks like this

H2O + CO2 ⇌ H2CO3 ⇌ H+ + HCO3- ⇌ 2H+ + CO3(-2)

Therefore when you put sodium bicarbonate (a salt of Na+ and HCO3-) into something like vinegar it completely dissociates to give the two ions, the Na+ does nothing, but the bicarbonate anion will become protonated and then decompose in water and carbon dioxide. Every High School kid sees a demonstration of baking soda plus vinegar giving you a cool bubbling effect. Thats from the evolution of carbon dioxide gas. If you have a basic solution the equilibrium gets pushed to the opposite side, the bicarbonate is deprotonated.

PEZenfuego said:
My understanding is that once Na+ dissociates it is done reacting. It just sticks around as a spectator. So why when I combine H2CO3 and NaOH do we get sodium bicarbonate? Don't the sodium ions want to dissociate from the hydroxide and call it a day?
What we get is really a matter of what the pH of solution is. Do you know about pKa's and Keq's yet? Hydroxide can deprotonate the second acidic hydrogen of bicarbonate yielding a CO3(-2). If all ions stay in solution (meaning no precipitate forms) you don't really get Na2CO3 as water will solvate both of the ions and they will be seperate from each other. However if you, say, evaporate all the solvent and recover crystals these crystals will now be Na2CO3 and will only act as a base if re-dissolved in new solvent.

So basically even though acids and bases can be considered as salts they don't behave exactly as salts. All salts completely dissociate (assuming adequate solubility) while acids that are not considered strong acids will not dissociate fully. NaCl behaves as a salt and fully dissociates (the fact that nothing happens afterward is because they are the conjugates of strong acids/bases). Acetic acid does not fully dissociate and an equilibrium exists between the acid and the conjugate base.

Whenever you think of these things consider what you have in the flask. Salts will fully dissociate but then you may have the conjugate base of a weak acid which then may act as a base and become protonated (if the solution is acidic). Then you may have the case of amphoteric molecules which can "go both ways." Which way they go depends on conditions.

Take away is this: Salts always fully dissociate in solution (assuming they are soluble) but acids do not. Salts and acids are not exactly the same thing and behave differently. Sodium bicarbonate is a salt but when it fully dissociates it gives you something that is not a strong acid and can deprotonate or become protonated (because it "goes both ways").

Hope this helps.
 
  • #8
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Yes that helps a lot and I think I understand it. Let me see:

NaOH --> Na+ + OH-

H2CO3 ⇌ H+ + HCO3-

H+ + OH- ⇌ H20
-----------------------------
I am going to cancel things just to see things clearly. I realize this formula doesn't just happen all at once.

NaOH + H2CO3 --> Na+ + HCO3- + H20

WE DO NOT GET

NaOH + H2CO3 --> NaHCO3 + H20 because those ions are dissociated in the solution

Only when we evaporate the water, do the Na+ and HCO3- crystalize to form NaHCO3. This is the way I understood what you wrote. Did I understand it correctly?

So here is a resulting question:

I want to make sure that I understand the property of HCO3-. So if a solution of water has excess protons, the HCO3- will take them up and form H2CO3? If there are more OH- ions, then it will give up its H+ to form CO3 (2-)?

And taking it one step further (hopefully not too far)

Let's say that we have a neutral solution (water) and we are adding NaHCO3 and let's say I don't know if it will act as an acid or base.

It could go: HCO3- + H20 --> H2CO3 + OH- OR

HCO3- --> CO3 (2-) + H+

One is basic and the other is acidic.

Can I say that since the Ka of HCO3- (4.7x10-11) is smaller than the Kb (2.1x10-4), that it will go with the kb and act as a base? That it will perform the first of the two equations I wrote?
 
  • #9
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Yes that helps a lot and I think I understand it. Let me see:

NaOH --> Na+ + OH-

H2CO3 ⇌ H+ + HCO3-

H+ + OH- ⇌ H20
-----------------------------
I am going to cancel things just to see things clearly. I realize this formula doesn't just happen all at once.

NaOH + H2CO3 --> Na+ + HCO3- + H20

WE DO NOT GET

NaOH + H2CO3 --> NaHCO3 + H20 because those ions are dissociated in the solution
More or less yes. Just keep in mind that Carbonic acid is not stable and will decompose into water and carbon dioxide (and carbon dioxide is a gas that bubbles out of solution which drives the equilibrium towards full decomposition ala Le Chatalier). This is why sodium bicarbonate gets a bit messy. But in a hypothetical type of way you pretty much got it. Also keep in mind that if you add enough NaOH you can get Na2CO3.

Only when we evaporate the water, do the Na+ and HCO3- crystalize to form NaHCO3. This is the way I understood what you wrote. Did I understand it correctly?
You got it.

So here is a resulting question:

I want to make sure that I understand the property of HCO3-. So if a solution of water has excess protons, the HCO3- will take them up and form H2CO3? If there are more OH- ions, then it will give up its H+ to form CO3 (2-)?
Yes sir. You can write out the chemical equations then go through different situations in your mind and use Le Chatalier's principle or use Ka expressions (and ICE tables or H and H equation) etc to get numbers. Its a good exercise to do, will help a ton come exam time.

And taking it one step further (hopefully not too far)

Let's say that we have a neutral solution (water) and we are adding NaHCO3 and let's say I don't know if it will act as an acid or base.

It could go: HCO3- + H20 --> H2CO3 + OH- OR

HCO3- --> CO3 (2-) + H+

One is basic and the other is acidic.

Can I say that since the Ka of HCO3- (4.7x10-11) is smaller than the Kb (2.1x10-4), that it will go with the kb and act as a base? That it will perform the first of the two equations I wrote?
Off the top of my head I'll say yes but you may want to wait for confirmation from others or your text/professor etc on this.

But in general it seems as though you pretty much have this down.
 
  • #10
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Thank you so much Mr. Yanick. I really appreciate you going to all this trouble just to help with my curiosity. Are you a teacher? You are quite good at explaining things.
 
  • #11
Borek
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Let's say that we have a neutral solution (water) and we are adding NaHCO3 and let's say I don't know if it will act as an acid or base.

It could go: HCO3- + H20 --> H2CO3 + OH- OR

HCO3- --> CO3 (2-) + H+

One is basic and the other is acidic.
It goes both ways at the same time, and in about the same amount ("symmetry does the trick"). See pH of amphiprotic salt.
 
  • #12
397
21
Thank you so much Mr. Yanick. I really appreciate you going to all this trouble just to help with my curiosity. Are you a teacher? You are quite good at explaining things.
Your welcome. Glad I could help.

Oh and no, I am not a teacher, just a lowly student who has really come to love this stuff.

Thanks Borek, thats a great site.
 

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