[inferno666]

Suppose I have 2 moles of lidocaine base that I want to protonate to obtain lidocaine HCl. I have the lidocaine base dissolved in a suitable solvent and I want to add HCl(aq) to cause the salt to precipitate. Since I want to protonate the lidocaine base, then I need 2 moles of protons, correct? My textbook says that a 1M solution of HCl completely ionizes, but solutions of higher concentrations will not ionize completely. Therefore, if I have a 12M solution of HCl, then I must dilute it to a 1M solution and then add two liters of this diluted solution to my dissolved lidocaine base. Is this correct?

I guess what I am really asking is: when protonating an amide, the only HCl that really matters, in a sense, is the HCl that ionizes. Is this correct?

[AWK]

In this molecule there is a tertiary amino group and amide group. Only one is sufficiently basic to form salt.

[inferno666]

Sorry, I don't follow. I am a beginner and obviously still have a lot more to learn. Is there a simpler answer to my question?

[AWK]

Lidocaine forms salt with HCl  1:1 only.

[inferno666]

So you're saying that for 1 mole of lidocaine I need 1 mole of HCl? But specifically, I need 1 mole of IONIZED HCl right? Otherwise how else could the lidocaine base be protonated? So I couldn't use  12 M solution of HCl because itwouldn't fully ionize and hence I wouldn't know how much H+ is actually in the solution. Thus I have to dilute the solution to a 1M solution to ensure that there is 1 mole of H+ available (complete ionization). Is this correct? I.e. the only HCl that will be neutralized is the HCl that ionizes?

[AWK]

Have you ever heard about strong acids?

[inferno666]

Yes. A strong acid entirely ionizes in water. But I was under the impression that HCl ionizes entirely only if its  a 1M solution or less. I'm guessing I'm wrong about that?

[AWK]

Adding concentrated HCl to your solution you dilute it. When salt is formed (hydrochloride) then its concentration is close to zero.

[Arkcon]

Yes. A strong acid entirely ionizes in water. But I was under the impression that HCl ionizes entirely only if its  a 1M solution or less. I'm guessing I'm wrong about that?

Yes.  A 2M solution would be mostly ionized.  Crazy concentrations, like 5, or 6, or 8 M  won't ionize completely in water.

But here's a funny question for you.  I said higher concentrations don't ionize.  Consider this ... you have 3 moles of strong base in solution, you add 4 moles of strong acid.  If 4 moles ionizes only 80%, how much ionizable acid is left?

[inferno666]

Yes. A strong acid entirely ionizes in water. But I was under the impression that HCl ionizes entirely only if its  a 1M solution or less. I'm guessing I'm wrong about that?

Yes.  A 2M solution would be mostly ionized.  Crazy concentrations, like 5, or 6, or 8 M  won't ionize completely in water.

But here's a funny question for you.  I said higher concentrations don't ionize.  Consider this ... you have 3 moles of strong base in solution, you add 4 moles of strong acid.  If 4 moles ionizes only 80%, how much ionizable acid is left?

Sorry for my late reply. I have been very caught up with things...

I have asked other people and they said that the percentage of ionized acid is independent of its concentration. I keep getting contradicting answers to this question.


As for the question you posed, 3.2 moles of acid would ionize. Three moles of this acid would be neutralized by the base. That leaves 0.2 moles of ionized acid and 0.8 moles of unionized acid. HOWEVER, the concentration of acid in solution now is much lower and so the rest of the acid (the 0.8 moles) should ionize, leaving a total of 0.2 + 0.8 = 1.0 mole of ionized acid. Is this right?

[inferno666]

Adding concentrated HCl to your solution you dilute it. When salt is formed (hydrochloride) then its concentration is close to zero.

But I am adding my concentrated HCl to a solution of lidocaine base ans a nonpolar solvent. Let's say this solvent is n-hexane. I don't know how the n-hexane will affect the ionization of the HCl, so I can't assume that the HCl will ionize fully when its mixed with the lidocaine/hexane solution.

[AWK]

I want to add HCl(aq) to cause the salt to precipitate

.
You add water with HCl then you perform reaction in water. Your salt show much, much better solubility in water then in hexane hence you will not get precipitate without evaporation of hexane and at least part of water.

[Borek]

I have asked other people and they said that the percentage of ionized acid is independent of its concentration.

So they were wrong.

Please read any general chemistry book, acid/base equilibrium (pH calculation) section. Or try here: http://www.chembuddy.com/?left=pH-calculation&right=toc (although it may be a bit advanced for a beginner).

HCl is a very strong acid, so for most practical purposes we can assume it is entirely dissociated (this is not entirely true, but with a Ka in the 10⁷ range that's quite a good approximation). However, for weaker acids situation looks differently - their dissociation percentage is not constant, it is a function of concentration and can be easily calculated from Ka and C. Sorry to say that, but as long as you don't know theory behind and don't understand how it is done, we are wasting time for a useless and unnecessary discussion.

[Arkcon]

As for the question you posed, 3.2 moles of acid would ionize. Three moles of this acid would be neutralized by the base. That leaves 0.2 moles of ionized acid and 0.8 moles of unionized acid. HOWEVER, the concentration of acid in solution now is much lower and so the rest of the acid (the 0.8 moles) should ionize, leaving a total of 0.2 + 0.8 = 1.0 mole of ionized acid. Is this right?

Nope.

Picture this.

You have two beakers.  One with 4 M acid, one with 1M acid.  You add 3 moles of base to the first.  How much acid is in each?  How much of that acid is completely ionized? 

[inferno666]

As for the question you posed, 3.2 moles of acid would ionize. Three moles of this acid would be neutralized by the base. That leaves 0.2 moles of ionized acid and 0.8 moles of unionized acid. HOWEVER, the concentration of acid in solution now is much lower and so the rest of the acid (the 0.8 moles) should ionize, leaving a total of 0.2 + 0.8 = 1.0 mole of ionized acid. Is this right?

Nope.

Picture this.

You have two beakers.  One with 4 M acid, one with 1M acid.  You add 3 moles of base to the first.  How much acid is in each?  How much of that acid is completely ionized?

Sorry for the late reply. I replied Friday night but I dont see my post here...

So in the first beaker, 0.8*4= 3.2 moles of acid is ionized. When I add the 3 moles of base to this beaker then 3 moles of acid is neutralized. This leaves 0.2+0.8= 1 mole of acid in the first beaker.

I am assuming that all of the acid, the 1 mole, is ionized in the second beaker. But I cant say this with certainty because not enough information is given.