[Arkcon]

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

Sorry about that, I deleted it, because it was a quote of a quote of a quote, without any new text within, so far as I could see.  More of a bμmp, really.

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.

That is not how I wanted you to figure this one out.

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.

I have no idea why you would say that. 

You already stated, at the beginning of this thread, that you knew that 1 M HCl completely ionizes.  You are likely correct that 12 M HCl doesn't completely ionize.  But we tried to tel you that 2 or 3 molar likely ionizes completely.

At any case, no matter what, the first 1 Molar of any solution of HCl ionizes.  And once consumed, the second one is available to likewise ionize completely, and so your original plan, to add two volumes of 1 M HCl, is exactly what you want to do.

[docnet]

I think I have an answer for Arkon's question.

If 4 moles of strong acid is added to 3 moles of strong base, 3 moles of both acid AND base will be consumed, leaving 1 mole of strong acid in the solution, which should be assumed to ionize completely in this scenario.

The fact that 4 moles of acid only ionizes to 80% doesn't matter because the acid-base neutralization will consume 3 of those 4 moles... leaving 1 mole of acid left in the solution, which ionizes more completely in dilute concentrations (4 times more dilute in this case).

I hope someone can correct me if it's wrong.

[Arkcon]

I hoped the OP would get that eventually, but they insisted on never reaching that point repeatedly.

[docnet]

LOL