[Stamos666]

     Having some trouble with this problem:

     Formic acid (HCO2H) has Ka= 1.77x10(-4)

A. What are the molar concentrations of the hydronium ion and formate ion (HCO2) in a .1M aqueous solution of formic acid? B. What percentage of the formic acid is ionized?

B.  What percentage of the formic acid is ionized?

 I'm using Solomon and Fryhle's Organic Chemistry 9th Edition book. This is on page 100 if anyone else has it. As you can see, they NEVER teach you how to find percentage or molar concentration. Anyone know how to do this? I don't have my solution manual yet.


Also, I'm having trouble with Lewis acid/base problems. I understand the concept but some parts elude me. For example:

 Use the curved-arrow notation to write the reaction that would tkae place between dimethylamine (CH3)2NH and boron trifluoride.

I know that BF3 is the Lewis acid and the Lewis base is dimethylamine. However, there is a proton on the Lewis Base (the H). Where does that go? Also, does the dimethylamine donate the unshared electron pair from N? What is the end result?

I'm just confused when they put in the H in Lewis bases, it throws me off especially since in all the examples in the book, there is an H in the Acid which goes to the base.

Another striking example of this is THIS:
Methanol (CH3OH) reacts with BF3, what is the lewis acid-base reaction

Can anyone help?? Thank you so much in advance.

[orgoclear]

              HCOOH + H20 = HCOO- + H3O+
initial             0.1
at equilibrium   0.1(1-x)       0.1x       0.1x

Ka = [H3O+][HCOO-]/[HCOOH]

from this you can calculate x which is the dissociation. percentage dissociation = 100*x
conc. of HCOO- = 0.1x = conc. of H30+

[gfunk]

1) What does K_a mean?  What formula describes it?

2) Boron is normally trivalent, and therefore is electron-deficient.  In Lewis acid-base reactions, BX₃ is usually the electron acceptor.  Besides, the nitrogenous proton on dimethylamine isn't all that acidic (consider the basicity of LDA for example).  As well, the lone-pair electrons of nitrogen in amines make a good nucleophile (electron-rich).

As for CH₃OH and BF₃, think about the possible roles BF₃ can play.  Can it be a Lewis acid?  A Lewis base?  Go from there.

Good luck!

[Stamos666]

Sweet! Okay, so like I had previously thought, the lone pair of Nitrogen donates it's electrons to the trivalent boron. My main question is, however, what happens to the nitrogenous proton after nitrogen donates it's pair of electrons to Boron? Does it join Boron to become BF3H because it is attracted to the new electron pair on the Nitrogen?

As for CH3OH and BF3, I know that BF3 is a Lewis Acid and acts like BF3 in the first problem I gave. I'm assuming the oxygen gives its unshared pair of electrons to B. Again, as usual, my problem is...where does the H go?

[Squirmy]

deleted

[orgopete]

This question needs to be answered as all electron transfer steps. You have correctly identified the motivation for a reaction to take place. The electrons of nitrogen are not tightly held (compared to O,F,Cl,Br). Boron is planar with an exposed positively charged nucleus. The electrons of the boron are also being pulled away from that nucleus by the fluorine atoms. Therefore it can be anticipated that boron can attract the electrons of nitrogen.

Once the complex has formed, the electrons on nitrogen will be more tightly held due to boron pulling one pair of electrons away from the nitrogen. That will make the remaining electrons on the nitrogen to be more tightly held than before. The boron will now have a completed octet and because the shared number of electrons is greater than the number of nuclear protons, the boron is noted with a negative formal charge. Will boron be able to attract an additional pair of electrons from the nitrogen?

An aside, I don't know this so I am only guessing. In this instance, if BF3 were being added to the amine, I could imagine that another molecule of the excess amine during addition to abstract the proton from the ammonium nitrogen. Then the non-bonded electrons could form a double bond to the boron with an expulsion of fluoride to give F2B=N(CH3)2 + (CH3)2NH2(+) + F(-). This could also be written as a more favorably appearing resonance structure F2BN(CH3)2. I don't think that is what is being expected from this problem, but since a question was asked about whether a proton transfer could take place, I thought I would suggest a rational proton transfer reaction. Again, for this reaction to take place, it is the non-bonded electrons of a nitrogen would be attracted to the proton of the ammonium nitrogen of the F3BNH(CH3)2 complex.

[Stamos666]

Ah gotcha, I don't think the problem goes that in-depth and advanced though. I just need to know what the products look like in this Lewis Acid/Base reaction.

I thought it would be (CH3)2NH +BF3 ---> (CH3)2N + HBF3-

But I have a feeling that is wrong.

As for the second one:
Methanol (CH3OH) reacts with BF3, what is the lewis acid-base reaction

Does the Oxygen give up the electrons?

[orgopete]

No, this is wrong.
(CH3)2NH +BF3 ---> (CH3)2N + HBF3-

There isn't any rational for the complex (CH3)2HN(+)-B(-)F3 to transfer a proton to the boron. It already has an excess of electrons, hence its negative formal charge. The complex is the product.

For methanol, it is the same addition of a pair of electrons from the oxygen.