Basic Chem Question

[Marrowist]

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Can someone elaborate on this? The answer is E, but I don't understand why this would be true. Thanks.

1. In an aqueous solution initially containing 0.10 mol/L of lactic acid to which 0.05 mol of NaOH has been added per liter of solution, _____.
   
   1. [H+] > [lactate ions]
      
      
   2. [lactate ions] > [lactic acid]
      
      
   3. [H+] ≈ [lactic acid]
      
      
   4. [H+] ≈ [lactate ions]
      
      
   5. [lactic acid] ≈ [lactate ions]
      

 

[Doogie.Howser]

The key here is to notice that you're adding half the amount of acid present. NaOH, a strong base, is the limiting reagent and the reaction will proceed to rid of the 0.05 mol of NaOH, leaving you with a solution of water with equal concentrations of lactic acid and lactate. If you're still stuck, write out an ICE table to see you'll be left with 0.05 of lactic, 0.05 of lactate.

 

[Zenabi90]

Try to think in simpler terms. Many get tripped up with decimals and unfamiliar acids in word form rather than in a chemical formula.

Lactic acid is just a monoprotic acid. Donates one proton. So if I were to compare it simplistically to HCl where its [H+][Cl-], then lactic acid is the same as [H+][Lactate-] or HLactate.

Because everything is "per liter" you can just read as whole numbers, meaning 10 and 5 instead of 0.10 and 0.05.

So You have 10 HLactate. 5 NaOH come along and rip away the H+ from five of the HLactate. Now you have 5 whole, unmolested HLactate and 5 [Lactate-] loners. In other words, 5 Lactic Acid and 5 Lactate ions. They are roughly equal, so E.

 

[Marrowist]

I keep getting stuck on he fact that acetic acid is a weak acid and will not disassociate completely... the picture is starting to come together. Thanks for all input.


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[Zenabi90]

I keep getting stuck on he fact that acetic acid is a weak acid and will not disassociate completely... the picture is starting to come together. Thanks for all input.

Just want to note that your original post says LACTIC acid, not ACETIC acid.

Also, NaOH is a pretty strong base, so even if in aqueous environment acetic acid alone doesn't completely dissociate, the NaOH is still gonna rip that proton right off.

 

[Marrowist]

Just want to note that your original post says LACTIC acid, not ACETIC acid.

Also, NaOH is a pretty strong base, so even if in aqueous environment acetic acid alone doesn't completely dissociate, the NaOH is still gonna rip that proton right off.

Yes, Lactic Acid. Sorry bout that.


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[aldol16]

Lactic acid is a weak acid. So the majority of it is gonna exist as the protonated form in aqueous solution. You add 0.5 equivalent of strong base and that strong base is gonna pull 0.5 equivalents of protons off of the lactic acid molecules. So you'll be left with essentially ~0.5 M of lactic acid and 0.5 M of lactate. This makes for approximately equal ratios of lactic acid to lactate. The key here is that lactic acid is a weak acid and doesn't dissociate to an appreciable extent on its own, without strong base present.

 

[rlaboss]

Lactic acid is a weak acid. So the majority of it is gonna exist as the protonated form in aqueous solution. You add 0.5 equivalent of strong base and that strong base is gonna pull 0.5 equivalents of protons off of the lactic acid molecules. So you'll be left with essentially ~0.5 M of lactic acid and 0.5 M of lactate. This makes for approximately equal ratios of lactic acid to lactate. The key here is that lactic acid is a weak acid and doesn't dissociate to an appreciable extent on its own, without strong base present.

When I originally saw this I thought the because the pH of water is 7 and the pKa of lactic acid is 3.86, I thought that most of the lactic acid must be deprotonated. But this would have been under the assumption that the pH of the solution was still 7, even with the lactic acid. After some math I discovered the pH was actaully 1 with 0.10 mol lactic acid:

Ka = 10^-pKa
Ka = 10^-3.86 = 0.000138038426 = [H+][lactate] / [lactic acid]
[H+] = [lactate] = "X"
[lactic acid] = 0.10 - X
0.000138038426 = (X^2) / (0.10-X)
0.0000138038426 - 0.000138038426(X) = X^2
0 = X^2 + 0.000138038426(X) - 0.0000138038426
--> solve for quadratic --> X = 0.003646974 (Obviously not expected to do on mcat)
pH = 10^(-[H+]) = 10^-0.003646974
pH = 1 which is << 3.86

For the MCAT, do we just need to be able to identify weak acids and strong acids and for strong acids assume they are completely dissociated in water and for weak acids assume they are only negligibly deprotonated in water?

 

[Zenabi90]

For the MCAT, do we just need to be able to identify weak acids and strong acids and for strong acids assume they are completely dissociated in water and for weak acids assume they are only negligibly deprotonated in water?

It wouldn't hurt to memorize the common strong acids and bases.

Common strong acids (7): HCl, HNO3, H2SO4 (HSO4- is a weak acid), HBr, HI, HClO4, HClO3
Common strong bases: hydroxides (-OH) of Group I/II metals. LiOH, NaOH, KOH, etc.

These are the most common and anything else I would just assume is a weak acid/base. Most efficient use of your mental storage imo.