[r4bbit_]

HI, our school is making us determine an unknown acid's identity
So far I've done everything possible and the following r my results:

molar mass of 67 g/mol
monoprotic
Pka of 4.4
soluble in water
weak acid
solid at room temperature
melting point of 135
boiling point around 140
produces ester with ethanol

But i just cannot find the acid, if anyone knows where i should look or anything please reply!

[azmanam]

Where did you get 67 gmol^-1 from?  I've tried several common bonding patterns, and I can't get a molecule down to 67 gmol^-1.  The one combination I got to 67.9 would have a pK_a much higher than 4.4

Is it necessarily an organic acid?

Have you tried the flame test on the solid?  I would expect a positive flame test if my 67.9 guess were to be correct (but it's not correct).

[azmanam]

OK, I did a bit more searching and I found a compound that meets the following of your criteria:

monoprotic, soluble, weak acid, solid, mp, bp, (that's a big clue for me) and ester formation. 

the pKa of this compound is a bit lower than yours, but perhaps within experimental error depending on the level of sophistication of your equipment. 

The major criteria my compound fails is your molar mass.  Mine is more than 2x your value.  Will you check your MW again?

[enahs]

Please explain in detail how you did, and your calculations for the MW. I think it might be wrong.

[r4bbit_]

Well, what I did for MW is that I titrated using standardized NaOH with a very veri precise molarity. So I titrated it using phenolpthalene. And since the acid is monoprotic on the titration curve, moles of my acid = moles of Naoh Therefore i simply divded the mass.

[r4bbit_]

More specifically, for the titration curve it looks something like this
  |                            o  o  o  o o o
  |                       o
  |                   o
  |                   o
  |                   o
  |                 o
  |              o
  |          o
  |      o
  | .o
  |_______________________________

And since only one plateau is present, it showed me that my acid is monoprotic. Therefore 1:1 ratio with the NaOH. Also, since endpoint is also shown, i also applied Henderson Hasselbach's equation to find the Pka (midpoint of the endpoint).

If you have any suggestions for what the acid may be, feel free to let me know and I'll look it up. Thank you fello chemists.

[r4bbit_]

Yup, I also did a flame test, and the colour was bright orange, the solid ignited in about 10 seconds.
So I figured it had to be organic. My best guess so far would either be oxalic acid (similar numbers but not entirely) or benzoic acid (huge molar mass thats it).

[azmanam]

moles of my acid = moles of NaOH Therefore i simply divided the mass.

Interesting procedure.

Will you show us your calculations?  I'm fairly confident your MW is wrong.  67 is a pretty small molar mass for an organic acid.  The two smallest monoprotic organic acid, formic and acetic acid, have MW of 46 and 60 respectively.

Oxalic acid is diprotic, so that's out.  Benzoic acid is closer to my guess, but looking up the physical data - the melting point and boiling point are off.  I like where you're going, though.

You don't have access to an NMR or mass spec, do you?

[r4bbit_]

Heh, no I was looking for a NMR though, but we didn't have one.
Yeah, my values matched acetic acid PERFECTLY actually because I performed more than 2 trials for each test, and the averages actually turned out exactly like acetic acid's. For molar mass it does not necessarily have to be 67 g/mol because in my calcs, some of em were actually 62. So I would say 60~75 for MW b/c I also did uncertainty in my calcs:

volume NaOH   = (17.20±0.05)-(1.60±0.05)
      = 15.60±0.10 mL

mol NaOH    = [0.097140877±5.308764631×〖10〗^(-4)  mol/L]×(0.01560±0.00010 L)
      = (0.097140877 mol/L×0.01560 L) ± [((5.308764631×〖10〗^(-4))/0.097140877+0.00010/0.01560)×100%]
      = (0.001515397681)±(1.799575243×〖10〗^(-5)) mol

mol unknown acid = (0.001515397681)±(1.799575243×〖10〗^(-5)) mol
   
molar mass = (0.104±0.001 g)/((0.001515397681)±(1.799575243×〖10〗^(-5)) mol)
   = (0.104 g)/(0.001515397681 mol)±[(0.001/0.104+(1.799575243×〖10〗^(- 5))/0.001515397681)×100%]
= 68.62884991 ±[2.149065719%]
≈ 68.63 ± 1.47 g/mol 

That's calculation for one of the trials, hard to follow =(.

[azmanam]

Wow.  That's a lot of sig figs.  How'd you get those numbers?

First, I get 1.5x10^-7 mols NaOH, not x10^-8 as your calculations appear to.  Is that volume NaOH to phenolphthalein end point? or total NaOH for total titration?  Those two values mean different things.

Second, it 0.104 g the mass of the acid?  Is that from direct massing of the sample before titration?

Third, 0.104 g / 1.5x10^-7 mol = 68700 gmol^-1 as molar mass.  yikes.

Can you walk me through your calcs in words?

[r4bbit_]

Yup, sure.

Firstly, I calculated how much NaOH I poured into the erlenmeyer using my final burette reading and initial burette reading, which is 15.6 +/-0.1 mL.

After knowing the volume of NaOH, I multiplied it by concentrion to find moles.
The concentration of NaOH is around 0.1 M but not quite, it is 0.097±5.3×〖10〗^(-4)  mol/L.
This works out to about 0.00151539 mols of NaOH

Next, since it is monoprotic acid, (from titration curve) my moles of base should = moles of acid. Hence the amount of moles of acid is also 0.00151539... And yes 0.104 g is how much acid I measured before the titration. Therefore since 0.00151539 moles reacted with the base in a 1:1 ratio. 0.104g/0.00151539 = will give me the molar mass. which works out to be 68.

[azmanam]

oh. ok.  I was reading your 10^x numbers as part of the number, not the error.  My bad.  I'm with you now.

Ok.  I think we're on the right track.  Using your numbers, and my guess, your calculated molar mass is exactly one half of the molar mass of my guess.

Here's what I think.  Aside from being a base for your phenolphthalein titration, sodium hydroxide is also going to react as a nucleophile and chemically change your compound - eating up one molar equivalent in the process...  I think this is why you need 15 mL instead of 7 or 8 like I predict you should have needed.

Try your calculations again assuming you used 7.76 mL of base.  See where that gets you.  Did you notice a smell change after the titration (you may not have...)

This is excellent material for your lab report.  I'm going to try not to give you too obvious of hints so that you can honestly say you solved this yourself - your instructor should be very impressed here.  btw, did your instructor suggest the pK_a test you used?

Edited...

[r4bbit_]

When I halved the volume like you said, it works out to be 135 g/mol for MW, which would make everything fit for the benzoic acid guess (if that is the one you are referring to). However, I was a little confused about what you meant. Here is what I think you meant, please correct me:

1. NaOH should act as a nucleophile, so that means it should either react in Sn1 or Sn2 with benzoic acid (which carbon specificically im not sure). Therefore it is able to react with nucleophilic substitution at the same time as it goes into neutralization with NaoH. This is very vague, please clarify or correct me.

2. I was actually thinking about Le chatelier's principle. From my sheet I actually had the first pink appearing when it was 7 mL or 8 mL, but when i twirled the erlenmeyer, it disappeared. So I figured that perhaps that was the real endpoint, but since the amount of hydronium ions has decrease (in the neutralization) according to le chatelier, the acid must dissociate (with the Pka constant) to bring up more hydronium to be neutralized.

[r4bbit_]

regarding the pKa value, I actually figured that out myself with a titration curve. My instructor only supplied me with the acid, he's strict like that.
The pKa value of 4.44 something is averaged, but I saw benzoic acid's pka, it is fairly close.

[r4bbit_]

O, sodium benzoate. NaOH + Benzoic acid also produces sodium benzoate. Therefore each drop of my base that is added to the acid, will also react with benzoic acid to produce to make sodium benzoate as opposed to reacting in the neutralization reaction, thus I actually have 2 reactions going on at the same time...Is that what you are referring to?