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Topic: Solubility  (Read 11393 times)

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Offline a confused chiral girl

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Solubility
« on: July 25, 2007, 01:07:56 AM »
Hi there everyone!

I have some trouble with predicting which one is more soluble in which-  Benzoic Acid (pKa ~ 4) or Pyridinium (pKa ~ 5) in an aqueous 0.1M NaOH or 0.1M HCl.

I predicted pyridinium to be soluble in NaOH because pyridinium has an cationic group and is more basic with a pKa of 5, and more basic is more soluble in basic conditions. Benzoic acid is soluble in HCl. but the correct answer should be reversed of what I have, and I don't understnad the solution behind it.  ???

thank you in advance!   ;)

Offline AWK

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Re: Solubility
« Reply #1 on: July 25, 2007, 02:13:18 AM »
You try to compare compound - benzoic acid with pyridinium unknown salt.
Benzoic asid is poorly soluble in water. Its sodium salt is much more soluble. Solubility of benzoic acid in strong acid solution will be lower from that in water
Concerning Pyridinium salt - it is good soluble in water, in HCl also, but when you have pyridinium hydrochloride and put it into solution of HCl, solubility of Pyr.HCl can be slightly lower. When you mix pyridinium salt with NaOH, a neutralization reaction will occur and free pyridine will form. Pyridine itself is miscible with water, hence you cannot determine its solubility.
Poor question!
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Offline a confused chiral girl

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Re: Solubility
« Reply #2 on: July 25, 2007, 03:14:09 AM »
 :o really???...
this is a poor question...? but I'm not trying to determine the solubility though

I'm only trying to see which one is more soluble in which 0.1M solution..

Offline AWK

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Re: Solubility
« Reply #3 on: July 25, 2007, 03:27:18 AM »
You cannot compare benzoic acid and pyridine. But you can compare solubility of benzoic acid in basic (benzoic acid more soluble) and acidic media.
For pyridine you cannot do such comparison
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Offline Raj Prabhugari

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Re: dilution
« Reply #4 on: November 08, 2009, 01:52:45 PM »
hi there...

i have glycolic acid of 70% concentration and i want to dilute it to 20%,25%,30% etc with distilled water.....please help in calculation part??
 could you help me out,
i have tried using N1V1=N2V2 formula by taking N1 as 70% and N2 as 20%, is this a right way of doing it?

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