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Author Topic: Solubility of Benzoic Acid and Ethyl 4-Aminobenzoate (Mechanisms/Work included)  (Read 29196 times)

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I just want to know if my approach and thinking is correct.  Please Do NOT flat out tell me the answer.
Reaction 1 = benzoic acid + 1.0 M NaOH
-There is water in this solution as well I presume.  NaOH forms hydroxide ions which deprotonate benzoic acid making it polar and thus soluble in water which is a polar compound. 

Reaction 2 = conjugate base of benzoic acid + 6.0 M HCl
-How does hydronium form?  Is it just reacting with NaOH to reach acid/base equilibrium?

Reaction 3 = Ethyl 4-Aminobenzoate + 1.0 M HCl
-Ethyl 4-Aminobenzoate is protonated by Hydronium which forms when HCl and H2O reach acid base equilibrium.  The protonated compound is soluble in water?

Reaction 4 = Deprotonation of Amoniumbenzoate + 6.0 M NaOH
-When NaOH is added, lots of OH forms and makes the solution basic and the compound is deprotonated making precipitate out of solution?



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1) Polar and ionic are different. Also, remember that Na+ is floating around and needs to go somewhere.

2) If you are considering 6M HCl, the concentration of OH- is so low as to be laughably inconsiderable. Remember, since its 6M, the HCl is a solution in water. Work it form there.

3 & 4) Seems ok
Individual results may vary

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