Solving Solubility of Unknown Compound in Base

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The solubility test of an unknown compound in 5% NaOH suggests that if the compound is insoluble, the supernatant should be treated with 5% HCl until neutral. The resulting precipitate or turbidity indicates that the unknown compound is likely an organic acid. The reaction involves the formation of a soluble conjugate base (A-) in the basic solution, which reverts to the insoluble acid (HA) upon acidification. This process explains the observed turbidity or precipitation. Understanding these reactions is crucial for accurately identifying the unknown compound.
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In my lab, I'm trying to identify an unknown compound. So I did a solubility test of the unknown into 5% NaOH. According to the manual it said that if it appears insoluble,take the supernatant out and add 5% HCl until neutral. If percipitate or turbidity results, it means the unknown is an acid.
So far I thought this:
HA (unknown) + OH- <-> H20 + A-
I'm assuming the conjugate base was in the supernatant, but I'm not sure.
A- + HCl <-> Cl- + HA
I'm not sure where the percipitate or turbidity is from. Please help me!
 
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In the first part, the organic acid present dissolved (why you obtained the very soluble A anion). Later, you added H+ which recombined with the A- giving you back the fairly insoluble organic acid again, the HA. The HA causes this turbidity or precipitation.
 
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