Question on AAMC OG

[salemstein]

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I got really confused by question 24 for the C/P section (page 221). The correct answer says you can dissolve the precipitate M(C17H32COO)2 (solid) using HCl, where M originally came from a 2+ metalic cation and RCOO came from a weak base. I was wondering why cant you use a base like NaOH to dissolve it? How do you even tell if the precipitate is acidic or basic?

 

[NextStepTutor_3]

Hi there, good question! As far as the precipitate goes, you've already almost figured it out. Break it down into its component cation and anion: M2+ can't be acidic (as far as the Bronsted-Lowry definition goes), as it doesn't possess a proton at all. C17H32COO- is a weak base - it's the conjugate base of a carboxylic acid. Since our precipitate contains a basic ion and no acidic ions, we call it a basic precipitate. We'd do the same thing for any simpler precipitate, like Ca(OH)2 - calcium ion isn't acidic or basic, while OH- is basic, so that's a basic precipitate overall.

Now, the question is: why do we need to use an acid to dissolve our precipitate? To understand this, think about the dissociation expression for the precipitate:

M(C17H32COO)2 (s) <-----> M2+ + 2 C17H32COO-

At any given moment, at least a tiny amount of M2+ cations and carboxylate anions exist in solution; that's true for any solid precipitate. So, what happens when we add HCl? The C17H32COO- ion is protonated to form C17H32COOH, effectively removing C17H32COO- from solution. According to Le Chatelier's principle, this drives the reaction to the right to regenerate more ions. If we continue to add HCl, more and more of the solid precipitate will dissolve in this manner.

Now, the opposite tends to be true if you add a base to a basic precipitate. Consider our above example of calcium hydroxide: Ca(OH)2 (s) <----> Ca2+ + 2 OH-. If we were to try to dissolve this precipitate using NaOH, we would be adding hydroxide anions to the solution, forcing the reaction backwards to remove those excess ions (and forming additional precipitate). Remember, basic solids dissolve best in acidic solution, while acidic solids dissolve best in basic solution.

 

[salemstein]

Thanks for the help!

 

[swswsw]

Hi there, good question! As far as the precipitate goes, you've already almost figured it out. Break it down into its component cation and anion: M2+ can't be acidic (as far as the Bronsted-Lowry definition goes), as it doesn't possess a proton at all. C17H32COO- is a weak base - it's the conjugate base of a carboxylic acid. Since our precipitate contains a basic ion and no acidic ions, we call it a basic precipitate. We'd do the same thing for any simpler precipitate, like Ca(OH)2 - calcium ion isn't acidic or basic, while OH- is basic, so that's a basic precipitate overall.

Now, the question is: why do we need to use an acid to dissolve our precipitate? To understand this, think about the dissociation expression for the precipitate:

M(C17H32COO)2 (s) <-----> M2+ + 2 C17H32COO-

At any given moment, at least a tiny amount of M2+ cations and carboxylate anions exist in solution; that's true for any solid precipitate. So, what happens when we add HCl? The C17H32COO- ion is protonated to form C17H32COOH, effectively removing C17H32COO- from solution. According to Le Chatelier's principle, this drives the reaction to the right to regenerate more ions. If we continue to add HCl, more and more of the solid precipitate will dissolve in this manner.

Now, the opposite tends to be true if you add a base to a basic precipitate. Consider our above example of calcium hydroxide: Ca(OH)2 (s) <----> Ca2+ + 2 OH-. If we were to try to dissolve this precipitate using NaOH, we would be adding hydroxide anions to the solution, forcing the reaction backwards to remove those excess ions (and forming additional precipitate). Remember, basic solids dissolve best in acidic solution, while acidic solids dissolve best in basic solution.

I totally understand everything in your explanation, but what I dont get is why everyone always says "like dissolves like"- is that a different situation that I'm thinking of/when is that applicable/how is that different than the situation here? Thanks!