Like always dissolves like... right?

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MedChad

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So we have all heard that like dissolves like, but I have a question from aamc cbt 5 that confuses me when applying this concept.

Which of the following will occur if the acidity of a saturated solution of ca(oh)2 is increased?

a) Ksp will decrease and additional ca(oh)2 will dissolve.
b) The Ksp will be unchanged and additional ca(oh)2 will dissolve
c) The ksp will decrease and additional ca(oh)2 will precipitate
d) The Ksp will be unchanged and additional ca(oh)2 will precipitate

I understand the Ksp will remain unchanged, but I don't conceptually understand the latter part of the answer. Thanks in advance for the help guys!!

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Basically the way I thought about it was the CaOH2 will dissolve in solution making it slightly basic because all the OH-. By lowering the pH or adding H+ ions into solution, the OH- in solution will react with the H+ creating H2O, allowing more OH- to dissolve in solution. There might be more to it than that, but that's how I thought about it.
 
I understand your explanation perfectly, as it was similar to what aamc said. But doesn't that go against the general "like dissolves like" theory since your adding an acid and dissolving more of a base?
 
Like dissolves like, but this is an equilibrium. You're going to want to shift it right for more to dissolve. It will dissolve it, the question's not asking between dodecane and water, it's saying how do we dissolve MORE. The decrease in OH will shift Eq to right. KSp only changes with temperature, so A+C are right out.
 
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Is the answer B? Like dissolves like, yes, but acid dissolves base.
 
So we have all heard that like dissolves like, but I have a question from aamc cbt 5 that confuses me when applying this concept.

Which of the following will occur if the acidity of a saturated solution of ca(oh)2 is increased?

a) Ksp will decrease and additional ca(oh)2 will dissolve.
b) The Ksp will be unchanged and additional ca(oh)2 will dissolve
c) The ksp will decrease and additional ca(oh)2 will precipitate
d) The Ksp will be unchanged and additional ca(oh)2 will precipitate

I understand the Ksp will remain unchanged, but I don't conceptually understand the latter part of the answer. Thanks in advance for the help guys!!

Write the chemical equation out

Ca(OH)2(s) ->Ca2+(aq) + 2OH-(aq)

If you were to add acid, you will be decreasing the concentration of OH on the right. According to Le Chatelier's principle, the reaction will respond by moving to the right thereby dissolving more CaOH2 to make more 2OH. (Think about this as form of compensation)


Ksp is constant because it will only change with temperature.
 
Ok I think I got it now, and yes medpr the answer was B. I was just trying to get the relationship of like dissolves like and applying it to an acid and base like this problem. Thanks guys!
 
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