EK Lecture Question about Saturated Solutions

[Gauss44]

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90. The addition of a strong base to a saturated solution of Ca(OH)2 would:

A. Decrease the number of OH ions in solution
C. Cause Ca(OH)2 to precipitate

C is correct.

It seems that C would necessitate A. Can anyone explain why A is wrong?

 

[jayoh]

If you're adding a strong base, then it will be contributing OH- ions to the solution. So OH- becomes a common ion in the solution. Adding more OH- ions in this solution will lower the solubility of Ca(OH)2, forcing some to precipitate out. Since you're adding OH- as well as taking some out, it wouldn't reduce OH- concentration.

Think of it like le chateliers principle- for the equation
Ca(OH)2 -> Ca + 2OH

Adding more OH will push the equilibrium back towards Ca(OH)2.

Hope that helps!

 

[Schenker]

That's a good explanation for why C is correct.

To answer your question, here's why A is wrong: The concentration of OH- will increase:

1. Since we know that Ca(OH)2 will precipitate out (per jayoh's explanation), and that we're not adding any Ca2+ to the solution, [Ca2+] must decrease.
2. Since [Ca2+] decreases, [OH-] must increase since [OH-]^2 = Ksp/[Ca2+]. Ksp is a constant. (The solution is saturated in Ca(OH)2 before and after the addition of base, so we can use the expression for Ksp.)​

 

[Czarcasm]

I guess the way I understood it was: you're adding excess -OH ions to the solution. For every 2 -OH ions that react with Calcium cation to precipitate into Ca(OH)2, two additional -OH ions are replacing them. Hence a decrease in calcium cations, a neutral level of -OH ions, and an overall increase in the reverse reaction: Ca(OH)2, resulting in precipitation.

If you want to be technical about it, then yes -- you're right. But, choice C is still the better answer. For the MCAT, always choose the better answer. You'll encounter situations like these often. Best not to overthink it. 😉

 

[Schenker]

Hence a decrease in calcium cations, a neutral level of -OH ions, and an overall increase in the reverse reaction: Ca(OH)2, resulting in precipitation.

It's important to realize that the concentration of OH- ions increases and does not stay the same. Perhaps it doesn't matter for answering this question, but it matters for grasping the concepts of solution chemistry as well as acid/base chemistry. A Ca(OH)2 solution is not buffered, so adding a strong base will result in a pH increase (aka, an increase in [OH-]).

 

[Czarcasm]

It's important to realize that the concentration of OH- ions increases and does not stay the same. Perhaps it doesn't matter for answering this question, but it matters for grasping the concepts of solution chemistry as well as acid/base chemistry. A Ca(OH)2 solution is not buffered, so adding a strong base will result in a pH increase (aka, an increase in [OH-]).

Sorry OP, that's actually true. I'm a bit rusty with this since it's been a long time.