How to solve for solubility

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deleted647690

I have been going over many practice problems for Ksp and solubility, and I am still getting confused.
For example:

There is a passage in BR with a question that says, "The maximum [Ca2+] in a pH=14 solution would be which of the following values?"

They gave a Ksp for Ca(OH)2 at Ksp = 4.3*10^-6 M^3 in the passage


I solved it like this:

Since they gave a pH of 14, I knew the [OH-] was 1 because of 10^-poH

I took that as meaning that the initial concentration of OH- before dissolution of CaOH was 1

So I set up an ICE table

Ca(OH)2--->Ca + 2OH-
-- 0 1
-x +x +2x
So I got an equation: 4.3*10^-6 = x(1 + 2x)^2

Well, first of all, I thought you ignored 2x in this case? If I did that, I would get x^2 = 4.3*10^-6, which doesn't make sense for this problem. Why is this like this in this case?



So I tried to solve it without ignoring 2x, and I got 1*10^-3








In the answer, they solved it like this:

Ksp = [Ca2+][OH-]^2

Then for some reason, they just plugged the 1 M value for [OH-] concentration from the pH to this Ksp expression and got Ksp = [Ca2+]. Using the Ksp value given, they got an answer of 4.3*10^-6 M


So I'm confused because I followed the correct method for solving with an ICE table but I got it wrong. I guess it must be because the ICE table is meant for a different type of problem? I don't see why they just plugged the initial value right into the Ksp expression

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Do you mind sending a picture of the problem and the key? Ksp shouldn't have a unit in the first place.
If pH is 14, pOH should be 0, not 1.
 
View attachment 208857

I jumped a step....-log(1)=0 -->pOH = 0

I hope this makes sense!

Kind of sloppy hand writing....


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Do you mind sending a picture of the problem and the key? Ksp shouldn't have a unit in the first place.
If pH is 14, pOH should be 0, not 1.

You are correct that units are not associated with Ksp's, however unit analysis can often help figure out if you've arrived at the correct answer. Ksp's lack of units is by convention, and not technically correct mathematically. When you square, multiply, divide a dimensional value (one with units associated) you must do the exact same with the units.

In the test, if they give you some units tied to a given value, use it to your advantage! See the example above on how I did this exact thing 🙂

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Realized I didn't actually answer your question...

The equation you arrived at was wrong...

You wrote: Ksp = x^2
Correct equation: Ksp = x

Reducing x(1+2x)^2, ignoring 2x yields: x(1)^2 = x

In common ion solution problems, you can assume that the contribution of the dissolved common ion essentially comes from a single source. This means you can ignore the contribution of the solid you are trying to dissolve.

So if I am trying to dissolve BaSO4 in a solution of .1M sodium sulfate, the sulfate ion concentration doesn't actually change. LeChatelier's principle tells us that a little sodium sulfate precipitates and a little barium sulfate dissolves, but the overall ion concentration of sulfate remains constant at .1M

This is obviously an approximation, but one that everyone uses. Without this approximation, the calculations would involve all kinds of terrible math and imaginary values that we would have to sift through. Me personally, I have no idea how to solve 3rd and 4th order equations by hand...

So that is why the text ignored the "2x" and plugged in 1M for the OH- value.


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