Solubility of Metal Hydroxide in Solutions

1. Dec 15, 2013

Qube

1. The problem statement, all variables and given/known data

Of the following solutions, the solubility of calcium hydroxide (s) is least in:

1) 0.4 M calcium chloride
2) 0.2 M barium hydroxide
3) 0.25 M calcium nitrate
4) 0.3 M potassium hydroxide
5) 0.45 M sodium nitrate

2. Relevant equations

Either I haven't gotten to the relevant lecture yet or the prof just doesn't cover this topic in lecture. Either way, I found a U. Texas lecture online explaining that for the solubility of magnesium hydroxide, it is highest in a solution of hydrochloric acid as opposed to pure water, sodium hydroxide, or magnesium chloride, since the acidic hydrochloric acid will neutralize the hydroxide in the metal hydroxide, allowing more to dissolve.

3. The attempt at a solution

So by the above logic, I can eliminate choices 3 and 5, since these dissolve into parts that don't even appear on my acid-base table. They aren't going to be acidic or basic in solution; they're actually salts resulting from neutralization reactions. Choice 1 is also a salt, so we can eliminate that choice.

We are left with choices 2 and 4. Choice 2 has the higher hydroxide concentration (0.2 M * 2, the subscript of hydroxide) so it appears that solution would be the least favorable to calcium hydroxide dissolution since solution 2 is already very basic, which seems to be the exact opposite environment of an solution environment favorable to metal hydroxide dissolution.

2. Dec 15, 2013

Staff: Mentor

This is not about acid-base reactions, more like a common ion effect.

You already know about Ka, so you are ready to learn what the Ksp (solubility product) is.

3. Dec 15, 2013

Qube

Thanks for the pointer! I know I saw the Common-Ion Effect in my textbook as a heading; I'll be sure to read up on that :)!

4. Dec 15, 2013

Qube

Alright, this is what I got from learning about the common ion effect, which is effectively an application of Le Chatelier's principle.

The presence of a common ion shifts the solubility equilibrium toward the undissolved solid, because Le Chatelier's principle states that a disruption in equilibrium through an addition of material will cause a system to return to equilibrium by attempting to consume the excess material.

So in the case of calcium hydroxide, we have the following solubility equilibrium:

$$Ca(OH)_{2}(s) \leftrightharpoons Ca^{2+}(aq) + 2OH^{-}(aq)$$

Step 1: Let's consider the answer choices individually:

1) 0.4 M calcium chloride

Calcium is a common ion. Solubility would be low in this solution.

2) 0.2 M barium hydroxide

Hydroxide is a common ion. Solubility would be low.

3) 0.25 M calcium nitrate

Calcium is a common ion. Solubility would be low.

4) 0.3 M potassium hydroxide

Hydroxide is a common ion. Solubility would be low.

5) 0.45 M sodium nitrate

No common ions. Solubility would be higher than the other solutions.

Step 2: We can eliminate choice 5. Now we can examine the number of moles of the common ions added by the solutions. I'd expect that the more moles of common ions contributed by the solutions, the lower the solubility. We can find the number of moles by multiplying the molarity of the solution by the subscript or coefficient of the common ion.

1) 0.4 M calcium chloride

0.4 * 1 = 0.4

2) 0.2 M barium hydroxide

0.2 * 2 = 0.4

3) 0.25 M calcium nitrate

0.25 * 1 = 0.25

4) 0.3 M potassium hydroxide

0.3 * 1 = 0.3

We have now narrowed the answer choices down to 1 and 2.

1) 0.4 M calcium chloride

0.4 * 1 = 0.4

2) 0.2 M barium hydroxide

0.2 * 2 = 0.4

Step 3: Upon reinspection of the balanced equilibrium equation:

$$Ca(OH)_{2}(s) \leftrightharpoons Ca^{2+}(aq) + 2OH^{-}(aq)$$

We find that it takes twice as much hydroxide ions to shift the equilibrium to the left as it does calcium ions. So I conclude that solid calcium hydroxide will be least soluble in a solution of 0.4 M calcium chloride due to the common-ion effect and examination of equilibrium stoichiometry.

I'm fairly sure I did everything right up to the last part - not sure about the stoichiometric part.

Last edited: Dec 15, 2013
5. Dec 15, 2013

Staff: Mentor

To some extent - yes. But for exact results you have to plug numbers into Ksp.

6. Dec 15, 2013

Qube

Ksp is for calcium hydroxide, I think, correct?

If I have the Ksp for calcium chloride I just plug in the numbers for the common ions contributed, correct? E.g. for calcium chloride I could plug in the concentration of calcium ions contributed by CaCl2, correct?

7. Dec 16, 2013

Staff: Mentor

Yes, that's the one you will need.

I don't see anything wrong about what you wrote, but I strongly suggest you read about Ksp, as I have a gut feeling you don't know yet how it works.

8. Dec 16, 2013

Qube

Alright :). I don't know how the solubility constant works but I'll sure as heck find out tomorrow! Thanks again!

9. Dec 16, 2013

Qube

Okay ..... I think I figured it out now :). Thanks again to you Borek for pointing me into the right direction. Here is the ICE table I set up:

http://i.minus.com/jNGbtzDF3Srz4.jpg [Broken]

Anyway with the ICE table I set up, I solve for Ksp as shown beneath each ICE table by plugging in Ksp = 6.5*10^-6.

I get, however, the opposite answer as I got before; that the solution is least soluble in barium hydroxide rather than calcium chloride.

What's up? Was there something wrong with my previous reasoning?

Last edited by a moderator: May 6, 2017
10. Dec 16, 2013

Staff: Mentor

I am not sure I understand your approach.

Actually no need for the ICE table - while it is not wrong to use it, you can safely assume solubility of the calcium hydroxide is so low it won't substantially change the concentration of the common ion already present in the solution. See the end of the post.

Depending on which ion is already present in the solution, solubility will be either

$$s = \frac {K_{sp}} {[Ca^{2+}]}$$

or

$$s = \frac {K_{sp}} {[OH^-]^2}$$

Note concentration being squared in the second case - that makes a huge difference. Now it is just a matter of plugging one of the concentrations into the equation to calculate solubility.

After calculating solubility you can check by how much concentration of the ion present in the solution changed. Say, for 0.2 M Ba(OH)2 [OH-]=0.4 M, according to wikipedia Ksp=4.68×10−6, so the solubility is

$$s = \frac {4.68 \times 10^{-6}}{0.4^2} = 3 \times 10^{-5}$$

Concentration of OH- grows by 2×3×10-5 - compared to initial 0.4 M. It is as if didn't change.

11. Dec 16, 2013

Qube

Thank you!

It seems as if then the solubility of calcium hydroxide then is less soluble in the 0.4 M calcium chloride solution because for that solution, I get s = 1.6 * 10^-5.

On the other hand for the 0.2 M barium hydroxide solution I get s = 4.0 * 10^-5.

S (CaCl2) < S of barium hydroxide so it seems that 1 is the correct answer.

----

Edit:

I plugged in the Ksp for calcium hydroxide into each; I'm not sure if this is correct. Sorry, but it seems as if I'm still confused about solubility products!

12. Dec 17, 2013

Staff: Mentor

Sigh.

Sorry, I was apparently not thinking yesterday. My formula for solubility in the solution containing calcium ions was wrong (which is obvious when you think about units - both formulas can't be right at the same time).

Instead of listing just formulas I should explain where they come from, they are pretty easy to derive. When solubility is s, s=[Ca2+] and s=1/2[OH-] (where we refer to concentration of ions from the dissolution, not to the total concentration if one of the ions was already present in the solution). I hope that is obvious.

Now, we know

$$K_{sp} = [Ca^{2+}][OH^-]^2$$

so it is a matter of plugging in known concentration of ions already present (perhaps - if the solubility is high - using ICE table as you did), solving for the other ion concentration, and expressing solubility using this other ion. For solution containing known concentration of calcium ions we have

$$[OH^-] = \sqrt{\frac{K_{sp}}{[Ca^{2+}]}}$$

so

$$s = \frac 1 2 \sqrt{\frac{K_{sp}}{[Ca^{2+}]}}$$

For solution containing [OH-] ions formula I gave was OK.