# Homework Help: Creating a solution with a pOH of 10.54

1. Nov 23, 2014

### jumbogala

1. The problem statement, all variables and given/known data
Is it possible to make an aqueous solution with strontium hydroxide, Sr(OH)2, that gives a pOH of 10.54? Explain why or why not.

2. Relevant equations
[OH-] = 10-pOH

3. The attempt at a solution

[OH-] = 10-pOH = 10-10.54 = 2.88 x 10-11 mol/L OH-

Every mole of Sr(OH)2 produces 2 moles of OH-. So the concentration of Sr(OH)2 is (2.88 x 10-11 mol/L) / 2 = 1.4 x 10-11 mol/L Sr(OH)2

My data table says Sr(OH)2 falls in the category of 'slightly soluble' (solubility less than 0.1 mol/L). Since the amount of Sr(OH)2 needed to make this solution is so tiny, it seems like this should work. Am I on the right track? I am confused about what this question is getting at.

2. Nov 23, 2014

### Staff: Mentor

What is pH of this solution?

3. Nov 23, 2014

### jumbogala

The pH is 3.46. So it's an acidic solution.

If you use any Sr(OH)2 at all, the solution should be basic. Is that what the question is getting at?

4. Nov 23, 2014

### Staff: Mentor

That would be my understanding.

5. Nov 23, 2014

### jumbogala

Okay. I see that it doesn't work in theory now. However, I don't understand why it won't work in the lab. Suppose you start with pure water. Then you add a tiny, tiny amount of Sr(OH)2 to it. This should increase the [OH-] a bit. So now the pH should be slightly more than 7. It seems like this part should be possible to do in real life.

So suppose the concentration of the Sr(OH)2 solution is 1.4 x 10 -11 mol/L.

Doing the math above, we find that the pOH is 10.85, and the pH is 3.15. Which makes no sense. What's going on?

6. Nov 23, 2014

### Staff: Mentor

Show the math you refer to.

7. Nov 23, 2014

### jumbogala

Okay.

pOH = -log(1.4 x 10 -11 mol/L) = 10.85

pH = 14.00 - 10.85 = 3.15

Last edited: Nov 23, 2014
8. Nov 23, 2014

### Staff: Mentor

Where is OH- from water autodissociation?

9. Nov 23, 2014

### jumbogala

So, because pure water has [OH-] of 1.0 x 10-7 mol/L, adding a small amount of strontium hydroxide to it would actually make the [OH-] go up. So the actual concentration would be something like 1.0001 x 10-7 mol/L, which would have a pOH of slightly less than 7.

So the point is, because the OH- concentration is already at 1.0 x 10-7 mol/L, you can't have an OH- concentration on the order of 10-11 (unless you add a lot of an acidic solution).

10. Nov 23, 2014

### Staff: Mentor

11. Nov 23, 2014

### jumbogala

Got it. Thank you!