Creating a solution with a pOH of 10.54

[jumbogala]

Homework Statement

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.

Homework Equations

[OH^-] = 10^-pOH

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.

 

[Borek]

pOH of 10.54

What is pH of this solution?

 

[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?

 

[Borek]

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

That would be my understanding.

 

[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?

 

[Borek]

Show the math you refer to.

 

[jumbogala]

Okay.

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

pH = 14.00 - 10.85 = 3.15

 

[Borek]

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

Where is OH^- from water autodissociation?

 

[jumbogala]

Where is OH^- from water autodissociation?

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).

 

[Borek]

Yes. Compare http://www.chembuddy.com/?left=pH-calculation&right=pH-strong-acid-base

 

[jumbogala]

Got it. Thank you!