Creating a solution with a pOH of 10.54

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Discussion Overview

The discussion revolves around the feasibility of creating an aqueous solution of strontium hydroxide, Sr(OH)2, that results in a pOH of 10.54. Participants explore the implications of this pOH value, the resulting pH, and the solubility of Sr(OH)2 in water.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the concentration of OH- needed for a pOH of 10.54 and derives a corresponding concentration for Sr(OH)2, questioning if this is feasible given its slight solubility.
  • Another participant points out that a pOH of 10.54 would imply an acidic solution, which contradicts the basic nature expected from adding Sr(OH)2.
  • A participant expresses confusion about the practical implications of adding a tiny amount of Sr(OH)2 to pure water, suggesting that it should increase the OH- concentration slightly.
  • Further calculations are presented, showing that the derived pOH and pH values do not align with expectations, leading to questions about the underlying chemistry.
  • Discussion includes the consideration of OH- contributions from water autodissociation, indicating that the initial concentration of OH- in pure water complicates the scenario.

Areas of Agreement / Disagreement

Participants generally agree that a pOH of 10.54 is inconsistent with the expected behavior of a solution containing Sr(OH)2, but there is no consensus on the practical implications or the reasons behind the conflicting calculations.

Contextual Notes

Participants note that the solubility of Sr(OH)2 and the contributions of OH- from water autodissociation are critical factors in understanding the situation, but these aspects remain unresolved in the discussion.

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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.
 
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jumbogala said:
pOH of 10.54

What is pH of this solution?
 
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?
 
jumbogala said:
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.
 
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?
 
Show the math you refer to.
 
Okay.

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

pH = 14.00 - 10.85 = 3.15
 
Last edited:
jumbogala said:
pOH = -log(1.4 x 10 -11 mol/L) = 10.85

Where is OH- from water autodissociation?
 
Borek said:
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).
 
  • #11
Got it. Thank you!
 

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