Solubility and solubility product

  • #1

Homework Statement


Which of the following expressions represents the solubility product for Cu(OH)2?
(A) Ksp=[Cu2+][OH-]2
(B) Ksp=[Cu2+]2[OH-]
(C) Ksp=[Cu2+]2[OH-]2
(D) Ksp=[Cu2+][OH-]

Homework Equations


Ksp= [A][ B]

The Attempt at a Solution


Okay, so I understand equilibrium expressions and in this case, solids and pure liquids are left out of the expression. My question is regarding the coefficients and exponents of solubility expressions. In chemical equilibrium expression is the concentration of the products are over the reactants (with exponents being the coefficients of each substance). Knowing that solubility rules are pretty much the same, I know that answer A is correct but my books reasoning for the following reaction does not make sense:

BaF2↔Ba2+aq+2F-aq
=Ksp=[Ba2+][F-]2

However, they write the equation as:

2.4 x 10-5 = (x)(2x)2 (and then they solve for x).

This expression is not consistent, say, with answer choice A in another problem. How did they get 2x in this instance for BaF2? Shouldn't the first problem above be Ksp=[Cu2+][2OH]2?[/B][/B][/B]
 
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Answers and Replies

  • #2
Borek
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Assuming x moles of BeF2 dissolved in 1 L of water, what is the concentration of Be2+? What is the concentration of F-? Can you plug them into Ksp?
 
  • #3
We know that BeF2 dissociates to 1 mol of Be2+ and 2 mol of F-. but why must the coefficient of the F- be both 2x and squared? In equilibrium expressions and assuming that the reaction is elementary, the coefficients are the squares but not both. I hope I am making sense here.
 
  • #4
Borek
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You are squaring CONCENTRATION. Concentration is 2x, so its square is (2x)2.
 
  • #5
Ok so in the original question with copper II hydroxide, answer A is wrong because we get 2OH for dissociation. Yes?
 
  • #6
Borek
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Nope, answer A is the correct one.

You are confused about what [OH-] is and what x is. They are not the same thing.

[OH-] is just the concentration of the OH-.

x is the solubility of the copper hydroxide. Concentration of OH- is not equal to the solubility, as I explained to you in my previous post. Concentration of OH- is twice the solubility of the copper hydroxide, as per each mole of copper hydroxide dissolved there are two moles of OH- produced. This is a simple result of the dissolution reaction stoichiometry:

Cu(OH)2 ↔ Cu2+ + 2OH-
 
  • #7
Ok, thank you. I think my head is finally wrapping around this concept. What I was misinterpreting was the concentration of the hydroxide. For every one mole of Ca2+ there are actually 2 moles of OH-.
 

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