solution question ek 1001 #533

[MedGrl@2022]

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Hey everyone,

I am in the end of the 6th week of EK Homestudy and I am studying solutions. I came across this problem #533 in the chemistry 1001 book:

"If we consider compounds with a solubility of less than 0.01 mol/L to be insoluble, and we use the Ksp for Ca(OH)2 as given in the table (Ksp= 1.3x10^-6) and make standard calculations, we find that Ca(OH)2 is, in fact, soluble by the definition given. How can this discrepancy be reconciled?

A. Ca2+ ions bond with OH- ions to form ion pairs in solution.
B. The reaction is exothermic, raising the temperature of the solution and increasing solubility.
C. Calcium hydroxide dissociates into three particles, not two.
D. Calcium hydroxide creates a basic solution, shifting the equilibrium expression to the right.”

According to EK “A is correct. Ion pairing removes ions from the equilibrium expression shifting the equilibrium to the right. B is wrong because we don’t consider whether or not the reaction is exothermic when finding the solubility with Ksp. C was already taken into account by the equilibrium expression. D is already taken into account by the equilibrium expression and a basic solution would shift the equilibrium to the left anyway.”

So here are my questions: What is ion pairing? Where is this discussed (I have searched through a few MCAT sources and did some Google searching)? I am assuming that it means that Ca from the Ca(OH)2 solid is pairing with the OH molecules from H2O. Wouldn’t this cause Ca(OH)2 to reform?

In addition, Ca(OH)2 would result in a basic solution, due to the OH- ions that would result, right? Why would creating a basic solution shift the equilibrium to the left? Will I learn about this when I get to the acid bases chapter in EK? Also, what chapter do they cover this stuff (regarding solutions) in the Berkeley review books?

Please let me know if you may know the answers to the questions. I appreciate all your help.

Sincerely,

Verónica

 

[gettheleadout]

Ion pairing is just a weak association of partially solvated ions in solution. Because paired ions are not fully solvated, however, they don't completely contribute to the ion concentration in the solubility product.

B is wrong because if the reaction were exothermic, though dissolution of Ca(OH)2 would raise the solution temperature, it wouldn't act to increase solubility (because it's exothermic). Lower initial solution temperature or constant removal of heat from solution would increase solubility in this case.

C is correct but irrelevant to the question. The dissociation of 2 OH- vs only 1 Ca2+ is already factored in the solubility product in the form of exponents.

D is wrong, because while Ca(OH)2 dissolution does create a basic solution, it wouldn't increase solubility. Addition of acid to the solution to neutralize the base released from Ca(OH)2 would increase solubility in this case.

As a note, the Ca2+ is pairing with OH- ions in solution, which could be either the actual atoms from dissociated Ca(OH)2 or from water which has reacted with said OH-.

As to your question, the reasoning behind D requires that one know that acids and bases are a major exception to "like dissolves like." Acids are more soluble in basic solution, and vice versa.

Moving to Q&A.

 

[MedGrl@2022]

Ion pairing is just a weak association of partially solvated ions in solution. Because paired ions are not fully solvated, however, they don't completely contribute to the ion concentration in the solubility product.

What do you mean by partially solvated ions in the solution? Does that mean that Ca(OH)2 does not fully dissociate and only one OH- dissociates and there is CaOH+? Is the CaOH+ the paired ion that does not contribute to the ion concentration in the solubility product?

Thank you for all your help.

Sincerely,

Verónica

 

[gettheleadout]

What do you mean by partially solvated ions in the solution? Does that mean that Ca(OH)2 does not fully dissociate and only one OH- dissociates and there is CaOH+? Is the CaOH+ the paired ion that does not contribute to the ion concentration in the solubility product?

Thank you for all your help.

Sincerely,

Verónica

When I said partially solvated I meant not having complete, independent spheres of solvation. See this article for a depiction of solvent-shared and contact ion pairs.

 

[MedGrl@2022]

When I said partially solvated I meant not having complete, independent spheres of solvation. See this article for a depiction of solvent-shared and contact ion pairs.

Okay interesting... so basically it is almost if the Ca(OH)2 ions don't completely dissociate thus there may also be some Ca(OH)- ions and Ca(OH)2 and are still solvated by H2O. These are not represented by the Ksp equilibrium constant and contribute to the solvation of Ca(OH)2 thus, Ca(OH)2 is more soluble than its Ksp would represent. Is this correct?

As I was doing more EK 1001 problems #541 mentions that some anions and cations undergo hydrolysis which also causing the actual solubility of a substance to deviate from the solubility product. In this case water is further breaking down the anion and/or cation and this reaction is not taken into account by Ksp. Is this right? I just want to make sure I am understanding everything correctly.

Thank you once again for all your help!

Sincerely,

Verónica

 

[gettheleadout]

Okay interesting... so basically it is almost if the Ca(OH)2 ions don't completely dissociate thus there may also be some Ca(OH)- ions and Ca(OH)2 and are still solvated by H2O. These are not represented by the Ksp equilibrium constant and contribute to the solvation of Ca(OH)2 thus, Ca(OH)2 is more soluble than its Ksp would represent. Is this correct?

As I was doing more EK 1001 problems #541 mentions that some anions and cations undergo hydrolysis which also causing the actual solubility of a substance to deviate from the solubility product. In this case water is further breaking down the anion and/or cation and this reaction is not taken into account by Ksp. Is this right? I just want to make sure I am understanding everything correctly.

Thank you once again for all your help!

Sincerely,

Verónica

Yes, that sounds right. For #541 it seems like a good example would be an ammonium salt. The salt dissociates, but the [NH4] isn't directly proportional to the amount of salt dissociated because some ammonium undergoes ionization to ammonia in water, shifting the equilibrium toward greater solubility of the salt. Acetate would also undergo some hydrolysis and similarly shift a solubility equilibrium.

 

[MedGrl@2022]

Yes, that sounds right. For #541 it seems like a good example would be an ammonium salt. The salt dissociates, but the [NH4] isn't directly proportional to the amount of salt dissociated because some ammonium undergoes ionization to ammonia in water, shifting the equilibrium toward greater solubility of the salt. Acetate would also undergo some hydrolysis and similarly shift a solubility equilibrium.

Just curious, in the ionization of NH4+, does H2O donate an electron to NH4+ which makes NH4 an unstable molecule and thus a hydrogen comes off and NH4 becomes NH3?

 

[gettheleadout]

Just curious, in the ionization of NH4+, does H2O donate an electron to NH4+ which makes NH4 an unstable molecule and thus a hydrogen comes off and NH4 becomes NH3?

If you're suggesting that neutral NH4 is an intermediate, then no. The exchange is due to the attraction between the partial positive charge of the H on NH4+ and the basic lone pairs of the H2O. Whether there exists a transition state where you have NH3---H-OH2 I don't know. I guess that would be necessary, but to think of it as electron donation from H2O making NH4 unstable is incorrect. NH4+ is already somewhat unstable, that's precisely why it reacts to donate the proton and become a stable, neutral NH3.

 

[MedGrl@2022]

If you're suggesting that neutral NH4 is an intermediate, then no. The exchange is due to the attraction between the partial positive charge of the H on NH4+ and the basic lone pairs of the H2O. Whether there exists a transition state where you have NH3---H-OH2 I don't know. I guess that would be necessary, but to think of it as electron donation from H2O making NH4 unstable is incorrect. NH4+ is already somewhat unstable, that's precisely why it reacts to donate the proton and become a stable, neutral NH3.

I guess it was just interesting that you used the phrase "ionization"... I always associated that term with gaining or losing an electron. However, according to wikipedia it is also the addition or removal of an ion: http://en.wikipedia.org/wiki/

Thank you for all your help! 🙂

 

[gettheleadout]

I guess it was just interesting that you used the phrase "ionization"... I always associated that term with gaining or losing an electron. Is the term "ionization" frequently used in other ways too?

Thank you for all your help! 🙂

Ionization simply means gaining a net charge. In the context of NH4+, which is already an ion, I'm referring to the ionization of its proton, which is neutral (though partially positive) while bound within to NH4+.