Ksp and Molar Solubility?

[Heplayer92]

Can someone help explain why the answers to these 2 questions are different? I'm not really getting it from the answer sheet.

Thank you for your time!

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[gettheleadout]

Can someone help explain why the answers to these 2 questions are different? I'm not really getting it from the answer sheet.

Thank you for your time!

Alright, so the definition given in the TPR answer explanation is a good place to start. The solubility product (Ksp) is the equilibrium constant for the dissociation of soluble ionic salt in water. So, for a given dissociation reaction A2B (s) <--> 2 A (aq) + B (aq) we find (since the dissociation process is assumed to be elementary, i.e one step) that Ksp = ([A]^2){B}. Thus, we just an understanding of equilibrium constant expressions, we can find one representation of Ksp.

At this point I'd like to shift gears and present a definition for this new term, "molar solubility." Molar solubility is a measure of the solubility of a species, equal to the number of moles species that can be dissolved in enough water to make 1 L of saturated solution. Now, recall that we used the format of an equilibrium expression to derive the Ksp expression above. This means that we are considering the given dissociation reaction at its equilibrium, which for a dissolution reaction of solid ionic salt in water is the point of saturation of the solution. Thus, connecting the context of the Ksp expression to the definition of molar solubility, we can draw a new connection. Look at the mole ratios between A2B : A and A2B : B, they are 1 : 2 and 1 : 1 respectively. Now, since the equilibrium represents a point of saturation, we know that the number of moles of solid A2B we added equals the molar solubility by definition. By the 1 : 1 mole ratio between A2B : B we can then infer that the equilibrium concentration of B is also equal to the molar solubility. Similarly, because the equilibrium concentration of A is twice that of B (A : B = 2 : 1), the eq. concentration of A is twice the molar solubility.

Now, on to representing molar solubility and solving for it. I typically use "x", but the AAMC question uses "S" to represent molar solubility so we'll go with that: molar solubility = S. We've established above that {B}_eq (represented as just {B} in the Ksp expression we wrote above) is equal to the molar solubility. Thus, {B}_eq = {B} = S. By the mole ratio between the products, [A] = 2 {B} = 2 S. We can substitute these values into the Ksp expression like so:

Ksp = ([A]^2){B} = ([2 S]^2){S} = 4 S^3.

You'll get different (but predictable) simplified Ksp expressions as representations of molar solubility depending on the number of ionic components of the solid salt. Two component ions gives Ksp = S^2, three component ions gives Ksp = 4 S^3, four component ions gives Ksp = 27 S^4, and five component ions gives Ksp = 108 S^5.

Note: I had to use carriages {} to enclose some variables to represent concentration due to the post formatting interpreting a bracketed B or S as a font style tag. Carriages used are equivalent to brackets.

 

[PlumbLine]

Refer to the first sentence of the AAMC explanation, which agrees to the correct answer on the TPR question.

The only difference is that in the AAMC question you had to take it one step further and calculate the actual concentration of each ion, given S, and then plug that into the Ksp equation. Does that make sense?


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[Heplayer92]

Alright, so the definition given in the TPR answer explanation is a good place to start. The solubility product (Ksp) is the equilibrium constant for the dissociation of soluble ionic salt in water. So, for a given dissociation reaction A2B (s) <--> 2 A (aq) + B (aq) we find (since the dissociation process is assumed to be elementary, i.e one step) that Ksp = ([A]^2){B}. Thus, we just an understanding of equilibrium constant expressions, we can find one representation of Ksp.

At this point I'd like to shift gears and present a definition for this new term, "molar solubility." Molar solubility is a measure of the solubility of a species, equal to the number of moles species that can be dissolved in enough water to make 1 L of saturated solution. Now, recall that we used the format of an equilibrium expression to derive the Ksp expression above. This means that we are considering the given dissociation reaction at its equilibrium, which for a dissolution reaction of solid ionic salt in water is the point of saturation of the solution. Thus, connecting the context of the Ksp expression to the definition of molar solubility, we can draw a new connection. Look at the mole ratios between A2B : A and A2B : B, they are 1 : 2 and 1 : 1 respectively. Now, since the equilibrium represents a point of saturation, we know that the number of moles of solid A2B we added equals the molar solubility by definition. By the 1 : 1 mole ratio between A2B : B we can then infer that the equilibrium concentration of B is also equal to the molar solubility. Similarly, because the equilibrium concentration of A is twice that of B (A : B = 2 : 1), the eq. concentration of A is twice the molar solubility.

Now, on to representing molar solubility and solving for it. I typically use "x", but the AAMC question uses "S" to represent molar solubility so we'll go with that: molar solubility = S. We've established above that {B}_eq (represented as just {B} in the Ksp expression we wrote above) is equal to the molar solubility. Thus, {B}_eq = {B} = S. By the mole ratio between the products, [A] = 2 {B} = 2 S. We can substitute these values into the Ksp expression like so:

Ksp = ([A]^2){B} = ([2 S]^2){S} = 4 S^3.

You'll get different (but predictable) simplified Ksp expressions as representations of molar solubility depending on the number of ionic components of the solid salt. Two component ions gives Ksp = S^2, three component ions gives Ksp = 4 S^3, four component ions gives Ksp = 27 S^4, and five component ions gives Ksp = 108 S^5.

Note: I had to use carriages {} to enclose some variables to represent concentration due to the post formatting interpreting a bracketed B or S as a font style tag. Carriages used are equivalent to brackets.

Wow, thanks for taking the time to explain all of that =D Appreciate it friend! A big part that also threw me off was the wording. "What is the Ksp" versus "Its Ksp value is:" lol. I guess if they give the "molar solubility" they will want the 4S^3 or 27S^4 etc instead of the straight up Ksp equilibrium. Thanks again!

 

[Heplayer92]

Refer to the first sentence of the AAMC explanation, which agrees to the correct answer on the TPR question.

The only difference is that in the AAMC question you had to take it one step further and calculate the actual concentration of each ion, given S, and then plug that into the Ksp equation. Does that make sense?


Posted using SDN Mobile

Ooh, the "taking it a step further to calculate the actual concentration of each ion" makes a lot of sense. Thanks for pointing that out!