TBR CH3 molar solubility?

[Status Sciaticus]

Advertisement - Members don't see this ad

Okay so i just finished the Eq chapter in TBR and i was stumped on one specific part.

Some of the questions when asking for solubility for a compound like M2X use the 4x^3 method which i get, but them some of them dont...

How exactly do i know when to use which?

look at 3.20/3.21 vs 3.28/3.29 to see what I'm talking about

thanks!

 

[Meredith92]

They're doing the exact same method except for 3.28 and 3.39 there was already ions present in the solution so you have to do an ICE chart to solve it

Does tht answer your question?

 

[chonc]

3.20) You know that MX2 means 3 total components:
1 metal cation --> x
2 anions --> 2x^2

Multiply these together to get Ksp, which you are given:
x(2x^2) = 4x^3 = 1.08e-7

The question asks for molar solubility, which is the value x. It is the moles of solute that dissolve in enough solvent to make 1L.

You decide which Ksp calculation to use based on the ratio of metal cation: anion. For something like NaCl, you would use MX, which forms x^2 = Ksp, as given in table 3.7

 

[Status Sciaticus]

They're doing the exact same method except for 3.28 and 3.39 there was already ions present in the solution so you have to do an ICE chart to solve it

Does tht answer your question?

sorry it doesnt.

what im trying to say is look at 3.20/3.21 versus 3.28/3.29

3.20) You know that MX2 means 3 total components:
1 metal cation --> x
2 anions --> 2x^2

Multiply these together to get Ksp, which you are given:
x(2x^2) = 4x^3 = 1.08e-7

The question asks for molar solubility, which is the value x. It is the moles of solute that dissolve in enough solvent to make 1L.

You decide which Ksp calculation to use based on the ratio of metal cation: anion. For something like NaCl, you would use MX, which forms x^2 = Ksp, as given in table 3.7

yes i get that. but look at 3.28 and 3.29. why cant I use the M2X = 4x^3 method for those two problems?

 

[Shjanzey]

sorry it doesnt.

what im trying to say is look at 3.20/3.21 versus 3.28/3.29



yes i get that. but look at 3.28 and 3.29. why cant I use the M2X = 4x^3 method for those two problems?

Can you post the molecular compounds used in those 2 problems?

 

[ouroboros39]

They are using the [X][2X]^2 method, essentialy, (4X^3), but like the other user said, since you already have one of the ions in solution, e.g. the first case is putting CaCl2 in NaCl(aq). So your equation for Ksp is [X][.01+2X]^2 (.01 is the starting concentration of Cl- ion). They dont factor this out, but if you did, you would find somewhere there is a 4X^3 term, plus a few more terms.

Instead of doing this, we make an assumption that relative to the starting concentration, the amount of new Cl- ion (represented by 2X) that will be added will be a negligable amount, so we can assume it is 0. The new equation is X*(.01)*(.01)=Ksp. This makes it simpler for us than trying to solve the 4X^3 + other terms when you factor the equation out. It only works when the Ksp is a very small number and the initial ion concentration is large relative to the possible additional ion concentration.

Theres actually another equation which can test whether this approximation is valid, but most likely, if you find that you will be faced with something like4X^3+.0001X+etc in a Ksp context, you can assume the new ion concentration will be small.

Hopefully this is what you were asking.

 

[circulus vitios]

sorry it doesnt.

what im trying to say is look at 3.20/3.21 versus 3.28/3.29



yes i get that. but look at 3.28 and 3.29. why cant I use the M2X = 4x^3 method for those two problems?

3.20/3.21 are already at equilibrium. 3.28/3.29 are equilibrium problems. 3.28 has you putting CaCl2 into a solution of 0.01M NaCl. You find the molar solubility after equilibrium has been established. The same applies to 3.29. You must use an ICE table whenever equilibrium is not established.

Using 3.28 as an example, our Ksp expression looks like Ksp=2.5E-10=[Ca+2][Cl-]^2. We don't divide by [CaCl2] because it's a pure solid. Use an ICE table to find the equilibrium values for [Ca+2] and [Cl-]. Plug these values into the Ksp equation and solve for x, which is the molar solubility.

edit: Cliffs notes version: Use the "4x^3" method if the solution is at equilibrium. Use an ICE table to find the equilibrium expressions if the solution is not at equilibrium, and then plug them into the Ksp expression to solve for x (the molar solubility.)

 

[Meredith92]

Haha yeah this is what I was saying but just in less words- they're using the same exact formula but it changes if you have something already in solution.... And in that case you can do the estimation trick used with ICE tables that makes the assumption that tha "+x" term is small. I hope that answers your question!!

 

[Temperature101]

The equilibrium is a hell of a hard concept for me to wrap my head around...I am getting detroyed by that chapter on TBR and cant find any other practice material so I can do more reading/problems on it.

 

[Status Sciaticus]

thank you everyone! but how do you know 3.21/3.20 are at equilibrium if it doesnt say it in the question?

 

[Meredith92]

The amount that dissolves when there is nothing already present in the solution is the molar solubility at equilibrium. You can simply do the Ksp =4s^3 ( I think it was that in that problem) and solve for the molar solubility at equilibrium ... But if you add one of the spectator ions you throw off equilibrium so that you get a different molar Solubility. Just like any equilibrium problem when you add a product the equilibrium shifts to the reactant side. It's the exact same thing here. It shifts to the left so less dissolves ( aka your molar solubility decreases)


Does that help at all?

 

[circulus vitios]

The equilibrium is a hell of a hard concept for me to wrap my head around...I am getting detroyed by that chapter on TBR and cant find any other practice material so I can do more reading/problems on it.

Same here. 😳