# Chem: molar solubility, ion product, Ksp, etc...

#### hospitaldoctor1

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Could someone please explain to me how this question is carried out exactly, step by step... (showing work)
Thank you
How does one find the answer to this question specifically:

You have two beakers of solutions. One beaker has 4x10^-6 moles MgCl2 in 40 mL solution, with a Ksp of 4.1x10^-20. The other has 4x10^-6 moles NaCl, in 50mL, Ksp of 3.6x10^-11. What is the osmotic pressure, boiling point temperature, and vapor pressure for each? Which beaker is each property higher for?

I understand that this question involves IP, Ksp, molarity, etc... but I do not understand what to do exactly. Thanks a lot.

#### TheBoondocks

##### StreetFighter 4 Virtuoso
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Could someone please explain to me how this question is carried out exactly, step by step... (showing work)
Thank you
How does one find the answer to this question specifically:

You have two beakers of solutions. One beaker has 4x10^-6 moles MgCl2 in 40 mL solution, with a Ksp of 4.1x10^-20. The other has 4x10^-6 moles NaCl, in 50mL, Ksp of 3.6x10^-11. What is the osmotic pressure, boiling point temperature, and vapor pressure for each? Which beaker is each property higher for?

I understand that this question involves IP, Ksp, molarity, etc... but I do not understand what to do exactly. Thanks a lot.
Molar solubility is concentration. It's the number of moles/Liter of solution that dissolves. so 4e-6/ 4e-2=4e-4 M/L of MgCl2. 4x^3= 4.1* 10^-20. Divide by 4= 1.025e-20=x^3= 10.25e-21= 2.whatever e-7. What does this mean? We have more than can be dissolved so most of it will remain solid.

4e-6/5e-2= .8e-4= 8e-5 Moles/Liter of NaCl. Well, 36e-12=x^2= 6e-6 for molar solubility. You have 6e-6 of both Na+ and Cl-. The total is 12 e-6 or 1.2 e-5 ions. More NaCl will dissolve than MgCl2. The amount of Cl in MgCl2 is 2*2e-7=4e-7 along with 2e-7 Mg +2. You have a total of 6e-7 ions of MgCl2. So, you have more ions dissolved in NaCl solution than the MgCl2 not surprisingly. You would need avogrado's number in order to calculate moles of actual solute but this is unnecessary.

For the calculations we have more solvent in the Nacl solution by 1.25 but the amount of ions is greater in NaCl by more than 1.25, over 10. Moles of solute/kg of solvent. With this in mind qualative comparisons can be made.

NaCl will have the higher boiling point, lower vapor pressure, and higher osmotic pressure. The opposite is true for MgCl2. I'm not sure how much this helps you but I tried.

#### TheBoondocks

##### StreetFighter 4 Virtuoso
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Molar solubility is concentration. It's the number of moles/Liter of solution that dissolves. so 4e-6/ 4e-2=4e-4 M/L of MgCl2. 4x^3= 4.1* 10^-20. Divide by 4= 1.025e-20=x^3= 10.25e-21= 2.whatever e-7. What does this mean? We have more than can be dissolved so most of it will remain solid.

4e-6/5e-2= .8e-4= 8e-5 Moles/Liter of NaCl. Well, 36e-12=x^2= 6e-6 for molar solubility. You have 6e-6 of both Na+ and Cl-. The total is 12 e-6 or 1.2 e-5 ions. More NaCl will dissolve than MgCl2. The amount of Cl in MgCl2 is 2*2e-7=4e-7 along with 2e-7 Mg +2. You have a total of 6e-7 ions of MgCl2. So, you have more ions dissolved in NaCl solution than the MgCl2 not surprisingly. You would need avogrado's number in order to calculate moles of actual solute but this is unnecessary.

For the calculations we have more solvent in the Nacl solution by 1.25 but the amount of ions is greater in NaCl by more than 1.25, over 10. Moles of solute/kg of solvent. With this in mind qualative comparisons can be made.

NaCl will have the higher boiling point, lower vapor pressure, and higher osmotic pressure. The opposite is true for MgCl2. I'm not sure how much this helps you but I tried.
Why would you need exact values? That's a lot of work and no Mcat question would ask this. If you actually need it, you would take the number of moles that actually dissolve which is the molar solubility* the volume to give you moles. I assume you you take the mass as 50 grams and 40 grams which comes out to .05 and .04 kg respectively.

Osmotic you would use molar solubility and MiRT. So, or MgCl2= (2e-7/4e-2)(3)(.0821)(298K assuming room temperature)
NaCl= (6e-6/5e-2)(2)*.0821)*298K)

boiling=.5(3)(moality of MgCl2)
.5(2)(moality of NaCl)

#### hospitaldoctor1

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What is the exact work?
I understand that for one of them, the IP is greater than Ksp
and the other, IP is less than Ksp...
what affect does this have? What happens?...
What do you do then?

#### Will Hunting

##### Aspiring Cardiologist
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5+ Year Member
What is the exact work?
I understand that for one of them, the IP is greater than Ksp
and the other, IP is less than Ksp...
what affect does this have? What happens?...
What do you do then?
the molar solubility for both is less than the the initial concentration. I think the work is shown. You take 4x^3=4.1e-20. Divide by 4= 1.025e-20=x^3 You have to move by a factor of one to divide by 3 so it becomes 10.25 e-20=x^3= A number greater than 2 but less than 3 and obviously closer to 2. 2^3=8 3^3=27.

So you get like 2.3e-7 for the molar solubility of MgCl2. Well, you have 4e-6 moles/40e-3 Liters. 40e-3=4e-2. so the concentration is 4e-4 which is greater than the amount of total solute. So, you get a precipitate.

Since you have 2.3e-7 moles that are actually dissolved, you divide this by the mass in kg which has the same numerical value as the volume 1g/ml= 1kg/L. so, you have 40 ml= .04grams. You will take 2.3e-7moles/4e-2 k = about .55e-5=5.5e-4 for molality. You then take 5.5e-4(3 ions from MgCl2)(.5=k for boiling). Since the equation is kb*i*m.

I'm not sure about vapor pressure because I know how to do that for two liquids but not with solutes. It would be reduced because it's more difficult for the water to leave.

#### Will Hunting

##### Aspiring Cardiologist
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the molar solubility for both is less than the the initial concentration. I think the work is shown. You take 4x^3=4.1e-20. Divide by 4= 1.025e-20=x^3 You have to move by a factor of one to divide by 3 so it becomes 10.25 e-20=x^3= A number greater than 2 but less than 3 and obviously closer to 2. 2^3=8 3^3=27.

So you get like 2.3e-7 for the molar solubility of MgCl2. Well, you have 4e-6 moles/40e-3 Liters. 40e-3=4e-2. so the concentration is 4e-4 which is greater than the amount of total solute. So, you get a precipitate.

Since you have 2.3e-7 moles that are actually dissolved, you divide this by the mass in kg which has the same numerical value as the volume 1g/ml= 1kg/L. so, you have 40 ml= .04grams. You will take 2.3e-7moles/4e-2 k = about .55e-5=5.5e-4 for molality. You then take 5.5e-4(3 ions from MgCl2)(.5=k for boiling). Since the equation is kb*i*m.

I'm not sure about vapor pressure because I know how to do that for two liquids but not with solutes. It would be reduced because it's more difficult for the water to leave.
forgot about osmotic. In osmotic you take the molarity which in this case we assume is the same as the molality since we don't know or are given the density of the solution. 5.5e-4(3)(.0821)298 K.

You do the same for NaCl. The qualitative answers by the other poster are correct. I doubt this is MCAT.

#### hospitaldoctor1

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Could you explain where you are getting all these numbers from?

Molar solubility is concentration. It's the number of moles/Liter of solution that dissolves. so 4e-6/ 4e-2=4e-4 M/L of MgCl2. 4x^3= 4.1* 10^-20. Divide by 4= 1.025e-20=x^3= 10.25e-21= 2.whatever e-7. What does this mean? We have more than can be dissolved so most of it will remain solid.

4e-6/5e-2= .8e-4= 8e-5 Moles/Liter of NaCl. Well, 36e-12=x^2= 6e-6 for molar solubility. You have 6e-6 of both Na+ and Cl-. The total is 12 e-6 or 1.2 e-5 ions. More NaCl will dissolve than MgCl2. The amount of Cl in MgCl2 is 2*2e-7=4e-7 along with 2e-7 Mg +2. You have a total of 6e-7 ions of MgCl2. So, you have more ions dissolved in NaCl solution than the MgCl2 not surprisingly. You would need avogrado's number in order to calculate moles of actual solute but this is unnecessary.

For the calculations we have more solvent in the Nacl solution by 1.25 but the amount of ions is greater in NaCl by more than 1.25, over 10. Moles of solute/kg of solvent. With this in mind qualative comparisons can be made.

NaCl will have the higher boiling point, lower vapor pressure, and higher osmotic pressure. The opposite is true for MgCl2. I'm not sure how much this helps you but I tried.

#### hospitaldoctor1

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What if you wanted to find the Conductance?

#### hospitaldoctor1

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Isnt 4e-6/ 4e-2=1e-4 M/L ??

Molar solubility is concentration. It's the number of moles/Liter of solution that dissolves. so 4e-6/ 4e-2=4e-4 M/L of MgCl2. 4x^3= 4.1* 10^-20. Divide by 4= 1.025e-20=x^3= 10.25e-21= 2.whatever e-7. What does this mean? We have more than can be dissolved so most of it will remain solid.

4e-6/5e-2= .8e-4= 8e-5 Moles/Liter of NaCl. Well, 36e-12=x^2= 6e-6 for molar solubility. You have 6e-6 of both Na+ and Cl-. The total is 12 e-6 or 1.2 e-5 ions. More NaCl will dissolve than MgCl2. The amount of Cl in MgCl2 is 2*2e-7=4e-7 along with 2e-7 Mg +2. You have a total of 6e-7 ions of MgCl2. So, you have more ions dissolved in NaCl solution than the MgCl2 not surprisingly. You would need avogrado's number in order to calculate moles of actual solute but this is unnecessary.

For the calculations we have more solvent in the Nacl solution by 1.25 but the amount of ions is greater in NaCl by more than 1.25, over 10. Moles of solute/kg of solvent. With this in mind qualative comparisons can be made.

NaCl will have the higher boiling point, lower vapor pressure, and higher osmotic pressure. The opposite is true for MgCl2. I'm not sure how much this helps you but I tried.

#### hospitaldoctor1

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boiling point Temp = molality x Kb... not molarilty... do you guys have your math right?
I am VERY confused...

#### TheBoondocks

##### StreetFighter 4 Virtuoso
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boiling point Temp = molality x Kb... not molarilty... do you guys have your math right?
I am VERY confused...
Dude, I know molality is what you want. However, for osmotic pressure it is Molarity. I did it pretty quickly. What the heck is IP? I thought my work was clear.

4e-6 Moles of MgCl2 in 40 ml. We must convert to Liters so 40 ml= 40e-3 which is equal to 4e-2 L. You need to calculate the molar solubility. The ksp is 4.1e-20=4x^3. Why, since you have two moles of CL for every mole of Mg, the equation calls that ksp = x(2x)^2
Now, (4.1e-20)/4=x^3= 1.025e-20=x^3
Now, we need to make the power a factor of 3 so we move the base over by one so it becomes 10.25e-21=x^3
x=2.25e-7 moles/Liter.
If the concentration of MgCl2 is greater than this then we will get a precipitate. Well, 4e-6 Moles/4e-2 Liters= 1e-4 M/L This is significantly greater than the molar solubility so most won't dissolve. The amount dissolved is the molar solubility.

To find the amount of moles in solution, we take 2.25e-7 * 4e-2= 9e-9moles. I messed this up before. Why? The molar solubility is for one liter so we have to calculate how much is in the reduced volume. Now we take 9e-9 moles/4e-2 Kg of solvent.

Why, well g/ml are equivalent so 4e-2 Liters=4e-2 Kg 9e-9/4e-2= 2.25e-7 molal for MgCl2.

The boiling elevation is .5(2.25e-7)(3)=3.375e-7 +100. The amount of insignificant.

osmotic pressure is .5(2.25e-7)(3)(.0821)(298k).

My major issue is that there is a minute difference between molarity and molality. We aren't given density so for all intents and purposes, the are the same.

I have no idea about vapor pressure as I only know Raoult's law for two liquids.

Nacl has a one to one ratio so ksp=x*x= 3.6e-11.
We have to move it a number divisible by 2, so 36e-12=x^2
x=6e-6 this is the molar solubility
The Molarity is 4e-6/5e-2= .8e-4=8e-5 M/L
The concentration is greater so it will precipitate. The amount dissolved is the molar solubility. The molality is equal to the the molar solublity as demonstrated above with the calculation of the number of moles in 40 ml of MgCl2.

The reason I use Molarity is that determines how much is dissolved. Since the molar solubility is 6e-6. If the concentration was 6e-8 then that would be the amount dissolved since it's less than the maximum. To solve for the varying points you need the MOLALITY of DISSOLVED whereas before you use the MOLARITY ASSUMING 100% DISSOCIATION.

Now, for boiling= 100 +.5(6e-6)(2)
Freezing =0-1.9(6e-6)(2)
Osmotic= 6e-6*2*(.0821)(298K). This is assuming room temperature.

The difference in molality is kg of solvent vs Liters of solution. You can distinguish if you don't have the density. In many of these problems the molarity and molality are the same unless they give you specific density to calculate molarity.

An ionic solution has little impact. Raoult's law is in effect. The normal vapor pressure of pure water is 20 torr. However, since the mole fraction is over 99% water, the vapor pressure is pretty much the same. The boiling point of water goes up by 1 degree Celcius for every Mole of NaCl!! You have an insignificant amount. HTH

#### Seraph 84

Hook, line, and sinker.

#### TheBoondocks

##### StreetFighter 4 Virtuoso
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Hook, line, and sinker.
I know it's homework. However, I benefited from it because I looked up vapor pressure and how to calculate it and gained better conceptual knowledge.

#### hospitaldoctor1

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IP is Ion Product, i.e. Q
Reaction Quotient for Ksp

#### hospitaldoctor1

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so qualitatively, how do you determine which one has a higher BP, FP, Osmotic, etc...?
Also, could you explain why you do this: To find the amount of moles in solution, we take 2.25e-7 * 4e-2= 9e-9moles.
Why multiply the Molarity and the Liters of solution?
If the solution was NOT supersatured, but instead it was UNsaturated, would you do the same?
thanks

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#### TheBoondocks

##### StreetFighter 4 Virtuoso
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so qualitatively, how do you determine which one has a higher BP, FP, Osmotic, etc...?
Also, could you explain why you do this: To find the amount of moles in solution, we take 2.25e-7 * 4e-2= 9e-9moles.
Why multiply the Molarity and the Liters of solution?
If the solution was NOT supersatured, but instead it was UNsaturated, would you do the same?
thanks
All of these properties are colligative. Since the reference is H20, the qualative is based on which solution has the most dissolved ions. At first thought you would think of MgCl2 since it has 3 ions dissolved compared to NaCl's two. However, you must consider solubility and how much actually dissolves.

That was unnecessary. I didn't need that. I did that because I wanted molality. The molar solubility for MgCl2 was 2.25e-7 M/L. I needed to calculate the Moles of solute dissolves/ Kg of solvent. So, in order to determine moles you take the maximum dissolved concentration (2.25e-7) times the volume 4e-2 Liters. Then you take that and divide by the mass, 4e-2 Kg. So, it's unnecessary since molality is a concentration unit. The molality is the same as the molar solubility Numerically. This is because Liters are equivalent to Kg.

To illustrate. Let's say we have 6 moles of NaCl in 2 Liters. Let's assume that the density is not significantly greater than one. So, you would take 6 Moles/ 2 Liters for a concentration of 3 M/L. Now, what about molality. Well we have 6 moles/ 2kg of solvent. So, they are the same numerically. In reality, the density of the solution is like 1.003 Kg/L so you would take 1L/1.003 KG *2kg. This would give a volume greater than 2 Liters and make the Molarity less than Molality which is almost always the case. Good catch.

Finally, if it weren't unsaturated, you wouldn't need to have down what I did. What you do is compare the total number of potential moles of solute/Liters of solution. If that number is less than the molar solubility then it is unsaturated and you just take that concentration. If that number is greater than molar solubility, it means that it is saturated and that you need to use the molar solubility as the amount dissolved. The concentration of dissolved solute in a saturated solution is always the molar solubility. If you had the 10 times the initial amount of NaCl and water, the concentration would be the same, but THE MOLES would not.

So, we saw that the potential concentration for MgCl2 was 1e-4. This is way greater than than 2.25e-7. So, you know to use the molar solubility. If however it were 3e-8, then you would use this because the solution isn't saturated and all of the solid will precipate.

Now, qualitatively, you have to determine total ions dissolved. For Nacl, you have 6e-6 M/L of both Na+ and Cl-. Well, here you need moles since they have different VOLUMES. If they had the same volumes, then you could do it by concentration. 6e-6 M/L *4e-2L= 24e-6 moles=2.4e-5 and you double it to get the total number of moles so it's 4.8e-6 moles of ions in the NaCl.

For the MgCl2, I already calculated moles, this is why I did it, but got confused. As above, it's 9e-7 moles of Mg and 2*9e-7 moles of Cl-. The total is 27e-7 moles of MgCl2= 2.7e-6. There is over 10 times more ions in the NaCl than MgCl2.

So, the boiling point will be higher in NaCl, the freezing point lower, the osmotic pressure higher in NaCl, and vapor pressure lower because there are more ions. The reason is that with more ions, it is harder for the water molecules to associate with each other since they are more attractice to the ionic species. This makes it harder for water molecules to escape thus requiring a higher temperature to boil. The surface area is reduced which is critical in the ability of a solvent to escape and become a gas. The reduced vapor pressure is what necessitates a higher boiling point.

Freezing poing is reduced because a lower temperature is required to force the water molecules back together. Osmotic pressure increases with more solutes because the driving force for water to enter is greater. HTH

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