Solubility and Intermolecular Forces - Video Tutorials & Practice Problems

Solubility and Intermolecular Forces Concept 1

Why was it so important that we identify a compound as polar or non-polar? Well, because we're going to say that compounds with the same intermolecular force or polarity will dissolve into each other to form a solution. Now, we're going to say that if you have a polar and a polar, they're going to mix together well. If you have a non-polar and a polar, their polarities are different so they won't be able to dissolve into each other to form a solution. Now, we're going to say, according to the theory of 'likes dissolve likes', basically the two compounds have to have the same intermolecular force. If they have the same intermolecular force, they have the same polarity. But they could also have different intermolecular forces. So let's say one compound had hydrogen bonding and the other one had dipole-dipole. That's okay because hydrogen bonding and dipole-dipole are both polar forces. So because they're both still polar, they'll be able to dissolve with one another. But let's say one had dipole-dipole and the other one had London dispersion. Dipole-dipole is polar. London dispersion is non-polar. Because of their differences in polarity, they will not mix. Also, we're going to say that there's a difference between a mixture and a solution. We're going to say mixtures. We've talked about this so many weeks ago. Mixtures come in two types. We have homogeneous or heterogeneous. We're going to say, homogeneous mixtures mix together. They dissolve into each other. And we're going to say that heterogeneous mixtures do not mix. Oil and water is a good example that we've talked about. They won't mix because why? Oils are non-polar solvents. They're non-polar. Water, on the other hand, is polar. As a result, polar and non-polar do not mix. That's why oil and water don't mix together at all. So, mixers come in these two types. A solution, all solutions are just homogeneous mixtures. So, remember the difference. Mixtures come in two types. They can either be heterogeneous where they mix together or homogeneous where they mix together or heterogeneous where they don't. All solutions are just homogeneous mixtures. In a solution, we can dissolve both things into each other, so they do mix.

Solubility and Intermolecular Forces Example 1

For this example, we have to identify the intermolecular forces present in both the solute and the solvent. Here it's safe to assume either one is the solute or the solvent. And predict whether a solution will form between the two.

So, again remember fundamentally, for a solution to form, both should be polar or both should be non-polar. If they are the same in polarity, they will mix together to form a solution. If we take a look at the first one, we have CCl₄ and P₄. Well, P₄ is the easy one. We say that anytime we have non-metals connected to themselves or by themselves, they are non-polar by default. If you are non-polar, your intermolecular forces are London dispersion. That's not as important. Fundamentally, we need to just know if it's polar or non-polar here.

CCl₄: We've drawn this already. Remember, C goes in the center. It has 4 valence electrons. We have 4 chlorines, each one has 7 valence electrons. The central element has no lone pairs so we use rule 1a: The central element must be connected to the same elements, which it is. And the central element must be less electronegative than the surrounding elements. So it follows rule 1a and 1b. Therefore, it is definitely non-polar.

Now, because both are non-polar, they will form a solution. We could also say that since they are both non-polar, they both exhibit London dispersion forces.

So, if we have to identify the intermolecular force, we'd say both are London dispersion.

Solubility and Intermolecular Forces Example 2

For B, in B, we have H directly connected to oxygen. So if H is connected to fun, that's H bonding. Remember, H bonding is a polar force. Now, we have C6H6. We've said this also. If your compound has only carbon and hydrogen, it's going to be nonpolar automatically and its force will be London dispersion. But fundamentally, one is polar, one is nonpolar. They have differences in polarity. Therefore, a solution does not form.

Solubility and Intermolecular Forces Example 3

For carbon, we have a lot of carbons and hydrogens here. We can ignore those parts because here's the major force. Hydrogen is connected to nitrogen. Hydrogen bonding is a stronger force than London dispersion. Because it has hydrogen connected to nitrogen here, we're going to say that this is hydrogen bonding. Hydrogen bonding is a polar force. And in this example, we have hydrogen connected to fluorine. This is also hydrogen bonding and it's also a polar force. Both are polar, so they'll definitely make a solution.

Now, one thing we need to talk about here is this compound in carbon has a lot of carbons with it. Let's say we're trying to dissolve a compound that had a lot of carbons in water. What we need to realize here is that the more carbons your compound has, the more nonpolar it becomes. This is an important concept for Chem 1 and also for those of you who are going to take organic 1 and organic 2. We're going to say the more carbons associated with the compound, the more nonpolar it becomes, and the less soluble it becomes in water. The less of it will dissolve to form a solution. We're going to say the cutoff is once you get up to 5 carbons or higher, that compound becomes very nonpolar and it becomes incredibly difficult to dissolve it in water.

So if we're looking at 2 compounds, CH₃CH₂CH₂CH₃ and another compound, and your professor wanted you to determine which one is more soluble in water. We're going to say both have the same number of carbons. Both have 4 carbons. So they are somewhat dissolvable in water. Because water, its intermolecular force is hydrogen bonding. And here, we have hydrogen connected to oxygen. Part of this compound does have hydrogen bonding. That's what makes it soluble somewhat in water.

Now, we're going to say that the second structure is more soluble because the second structure has more oxygens. So the more oxygens you have, the more nitrogens you have, the more hydrogens bonded to fluorine you have, the more hydrogen bonding you'll possess. And the more hydrogen bonding you possess, the more you'll be able to dissolve in water. So if we were comparing these 2, we'd say the second one is more dissolvable in water.

Now, let's say we had CH₃OH or CH₃CH₂OH. Both have oxygen, so both have hydrogen bonding. But the second structure has more carbons, so it's less soluble in water. The top one has hydrogen bonding because of the oxygen and it only has one carbon, so it would be more soluble. So just make a little note on this. The more carbons we have, the more nonpolar. The more oxygens, nitrogens, and hydrogens bonded to fluorine, the more hydrogen bonding you'll possess, and the more soluble you'll be in water.

Solubility and Intermolecular Forces Example 4

Now, let's look at this final one, d. We're going to say let's look at the easier one. We have H connected to N here. So it's going to be H bonding for its intermolecular force, which is a polar force. But now, we have to figure out what the heck is the force in IF₄^-. Here, it's not easy to see, so we have to draw it out. And we have to draw it because we know it's not ionic, we know it doesn't have H bonding. So it's not one of those 2. This compound is going to either be dipole-dipole if it's polar or London dispersion if it's non-polar. And the only way we can tell that is if we draw it out. So, I will go in the center because I is less electronegative. Iodine is in group 7a, so it has 7 valence electrons. Then we have 4 fluorines. Fluorines in group 7a, so it has 7 valence electrons. Fluorine only makes one bond, so there go the bonds. Now, also remember that minus 1 means what? Minus 1 means we gain another electron. So we're gonna add 1 more electron to the mix. Okay. So 1, 2, 3, 4, 5. There was one here also. We're going to say how many electrons do we have left? We have an unpaired electron here, here and these 2 are paired up. You see those 2 electrons that are separated? Bring them close together. We should pair up our lone pairs. So there they go. So we have 2 lone pairs. Now, we go over the rules. It has lone pairs around the central elements, so we're going to use the rules for rule 2. So first, the central element must be connected to the same elements. Iodine is only connected to fluorines. 2, the central element must be less electronegative. The central element is less electronegative, so it follows 2b. 2c, we use dipole arrows to point to the more electronegative element. So this dipole arrow points to fluorine. It gets cancelled out by this dipole arrow which heads in the opposite direction. This dipole arrow points to fluorine and it gets cancelled out by this arrow that points to the other fluorine. All our element dipole arrows cancel out. Then we have a lone pair dipole arrow that points this way and one that points the opposite way. So they also cancel each other out. Based on all the rules for rule 2, this compound is non-polar. And because this element is non-polar, it forces London dispersion. But here's the most important thing. 1 is non-polar and 1 is polar. Because there are differences in polarity, no solution is formed. So that's what we'd say for part d. Now, hopefully, you guys are working towards being able to draw these compounds faster and being able to identify the intermolecular force.

When it comes to solutions, for a solution to form, both compounds need to be polar or both need to be non-polar. Likes dissolve likes. Now, that we've seen this, I want you guys to attempt to do this practice question on your own. So in this one, we have to figure out which of the following statements is true. Meaning that one could be the correct answer or more than one answer could be the correct answers. So go over what we know about intermolecular forces, about polar and non-polar. And I'll give you guys a huge hint. If it ends with ane, e.g., methane, pentane, all that means is that compound is what we call an alkane. Alkanes are compounds with only carbon and hydrogen. And that should be a huge hint. If your compound has only carbons and hydrogens, what can you say about its polarity? Is it polar or non-polar?

Which of the following statements is/are true?

a) Methane will dissolve completely in acetone, CH₃COCH_3.

b) Hydrofluoric acid (HF) will form a heterogeneous mixture with tetrachloride, CCl₄.

c) Pentane will form a homogeneous mixture with CBr₄.

d) Methanethiol (CH₃SH) is miscible in fluoromethane (CH₃F).

• A. A
• B. B & C
• C. C & D
• D. B,C, & D