Was it so important that we identify a compound as polar or nonpolar? 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 nonpolar 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 nonpolar. 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 nonpolar solvents. They're nonpolar. Water, on the other hand, is polar. As a result, polar and nonpolar do not mix. That's why oil and water don't mix together at all. So, mixtures come in these two types. A solution, all solutions are just homogeneous mixtures. So, remember the difference. Mixtures 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.
- 1. The Chemical World9m
- 2. Measurement and Problem Solving2h 25m
- 3. Matter and Energy2h 15m
- Classification of Matter18m
- States of Matter8m
- Physical & Chemical Changes19m
- Chemical Properties8m
- Physical Properties5m
- Temperature (Simplified)9m
- Law of Conservation of Mass5m
- Nature of Energy5m
- First Law of Thermodynamics7m
- Endothermic & Exothermic Reactions7m
- Heat Capacity17m
- Thermal Equilibrium (Simplified)8m
- Intensive vs. Extensive Properties13m
- 4. Atoms and Elements2h 33m
- The Atom (Simplified)9m
- Subatomic Particles (Simplified)12m
- Isotopes17m
- Ions (Simplified)22m
- Atomic Mass (Simplified)17m
- Periodic Table: Element Symbols6m
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- Rutherford Gold Foil Experiment9m
- 5. Molecules and Compounds1h 50m
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- Naming Monoatomic Cations6m
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- Polyatomic Ions25m
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- Writing Formula Units of Ionic Compounds7m
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- Molecular Models4m
- Calculating Molar Mass9m
- 6. Chemical Composition1h 23m
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- 9. Electrons in Atoms and the Periodic Table2h 32m
- Wavelength and Frequency (Simplified)5m
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- Electronic Structure4m
- Electronic Structure: Shells5m
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- The Electron Configuration (Simplified)20m
- The Electron Configuration: Condensed4m
- Ions and the Octet Rule9m
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- Periodic Trend: Metallic Character4m
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- Electron Arrangements5m
- The Electron Configuration: Exceptions (Simplified)12m
- 10. Chemical Bonding2h 10m
- Lewis Dot Symbols (Simplified)7m
- Ionic Bonding6m
- Covalent Bonds6m
- Lewis Dot Structures: Neutral Compounds (Simplified)8m
- Bonding Preferences6m
- Multiple Bonds4m
- Lewis Dot Structures: Multiple Bonds10m
- Lewis Dot Structures: Ions (Simplified)8m
- Lewis Dot Structures: Exceptions (Simplified)12m
- Resonance Structures (Simplified)5m
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- Electron Geometry (Simplified)7m
- Molecular Geometry (Simplified)9m
- Bond Angles (Simplified)11m
- Dipole Moment (Simplified)14m
- Molecular Polarity (Simplified)7m
- 11 Gases2h 12m
- 12. Liquids, Solids, and Intermolecular Forces1h 11m
- 13. Solutions3h 1m
- 14. Acids and Bases2h 14m
- 15. Chemical Equilibrium1h 27m
- 16. Oxidation and Reduction1h 33m
- 17. Radioactivity and Nuclear Chemistry53m
Solubility and Intermolecular Forces: Study with Video Lessons, Practice Problems & Examples
Understanding the polarity of compounds is crucial for predicting solubility. Polar compounds dissolve in polar solvents, while non-polar compounds do not mix with polar solvents, exemplified by oil and water. This principle, known as "like dissolves like," indicates that compounds with similar intermolecular forces, such as hydrogen bonding and dipole-dipole interactions, can form solutions. Mixtures can be homogeneous or heterogeneous, with solutions being a type of homogeneous mixture where components are uniformly distributed.
Solubility deals with the dissolving of a solute in a solvent in order to create a solution.
Solubility and the Intermolecular Forcess
Solubility and Intermolecular Forces Concept 1
Video transcript
In order for a solvent to dissolve a solute both components have similar polarities.
Solubility and Intermolecular Forces Example 1
Video transcript
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. Remember, fundamentally, for a solution to form, both substances should be polar or both should be nonpolar. If they're the same in polarity, they will mix together to form a solution. If we take a look at the first example, we have CCl4 and P4. Well, P4 is the easy one. We say that anytime nonmetals are connected to themselves, they are nonpolar by default. Being nonpolar, their intermolecular force is London dispersion. That's not as important. Fundamentally, we need to just know if it's polar or nonpolar here. CCl4. We've drawn this already. Remember, C goes in the center. It has 4 valence electrons. We have 4 chlorines, each one with 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. Thus, it follows rules 1a and 1b. So, it is definitely nonpolar. Now, because both are nonpolar, they will form a solution. We could also say that since they're both nonpolar, they both exhibit London dispersion forces. If we have to identify the intermolecular force, we'd say both are London dispersion.
Solubility and Intermolecular Forces Example 2
Video transcript
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
Video transcript
For c, 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 will definitely make a solution.
Now, one thing we need to talk about here is that this compound in C 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 two compounds, CH3CH2CH2OH and CH2CH2CH2CH2OH. 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're somewhat dissolvable in water. Because water, its intermolecular force is hydrogen bonding. And here, we have hydrogen connected to oxygen, hydrogen connected to oxygen, 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 OH groups. So the more OHs you have, the more NHs you have, the more HFs 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 two, we'd say the second one is more dissolvable in water.
Now, let's say we had CH3OH or CH3CH2OH. Both have OH, 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 OH 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 OH, NH, and HF, the more hydrogen bonding you'll possess and the more soluble you'll be in water.
Solubility and Intermolecular Forces Example 4
Video transcript
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 force in IF4- is. Here, it's not easy to see so we have to draw it out. And drawing it is essential because we know it's not ionic; it doesn't have H bonding. So, it's not one of those two. This compound is going to either be dipole-dipole if it's polar or London dispersion if it's nonpolar. And the only way we can tell that is if we draw it out. So, I will be in the center because I (Iodine) is less electronegative. Iodine is in group 7a, so it has 7 valence electrons. Then, we have 4 fluorines. Fluorines in group 7a, therefore, it also has 7 valence electrons. Fluorine only makes one bond so there go the bonds. Now, also remember that minus one means what? Minus one means we gain another electron. 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. First, the central element must be connected to the same elements. Iodine is only connected to fluorines. Second, 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. This dipole arrow points to fluorine. It gets canceled out by this dipole arrow which heads in the opposite direction. This dipole arrow points to fluorine and it gets canceled out by this dipole 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. They also cancel each other out. Based on all the rules for rule 2, this compound is nonpolar.
And because this element is nonpolar, it forces London dispersion. But here's the most important thing. One is nonpolar and one 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 nonpolar. 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 are 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 nonpolar. And I'll give you guys a huge hint. If it ends with ane, like methane, pentane, all that means is that compound is what we call an alkane. Alkanes are compounds with only carbon and hydrogen. 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 nonpolar?
Which of the following statements is/are true?
a) Methane will dissolve completely in acetone, CH3COCH3.
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).
Here’s what students ask on this topic:
Why is it important to identify a compound as polar or non-polar?
Identifying a compound as polar or non-polar is crucial for predicting its solubility. Polar compounds dissolve in polar solvents, while non-polar compounds do not mix with polar solvents. This principle, known as 'like dissolves like,' indicates that compounds with similar intermolecular forces, such as hydrogen bonding and dipole-dipole interactions, can form solutions. Understanding polarity helps in predicting how substances will interact, which is essential in fields like chemistry, biology, and environmental science.
What is the 'like dissolves like' principle in chemistry?
The 'like dissolves like' principle in chemistry states that compounds with similar intermolecular forces will dissolve in each other. For example, polar compounds will dissolve in polar solvents, and non-polar compounds will dissolve in non-polar solvents. This principle helps predict solubility and is based on the idea that similar types of intermolecular forces, such as hydrogen bonding or dipole-dipole interactions, will allow substances to mix and form a solution.
What are the differences between homogeneous and heterogeneous mixtures?
Homogeneous mixtures have a uniform composition throughout, meaning the components are evenly distributed. An example is a saltwater solution. Heterogeneous mixtures, on the other hand, have a non-uniform composition, where the different components can be visibly distinguished. An example is a mixture of oil and water. All solutions are homogeneous mixtures, but not all homogeneous mixtures are solutions.
Why don't oil and water mix?
Oil and water don't mix because they have different polarities. Water is a polar solvent, meaning it has a partial positive and partial negative charge. Oil is non-polar, lacking these charges. According to the 'like dissolves like' principle, polar and non-polar substances do not mix because their intermolecular forces are incompatible. This is why oil and water form a heterogeneous mixture instead of a solution.
Can polar and non-polar compounds ever form a solution?
Generally, polar and non-polar compounds do not form solutions because their intermolecular forces are incompatible. Polar compounds have dipole-dipole interactions or hydrogen bonding, while non-polar compounds have London dispersion forces. These differences prevent them from mixing uniformly. However, in some cases, surfactants or emulsifiers can help stabilize mixtures of polar and non-polar substances, forming emulsions rather than true solutions.
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