Solubility Equilibrium

Summary

This video tutorial by mistercheung dives into the concept of solubility equilibrium, also known as KSP. The video thoroughly explains the processes involved at the molecular level when dissolving an ionic compound, such as table salt, in water and describes the importance of understanding the interactions between ionic and dipole charges. It details how to determine the solubility of compounds, the formation of equilibrium expressions, and the differences between soluble and insoluble substances. Additionally, the video uses conceptual examples and problems to illustrate how to calculate KSP values, molar solubility, and the application of ice charts in certain scenarios.

Highlights

• Understanding the molecular process of dissolving an ionic compound in water. 🌀
• Explaining the role of water's polarity in dissolving ionic compounds. 💧
• How ion-dipole attractions must be stronger than ion-ion attractions for solubility. ⚡
• KSP indicates the balance between dissolved and undissolved components in a solution. ⚖️
• Solubility determination without the need for memorizing solubility rules. 🔍
• Practical approaches showcasing calculations for KSP and molar solubility. 📚

Key Takeaways

• Solubility equilibrium involves understanding KSP, or the solubility product constant, to determine compound solubility. 🧪
• An ionic compound dissolves when water, a polar molecule, is able to separate its ions by overcoming their ionic bonds. 💦
• High KSP values indicate greater solubility, while low KSP values suggest a substance is likely insoluble. 📈
• Molar solubility measures the amount of solute that can dissolve in a solution after equilibrium is reached. ⚖️
• Using ice charts can help determine initial and equilibrium concentrations when molar solubility is known. 📝
• Real-life examples help in understanding solubility rules, especially when calculating KSP values. 🌍

Overview

Dive into the world of solubility equilibrium with mistercheung, where the complex process of dissolving ionic compounds is demystified. Learn how table salt can be broken down by water's polar molecules and what this means in terms of ion interaction. Through engaging animations and clear explanations, you'll gain a comprehensive understanding of these molecular dynamics.

The tutorial transitions into discussing the solubility product constant, KSP, shedding light on its significance in identifying compounds' solubility. Say goodbye to memorizing solubility rules; instead, understand how large KSP values point to soluble substances while small ones don't. With examples like silver chloride, the intricate balance of dissolution is laid bare.

Finally, mistercheung provides you with the calculation tools needed to tackle solubility problems—displaying how to set up equilibrium expressions and when to use ice charts. Real-world chemistry issues and their solutions give viewers practical insights, rounding off this thorough tutorial into solubility equilibrium.

Chapters

• 00:00 - 03:00: Introduction to Solubility Equilibrium The chapter provides an introduction to solubility equilibrium, also referred to as KSP. The discussion begins with a review of the dissolving process at the molecular level, using an animation of an ionic compound such as table salt. It emphasizes the repeating pattern of negative (chlorine anion) and positive (sodium cation) ions. Additionally, the polar nature of water molecules is mentioned, explaining how one end of the water molecule is more positive and the other more negative.
• 03:00 - 06:00: The Molecular Process of Dissolving The chapter explains the molecular process of dissolving, focusing on the polarity of water molecules. It highlights how the oxygen end of the water molecule is partially negatively charged, while the hydrogen end is partially positively charged due to the unequal sharing of electrons within chemical bonds. This polarity is essential for understanding how substances dissolve in water.
• 06:00 - 09:00: Understanding Solubility Terms The chapter 'Understanding Solubility Terms' explains the process of dissolving salt in water. It describes how water molecules interact with the ionic compound salt, causing it to dissociate ion by ion. The partially positive end of the hydrogen in the water molecule attracts the chloride anion, while the oxygen end interacts accordingly, highlighting the role of dipole interaction in solubility.
• 09:00 - 12:00: Dynamic Equilibrium in Solubility This chapter explains the concept of dynamic equilibrium in solubility, focusing on the interactions between water molecules and ions in a solution. It highlights the attraction between the negative dipole of the oxygen in a water molecule and the positive cation of an ionic compound, such as sodium in salt. The key point is that for an ionic compound to be soluble, the ion-ion attraction (which holds the compound together) must be weaker than the dipole-ion attraction between the water molecules and the ions.
• 12:00 - 15:00: Solubility Product Constant (Ksp) The chapter discusses the solubility product constant (Ksp), focusing on the concept of solubility and the factors affecting it. The explanation begins by comparing ion-dipole attractions and ion-ion attractions, stating that for a compound to dissolve in water, the ion-dipole attraction must be stronger than the ion-ion attraction. If the ion-ion attraction is stronger, the compound is considered insoluble or poorly soluble because the water molecules cannot dissociate the ions. The chapter highlights that solubility is defined as the maximum amount of solute that can dissolve in a solvent.
• 15:00 - 18:00: Calculating Molar Solubility The chapter explains the concept of molar solubility, emphasizing the difference between 'insoluble' and 'soluble' substances in a solution. Insoluble means less than 0.1 grams of solid dissolve in 100 milliliters of solution, indicating negligible solubility, whereas soluble refers to more than one gram dissolving in the same volume.
• 18:00 - 21:00: Applying ICE Chart for Solubility The chapter focuses on the application of the ICE (Initial, Change, Equilibrium) chart specifically for determining solubility. It discusses the solubility of slightly soluble and insoluble compounds, using silver chloride as an example. There is an explanation of dynamic equilibrium between dissolved ions and undissolved solid in such compounds.
• 21:00 - 25:00: Example Problem and Conclusion The chapter discusses the chemical and physical properties of dissolving, particularly focusing on silver chloride. It is emphasized that dissolving is a physical change, not a chemical one, as no new substances with new properties are formed. The process is described through an equilibrium expression, using the terms 'dissolved' and 'undissolved' rather than 'products' and 'reactants'.

Solubility Equilibrium Transcription

• 00:00 - 00:30 in this video tutorial we will be looking at solubility equilibrium also known as KSP before we move ahead let's briefly review over the process of dissolving at the molecular level so in this animation over here we have an example of a ionic compound let's pretend it's table salt where you have a repeating pattern of a negative and positive negative positive all right so this would be a chlorine anion negative ion and this would be a sodium cation positive ion now if you recall water is a polar molecule where one pole or one
• 00:30 - 01:00 end of the water molecule is partially negatively charged and the other Pole or the other end of the water molecule is partially positively charged this is due to the unequal sharing of electrons within the chemical bonds and so the oxygen atom tends to hold onto electrons when the rather than electrons spend more time around the oxygen atom on this side giving a negative charge or partially negative charge anyway and this end has a deficit of electrons the electrons do not spend as much time on this end making this n slightly positively charged and so when we
• 01:00 - 01:30 dissolve salt in water this is what happens the water mark is come in and pull the ionic compound apart they dissociate it ion by ion if we take a closer look over here you'll notice it's the partially positive end of the hydrogen grabs onto the chlorine anion which is negative and so opposite charges partial positive end of the dipole attracts the negative cat-eye or anion rather and pulls it apart similarly when the oxygen end of the
• 01:30 - 02:00 water molecule comes by it is attracted towards the positive cation and it pulls it apart so notice how the negative partial charge on the oxygen dipole is attracted to the positive cation of the sodium ion if it was salt for instance now please keep in mind that if a ionic compound is going to be soluble the ion ion attraction so the attraction between the ions attraction that holds this ionic compound together must be weaker than the dipole ion attraction all right
• 02:00 - 02:30 so in order for water to pull this and dissociate this thing apart the ion dipole attraction must be stronger than the ion ion attraction the whole together if the ion an attraction is stronger than the attraction between the iron and a dipole then we would say that the compound is insoluble it does not dissolve at least it doesn't dissolve very well anyway because the water molecule can't pull it apart so if you recall solubility is the maximum amount of solute that can
• 02:30 - 03:00 dissolve in a given quantity of solution at a certain temperature it is important to remember that the term insoluble doesn't mean that the salt can't dissolve at all it just means that very little bit can actually dissolve less than 0.1 grams of solid dissolving in 100 mils of solution all right so please give it a mind the soluble doesn't mean that doesn't dissolve it just means that the amount that does dissolve is typically considered to be negligible on the other hand soluble means more than one gram of solid dissolving in hundred mils of solution whereas anything in
• 03:00 - 03:30 between the point 1 grams and the 1 gram is considered to be slightly soluble now when dealing with insoluble compounds for example the silver chloride over here there exists a dynamic equilibrium between the dissolve component so the silver and the chlorine which is dissolved and the undissolved component the silver chloride which is a solid meaning that some of the silver chloride will break up into the aqueous silver and chlorine ions while some of the aqueous silver and chlorine ions will
• 03:30 - 04:00 recombine again back in the silver a salt silver chloride as such you can set up an equilibrium expression to describe this relationship this equilibrium but instead of using products over reactants we're going to call it dissolved over undissolved please keep in mind that the process of dissolution the process of dissolving is not a chemical reaction it's not a chemical change dissolving is a physical change you're not making any new substances with new properties all
• 04:00 - 04:30 you're doing is separating the molecules a little bit further away from each other and that's pretty much it you're not making any new compounds so please keep in mind it's not a chemical reaction dissolving is a physical change so as such we can't call products over reactants we need to call it dissolve over the undissolved component now remember when we set up this equilibrium expression because silver chloride is a solid we don't factor it in alright the concentration of pure solids and liquids are constant and should not be included in the equilibrium expression
• 04:30 - 05:00 thus we can simplify our equilibrium expression by getting rid of the denominator and leaving it as this so this is known as the solubility product constant KSP all right so KSP keq in the same equation but the by saying KSP were just being a little more specific we're referring to not an equally about chemical equilibrium but rather a solubility equilibrium instead when we have a large keq value that indicates the numerator is large in that case the
• 05:00 - 05:30 consecration of products is high well similarly a large KSP value tells us that the numerator is very large as well and what is in the numerator the dissolve component so that tells us that this compound dissolves very well on the other hand if you have a very low KSP value that tells me that the denominator is very large and what's in the denominator the undissolved component so I have a lot more undissolved that tells me that these substance is likely to be
• 05:30 - 06:00 insoluble so there's no need for you to have a solubility chart or to memorize the solubility rules anymore all we have to do is just give you your KSP values and that will tell you whether a substance is soluble or insoluble large KSP value soluble low KSP value insoluble now as with other equilibrium constants KSP is also temperature dependent unless otherwise stated most KSP values are going to be assumed to be in an aqueous solution at 25 degrees Celsius now molar solubility is another way for
• 06:00 - 06:30 us to express how well a substance will dissolve it describes how many mol of solute can dissolve in a set volume of solution typically measured in liters please keep in mind that solution is equal to the mass solute and solvent combined together when dealing with molar solubility questions a common mistake students will make is to set up an ice chart that isn't necessary molar solubility doesn't talk about how much can dissolve initially it describes how much can be dissolved at the end at
• 06:30 - 07:00 equilibrium so here the question states the molar solubility of barium hydroxide is 0.18 moles per liter meaning at equilibrium zero point 108 moles of VM hydroxide can dissolve in one liter of solution so now it's asking what is its KSP value so we write out our KSP expression yes B is equal to the products or in this case what is dissolved / what is undissolved of course because this is a solid we
• 07:00 - 07:30 don't have to factor it in just worried about the barium and the hydroxide over here now another common mistake I typically see students make is to write out the dissolution or dissolving equation as barium hydroxide produces a barium ion and then Oh H - no no what happens is Bayer hydroxide releases two hydroxides not an O H - that's combined together alright so please be careful that we do not go this route we have to go this way over here very hydroxide will break up into two hydroxides not an O H - that of
• 07:30 - 08:00 course becomes important because now we can see that there is a 1 to 1 to 2 ratio that way if I drop off or have a break apart 0.18 moles of Bayer might Rock side I also release 0.1 o 8 moles per liter of barium ions but then double that amount because we you release two hydroxides and don't forget to take the two coefficient and make it into an exponent and there you have your KSP expression so once again very important
• 08:00 - 08:30 that you remember bare my dioxide will break up into two hydroxide ions and do not write it as Oh H - otherwise we'll be missing out on this number here and double this amount as well from there it's just a matter of substituting in your values and solving for KSP with a KSP value of 0.005 that tells me that the denominator is larger and so there's more undissolved very hydroxide then
• 08:30 - 09:00 there is dissolved Bayer hydroxide telling me that bare hydroxide has a low solubility now what if we go in the opposite direction where I give you the KSP value and I ask for its molar solubility instead so again if I'm asking for molar solubility no need for an ice chart just equilibrium because most solubility by definition refers to how much can dissolve when all is said it done not initially now that how much can you dissolve when all is said and done at equilibrium you'll notice that I've left the solid empty
• 09:00 - 09:30 because solid concentrations do not change so I don't have to worry about them but I do need to know what are these values over here I decide to say let X represent the iron concentration and then that means the hydroxide concentration must be 3 X because this is a 1 to 1 to 3 ratio when you break iron 3 hydroxide apart please keep in mind that you do not have to do it this way I could easily say let's this represent X but then this would have to
• 09:30 - 10:00 be one third X because it's one third as much I don't recommend this method because personally I don't like fractions and I find that most students tend to make a lot more mistakes when involves fractions so keep it simple whoever has a coefficient of 1 let that one represent X and this one just becomes 3 times as much 3 X makes your calculation is a lot easier so we now substitute our values in X for the iron concentration 3 X for the hydroxide
• 10:00 - 10:30 concentration don't forget the coefficient of a 3 gives it a cubed and then your KSP value gets plugged in over here from there we solve for X and we get a value of nine point nine times 10 to power of negative 11 moles per liter and that represents the concentration of iron three-plus now since the iron three-plus concentration and the amount of iron 3 hydroxide that dissolves is a one to one ratio then the value for iron 3 hydroxide is also to nine point nine
• 10:30 - 11:00 10 to power 11 moles of either hydroxide can dissolve if you had a 1 liter solution alright let's try a problem where we do need an ice chart so in this situation over here we have 0.5 moles of calcium bromide dumped into a 1 liter solution of water now some students ask what happens if you dump powder into the water won't the volume of the water rise yes it will but keep in mind that rising water is typically going to be
• 11:00 - 11:30 negligible when you pour in a little bit of sugar into a cup of water do you notice the volume of the water rise significantly typically not so for problems that we will encounter in this course you may assume negligible volume increase unless otherwise specified in the question itself now the difference between this question the previous ones the reason why this one requires an ice chart while this one does not require an ice chart is because this one refers to molar solubility what is the molar solubility and by definition molar
• 11:30 - 12:00 solubility refers to what happens at equilibrium when the dissolution process is over whereas with this question it's giving you initial values it's saying point five moles of calcium bromide is dumped into a one liter solution it's not telling you what's happening at equilibrium it's telling you what is initially occurring then it says its degree of dissociation twelve percent meaning only 12% of the 0.5 molar sample will break up so therefore you have some information about your change so this all tells me that I'm going to need an ice chart to solve this problem so just
• 12:00 - 12:30 like all the other equilibrium questions we've done before just start filling in the chart since the calcium bromide is going to dissolve it's gonna be a minus x value this will be a plus X this will be a plus two X because you would have twice as many bromine ions being released all right and then at equilibrium you have a 0.5 minus X an X and a 2 X value because the question says 12% will break up while we do 12% 0.5 equals two point zero six then that
• 12:30 - 13:00 means I know that X is equal to 0.06 and I can go ahead and change my X values into 0.06 0.06 and two times point zero six that will leave us with equilibrium values of 0.4 4.0 six point one two and we can substitute these values in to our equilibrium expression notice how the point four four is ignored because it's a solid and of course don't forget the coefficient of a two must be factored in
• 13:00 - 13:30 as an exponent from there just solve and we get a KSP value of 8.6 for 10 to power negative 4 super small number and that indicates to us the denominator is bigger which is the undissolved component and that tells us the calcium bromide doesn't dissolve very well now Part B of the question asks what mass of calcium bromide is dissolve we know that we had point five moles originally and in the end we only have point four four left so putting four four moles of calcium bromide did
• 13:30 - 14:00 not dissolve mass of calcium bromine is mole times the molar mass which we can get from the periodic table and there you have it 88 grams of calcium bromide remain undissolved you