- Dissolution and precipitation
- Common polyatomic ions
- Introduction to solubility equilibria
- Worked example: Calculating solubility from Kₛₚ
- 2015 AP Chemistry free response 4
- The common-ion effect
- pH and solubility
- Solubility and complex ion formation
Introduction to solubility equilibria
The solubility product constant, Kₛₚ, is an equilibrium constant that reflects the extent to which an ionic compound dissolves in water. For compounds that dissolve to produce the same number of ions, we can directly compare their Kₛₚ values to determine their relative solubilities. If we know the solubility of a salt, we can use this information to calculate the Kₛₚ value for the compound. Created by Jay.
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- What would the unit for Ksp be?(9 votes)
- That varies depending on the substances. There is no set units for equilibrium constants and for that reason it is customary to just omit the units, and just let it be understood that all of the concentrations are in units of molarity (M) unless otherwise specified.(21 votes)
- I believe that I am getting mixed up with when to use mol vs M the initial concentration of the ICE table. I remember using mol in some of the problems in the titration lectures. Can you explain?(9 votes)
- where did you write the concentration of reactants?(1 vote)
- At8:51he states that you do not include the concentration of the reactants in the Solubility Equilibrium because it is a pure solid.(17 votes)
- How could PbCl2 be an ionic compound? the difference of electronegativity between the two compounds is not that great.(4 votes)
- If there is any electronegativity difference, then the bond will have some ionic character (even if it's not "fully" ionic). Bonding is not a dichotomy: it is not either covalent or ionic. Bonding happens on a sliding scale between 100% ionic (no electron sharing, electrostatic attraction only) and 100% covalent (electrons are shared exactly equally). PbCl2 falls between these two extremes.(7 votes)
- at 3.50mins you wrote 0.00079mol when the calculation was actually 7.91 mol, why did you do this?? it's already in grams and that's what we want the answer to be in so why make it into kg??(1 vote)
- The calculator shows 7.9108234448g -4 . You didn't notice the -4,which means multiplying 7.91 with 0.0001. And obviously, when dividing 0.22 by 278.1, you can't get 7.91.(9 votes)
- Do we have to put coefficient as exponent for Ka and Kb as well?(4 votes)
- I had the same question since it wasn't mentioned in previous videos!(1 vote)
- At about the1:00mark he says this is a saturated solution. But isn't a saturated solution one in which all of the solute has completely dissolved into the solvent?(2 votes)
- A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent. Additional solute will not dissolve in a saturated solution.
NOT "all of the solute has completely dissolved into the solvent" BUT Additional solute will not dissolve in the solvent (saturated solution) .(5 votes)
- will ksp change if temp change?(3 votes)
- Yes. How a temperature increase affects solubility depends on wheter the dissolving reaction is exo or endothermic. The majority of substances show an exothermic dissolving. In that case, increasing the temperature will decrease solubility.
There's a more mathematical approach, but I like to think about it this way:
A(solid) <---> A(aqueous) + Heat , so, if you increase temperature, the reaction will be able to consume heat energy from the enviroment more easily, what shifts it to the left, decreasing Ksp. On the other hand, if the temperature decreases, the reaction will, almost in a attempt to keep balance, shift to the right and generate heat energy.(3 votes)
- At around9:00, he leaves out the PbCl2 in the Ksp equation. If PbCl2 happened to be aq, how would you input it into the Ksp equation?(3 votes)
- The compound will always be (s), it will never be (aq), and will never be included in the Ksp expression.
It’s looking at how much of the solid splits up into ions and dissolves.(2 votes)
- (8:58) I don't understand whY we don't include solids and liquids in the equilibrium expression. If we are saying that it is because their concentrations do not change, isn't this false, as we have seen our solid dissolve into ions, meaning its concentration in the system will decrease?
Even if concentration does not change, is it simply a matter of convention - i.e. we do not include them, since it doesn't matter for comparing Q and K, since we can cancel out the unchanged concentrations from each side?(2 votes)
- Good question btw. So you are correct in that we don't include solids or pure liquids because their concentrations don't change while aqueous and gaseous species do change. This is because liquids and solids are practically incompressible. The molecules in liquids and solids are very close together, with very little room to squeeze any closer. This means their concentrations only depend on their densities which are constant. If part of our solid dissolves into ions, the density of the solid itself does not change. As long as some of the solid remains undissolved, we can disregard it in the equilibrium expression because density remains unchanged. Hope this helps.(4 votes)
- [Instructor] Let's say we have a beaker of distilled water at 25 degrees Celsius. And to the beaker, we add some barium sulfate. Barium sulfate is a white solid. A small amount of the barium sulfate dissolves in the water and forms Ba2+ ions in solution and sulfate ions in solution. So let me draw those in on our diagrams. So we're gonna form some Ba2+ ions and some sulfate anions. But most of the barium sulfate remains undissolved and so we'll draw that here sitting on the bottom of the beaker. So barium sulfate can dissolve to form Ba2+ ions and sulfate anions in solution. And it's possible for the Ba2+ ion to combine with the sulfate anion to form a precipitate, of barium sulfate. When the rate of dissolution is equal to the rate of precipitation, the system is at equilibrium. These types of equilibria are referred to as solubility equilibria. And when the system is at equilibrium, the concentrations of Ba2+ ions and sulfate anions solution are constant. And the amount of solid is constant too. And this forms a saturated solution. The balanced equation shows the dissolution of a salt barium sulfate. And from the balanced equation, we can write an equilibrium constant expression. So we would write the equilibrium constant K is equal to the concentration of Ba2+ and since there's a coefficient of one in the balanced equation, it'd be the concentration raised to the first power times the concentration of sulfate also raised to the first power. And since pure solids are left out of equilibrium constant expressions, we would not include the solid barium sulfate. For solubility equilibria, we would write Ksp where sp stands for solubility product. The solubility product constant Ksp has only one value for a given salt at a specific temperature. That temperature is usually 25 degrees Celsius. And Ksp indicates how much of that salt will dissolve. For example, at 25 degrees Celsius, the Ksp value for barium sulfate is 1.1 times 10 to the negative 10th. When the Ksp value is much less than one, that indicates the salt is not very soluble. So barium sulfate is not a soluble salt. If the Ksp value is greater than one, like it is for something like sodium chloride, that indicates a soluble salt that dissolves easily in water. The solubility of a substance refers to the amount of solid that dissolves to form a saturated solution. Usually the units for solubility are in grams per liter. Molar solubility refers to the number of moles of the solid that dissolve to form one litter of the saturated solution. And therefore the units would be moles per one liter or you could just write M. Ksp values can be used to predict the relative solubilities of salts that produce the same number of ions in solution. For example, silver chloride, silver bromide, and silver iodide all produce two ions in solution. Let's look at the dissolution equation for silver chloride to see why this is true. Solid silver chloride turns into Ag+ and Cl-. So that's one Ag+ ion and one Cl- ion for a total of two ions in solution. And we could write out similar equations for silver bromide and silver iodide, so they all produce two ions in solution. However, a salt like Lead(II)chloride produces three ions in solution. So Lead(II)chloride would give one Pb2+ ion and two chloride anions in solution. One plus two is three ions. Since Lead(II)chloride produces three ions in solution, we can determine its solubility relative to the other three by comparing Ksp values. Here are the Ksp values for the three salts at 25 degrees Celsius. For silver chloride, it's 1.8 times 10 to the negative 10th, for silver bromide, it's 5.0 times 10 to the negative 13th. And for silver iodide, it's 8.3 times 10 to the negative 17th. When comparing salts that produce the same number of ions, the higher the value of Ksp, the higher the solubility of the salt. And since silver chloride has the highest Ksp value of these three, silver chloride is the most soluble salts. For some insight into why this is true, let's look at the Ksp expression for silver chloride, which we can get from the balanced equation. The higher the value for Ksp, the higher the concentration of these ions at equilibrium, which means that more of the solid must have dissolved. Therefore, silver chloride has the highest solubility out of these three salts. Let's say we have some solid calcium fluoride that we add to pure water at 25 degrees Celsius. Eventually equilibrium is reached and the equilibrium concentration of Ca2+ ions is measured to be 2.1 times 10 to the negative 4th M. Our goal is to calculate the Ksp for calcium fluoride at 25 degrees Celsius. The first step is to write out the dissolution equation for calcium fluoride. So we would write CaF2 solid, and we know that calcium forms a 2+ cation, so we would write Ca2+ in aqueous solution and to balance everything we'd need two fluoride anions. So 2F- also in aqueous solution. The next step is to use the balanced equation to write the Ksp expression. So Ksp is equal to, there's a one as a coefficient in front of Ca2+ so it'd be the concentration of Ca2+ raise to the first power, times the concentration of fluoride anion and since there's a two as a coefficient, this is the concentration of fluoride anion squared. For a Ksp expression, these are equilibrium concentrations, and we already know the concentration of Ca2+ at equilibrium is 2.1 times 10 to the negative 4th. So that can be plugged in for the equilibrium concentration of Ca2+. So here's our expression with 2.1 times 10 to the negative 4th plugged in, and next we need to plug in the equilibrium concentration of fluoride anion. We're looking at the disillusion equation, the mole ratio of Ca2+ to fluoride anion is 1:2. Therefore, at equilibrium, there's twice as many fluoride ions in solution as there are Ca2+ ions. Therefore the equilibrium concentration of the fluoride anion would just be twice this concentration for Ca2+. So the equilibrium concentration of fluoride anion must be 4.2 times 10 to the negative 4th M. And when you do the math, you get that Ksp for calcium fluoride is equal to 3.7 times 10 to the negative 11th at 25 degrees Celsius. Ksp values can be difficult to measure and therefore different sources often give different values for Ksp at the same temperature. For example, for calcium fluoride at 25 degrees Celsius, one source had Ksp equal to 3.5 times 10 to the negative 11th. Another one had 3.9 times 10 to the negative 11th. Since we got 3.7 times 10 to the negative 11th, this sounds like a pretty good calculation for the numbers that we used for our problem. Finally, let's think about the molar solubility of calcium fluoride. So how many moles of our salt dissolve to form one litter of our saturated solution? Well, the mole ratio of Ca2+ ions to calcium fluoride is 1:1. Therefore, the concentration of Ca2+ ions in solution 2.1 times 10 to the negative 4th M, that number must also be the molar solubility of calcium fluoride. Therefore, for this problem, we could say that we use the molar solubility of calcium fluoride to calculate the Ksp value for calcium fluoride.