This episode is an application of earlier discussed concepts. In episode 33 we talked about disruptions affecting equilibrium: concentration, temperature and pressure.
This episode is an application of earlier discussed concepts. In episode 33 we talked about disruptions affecting equilibrium: concentration, temperature and pressure. In episode 34, solubility equilibria, we talked about the dissolution of a salt. In this episode we set our focus a bit differently: We look at environmental factors, like pH or other dissolved ions and how they affect the dissolution of a salt. pH was a concept introduced in Unit 4 with more in depth coverage in unit 8 It's all connected. So, let’s recap the recaps that lay the foundation for this episode, then introduce the new concepts and dive into how these factors affect the dissolution of a salt.
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Hi and welcome to the APsolute Recap: Chemistry Edition. Today’s episode will recap Factors Affecting the Dissolution of a Salt.
Lets Zoom out:
Unit 7 - Equilibrium
Topics - 7.12 and 7.13
Big idea - Scale, Proportion and Quantity
Introduction:
This episode is an application of earlier discussed concepts. In episode 33 we talked about disruptions affecting equilibrium: concentration, temperature and pressure. In episode 34, solubility equilibria, we talked about the dissolution of a salt. In this episode we set our focus a bit differently: We look at environmental factors, like pH or other dissolved ions and how they affect the dissolution of a salt. pH was a concept introduced in Unit 4 with more in depth coverage in unit 8 - I know! It's all connected. So, let’s recap the recaps that lay the foundation for this episode, then introduce the new concepts and dive into how these factors affect the dissolution of a salt.
Let’s zoom in:
In episode 33 we recapped that Le Chatelier’s Principle helps us predict how a system reestablishes its dynamic equilibrium after a disturbance. Increasing the concentration of reactants as well as decreasing the concentration of products increases the rate of the forward reaction. Decreasing the concentration of reactants as well as increasing the concentration of products increases the rate of the reverse reaction.
Since we are specifically looking at the dissolution of salt, let’s also recap a part of episode 34: Solubility is how much of an ionic compound will dissolve and form a saturated solution. In a solution you have an equilibrium between the solid salt and the aqueous ions. To quantify how much of a salt dissolves, we can use the Solubility-Product Constant, Ksp. Since the Ksp tells us the ratio of concentration of product over reactant, in this case aqueous ions over undissolved salt, a small Ksp indicates that the salt is pretty much insoluble. As a reminder: solids are not part of your Ksp calculation since they have a concentration of 1. So the Ksp tells us the concentration of the aqueous ions. This was really only a quick recap of the recap - if you need to recap a bit more, feel free to stop, listen to episodes 33 and 34 and then come back!
So now let’s take a closer look at the factors that affect the dissolution. We are dissolving the salt in a solution that has a common ion. What does that mean? It means that we don’t have pure water in which we are dissolving our salt, but rather a solution that already contains one of the ions that makes up the salt - a common ion. For example: Imagine you want to dissolve calcium sulfate in a 0.1 M solution of sodium sulfate. As we know, sodium sulfate is soluble, because NAAA - nitrates, alkali metals, acetates and ammonium are always soluble. So we don’t have to worry about that. But: Will the solubility of calcium sulfate be affected by the presence of sulfate ions? Will it dissolve more, less or the same?
We can answer this with Le Chatelier's Principle or Ksp. Using the dissolution equation, we can see that the aqueous ions are the product. Since you have already some product given, according to Le Chatelier's principle, this will increase the reverse reaction, towards the solid, and therefore less of the salt will dissolve. A second way of thinking about it, that I would encourage you to keep in mind for AP Chemistry, is tied to the Ksp. The Ksp is the product of the concentration of the aqueous ions taking into account their coefficients. Since you already have one of the ions in solution, the magnitude of Ksp will be reached faster and less of the salt will dissolve. The common-ion effect is an important principle that you will revisit in Unit 9: Acids and Bases!
Speaking of acids and bases: the pH also affects solubility. In unit 4 you’ve gotten an introduction to acids and bases: Bronsted-Lowry acids are proton donors, Bronsted-Lowry bases are proton acceptors. Generally speaking, acids increase the concentration of hydronium ions - which we measure as pH. Bases often, but not always, have hydroxide ions. In neutralization reactions, when we mix an acid and a base, the hydronium ions react with the hydroxide ions to form water. What does that have to do with the solubility of salt? Let’s look at two scenarios.
Scenario A - the salt contains hydroxide; Many slightly soluble salts actually contain hydroxide. For example, if we try to dissolve barium hydroxide in a basic solution - would more or less barium hydroxide dissolve, compared to pure water? Well, this is a common ion problem! The solvent as well as the solute contain hydroxide. As we’ve discussed before, less of the solid will dissolve, because we already have hydroxide ions present, which is our product in the dissolution. But what if we put the barium hydroxide in an acidic solution? The acidic solution contains hydronium ions, which will react with added hydroxide to form water. This reaction reduces the amount of hydroxide ions - it is like constantly removing a product. So, according to Le Chatelier’s principle, we know by removing a product the equilibrium will shift towards the products, which is our aqueous ions, and MORE solid will dissolve compared to pure water!
Scenario B: We have a salt with a weak conjugate base. An example for a conjugate base would be the fluoride ion, F-, which is the conjugate base for the weak acid HF. So, if we have calcium fluoride, the dissolution forms fluoride ions. In an acidic solution, the conjugate base fluoride ions react with the hydronium to form hydrogen fluoride and water. Therefore, similar to before, we are removing a product, the fluoride ion, and this will lead to a shift towards products, increasing the solubility. Salts with weak conjugate bases are more soluble in acidic solutions than in pure water.
To recap:
According to Le Chatelier’s Principle, decreasing the concentration of a product will shift the equilibrium towards the products. Solubility equilibria show the equilibrium between the solid salt and the aqueous ions. When we dissolve a slightly soluble salt in a solution containing a common ion, the solubility of the salt decreases, since we already have a product present.
PH also affects solubility. Salts containing hydroxide have a lower solubility in basic solutions due to the common ion effect, but a higher solubility in acidic solutions, due to a neutralization reaction. Salts containing a weak conjugate base are also more soluble in acidic solutions than in pure water due to a reaction between conjugate base and hydronium ions.
Coming up next on the APsolute RecAP Chemistry Edition: Unit 7 Selected FRQs
Today’s Question of the day is about conjugate acid-bases. True or false: The stronger the base, the weaker the conjugate acid.