The common ion effect will help us to understand a little bit more about solubility. And how we can both increase and decrease the solubility of certain compounds. So we wanna be able to understand it so we can apply it for more advanced problems on the MCAT. To understand the common ion effect we're gonna go ahead and look at an insoluble salt.

And it's equilibrium system in more of a Le Chatelier's Principle way at first. So we could calculate the amount of salt that ends up dissolving and becoming these ions using Ksp and a nice table. But first, let's just think about the Le Chatelier's Principle. If I added or removed reactant, what would be the consequences? Careful.

It's a solid, right? So actually nothing. But we have aqueous things as products. What would happen if I added some silver ion or added some chloride ion? That would cause us to shift left, and when you have quote/unquote too many products you're going to shift left.You're gonna go back to the reactants, you're gonna make some solid.

And you're going to get more precipitate. So this salt will not be as soluble, if some ions are already present and the basis for that is the Q being larger than the Ksp. Remember Q is the value that you calculate for some experimental conditions or some non equilibrium system. Or if you are checking if its equilibrium.

You are just asking the question, and you are gonna plug in for silver and chloride concentrations to solve for Q, and then you compare to Ksp. So that is the short version of what the common ion effect is, is that a salt solubility will decrease if one of its ions is already present in solution. Let's go ahead and look at how this would work out mathematically. So what would be the solubility of this salt and water given its Ksp value.

In the other videos, you might not have seen ice tables, but a lot of teachers use them. So I'm gonna pull one up here in just a make sure you guys are familiar. It's an easy way to work through some equilibrium situations. And it can be longer, once you get very comfortable. You could probably skip ahead, but if you are still uncomfortable with equilibrium, it's a good way to handle it.

So an ICE table has the Initial row, a change row, and an equilibrium row, that's why it's ICE, I-C-E. So, we're going to say that for equilibrium. Solids don't affect anything, right? So the solid is going to be non existent. But the aqueous ions will be important.

So, we're going to say, initially we have pure water. So there's gonna be zero, for each of these. The change, well, we don't know what it is, so, what do we call things we don't know? X, right? Good old algebra.

So in equilibrium, the initial was 0, we changed by x. So 0 plus x is x, right? And 0 plus x is x. Like I said, if you're very comfortable, you kind of know where it's going. You kind of know that it's x, but for some of the more complicated problems, it can be helpful.

So once we've done this. You guys have seen a couple problems from here where we have Ksp is equal to each of those multiplied. Working through the math I find that in pure water, the concentration will be 1.3x10 to the negative 5, but the common ion effect is gonna come into a play. Well let's see what happens if we add some chloride.

We're going to try to calculate now how much silver chloride we can dissolve into 2M NACL. So solids still don't matter. So that's great. The silver is still 0 to start because we're talking about a solution of sodium chloride.

There's no silver ions present, but there's sodium chloride. We will initially have 2N of the chloride ion. This change will still be x and this change will still also be x because remember, every AGCL unit every AGCL formula unit that dissolves, gives one silver in one chloride, so it's still a one to one ratio. Zero plus x is still x, but now You think 2+X is 2+X, right?

But remember, for Ksp, it's always an insoluble salt. We expect very little products, right? This is a very small Ksp and values of K less than one favor reactants, aka little product. So two plus a tiny number. Is just two, go back through your SIG fig rules, check it out.

Two plus a very tiny number is just two and that simplification makes all the difference. Now when we plug in, we're going to say it's x for silver and just 2. So we've just got 2x= 1.8x10^-10. Which means x= 9*10^-11. The solubility in to molar sodium chloride is going to be, what is that?

Times 10 to the negative 6th lower than it is in water. So that's a lot less that you can Because we have a common ion, because we have the chloride already in solution. Now we've confirmed mathematically that the salt solubility does decrease, if the ion is already present in solution. It's really just the shatters principle applied.

What about the reverse? What if I could remove one of the products and take that out of solution? that would shift us to the right. We'd actually make things more soluble. We'll get more of the solid dissolving and it turns out we can do that using a different kind of common ion effect.

So we know that the salt's solubility will increase in that case. So what kind of common ion effect are we talking about? We're talking about complex ion formation Kf is the equilibrium constant that describes this it's the K of formation like K S P for the solubility product. Complex Ion Formation is when you have a complex ion right that kind of makes sense.

So if I was to show you guys what this looks like here's some of them. these are just examples of some you can have nickel complex with six ammonia. And overall it's still an ion. So we have the formation of what looks like an ionic compound because we have something that's a cation and this other thing right and will ammonia is not in an ion.

But it looks like a ionic compound with the polyatomic ion that's actually not what it is at all. It's an overall ionic compound that is made up of several parts. The way this works by the way is through Louis acid in basis So you can review that electron donating and accepting later on. But what we have now is the ability to take one of the metal ions from an insoluble salt and remove it from solution.

Technically, nickel with ammonia or copper with ammonia are each of these options, is not the same as nickel alone or copper alone. So essentially we're lowering the concentration of one of the products. And when we do that, we can start to see that there's going to be an increase in solubility. Anytime a little bit of silver chloride dissolves, the AG+ is gonna be basically just hoovered up by the vacuum.

That is the formation of this complex ion. The reason why I say it's such a strong pull is because look at the value of Kf. Kf is 16 million. That's a pretty large value of K, right? Ksp is 1.8x10^-10, which is still very small. So once you make a little bit of this it's going to get grabbed up and turned into this complex ion.

So really, we're just pulling away the Ag positive until we've used up almost all of the ammonia. That's the shortcut that you would use. If they're asking about a solution that has some ammonia and some silver chloride. What we would do is say all of the silver chloride dissolves until all of the ammonia is used up.

So let's take a quick look at that. Let's make up a quick simple problem, not natural problem. But just a quick skill application. Could look something like this. 143 grams of silver chloride is added to 2lL of a 0.5 M NH3 solution, how much of the AgCl will remain as a precipitate?

All right. What we know is, if we didn't have any ammonia present. We would essentially just say 143 grams because so much of the AgCl. Is unable to dissolve the over to say that is so little of the AgCL is able to dissolve that it would be negligible. You just wouldn't even notice it.

But the ammonia is gonna pull out a lot of the silver so that we're going to dissolve as much as we can until the ammonia is used up. So, what does that look like? Well, we actually have to combine these two equations to figure out what's going on. I'm gonna say that AgCl + 2NH3, right?

The silver here is a product and the silver here is a reactant. So there's no net silver in the overall combined reaction. We'll produce Cl- + Ag NH3 2+ So know that to say 1 to 2 ratio of AgCl to NH3. I'm gonna set up an ice table, but this is one is gonna be a little bit different. Ice tables normally only have molarlity.

I'm actually gonna do moles and use that instead just to calculate the amounts that I will have after the reaction. So its actually an ice table cuZ I'm not suing equal I mean I am using equilibrium but its not an ice on the standard way that we use them. Cuz we would normally ignore the solid but Iam interested in the moles of the reactant.

So here intially for the moles of AgCl. I chose this number for a good reason its essentially 1M. For ammonia, 2L of 0.5 m of solution, well molarity is moles of our litres, which means I have 1M of ammonia as well. And I have none of either of these to start. But the change is gonna be we said to use up all the ammonia.

So I'm gonna say -1 mol, so I will have 0 leftover. For every 2, we use 1. For every 1 mole, I'll use one-half. Which means 1 minus a half is gonna be one-half mole leftover. So I will have one half more of AgCL at equilibrium. Which means I will have, well if one half of this reacted and it's a one to one ratio will have half a more of this and one half mole of this.

Now this problem the way that I set it up, they don't actually want this information, but just to show you in case you needed to find out Zero plus a half is a half and again zero plus half is what half. But the idea here the core of it was that I used up all of like and all of the thing that was grabbing the silver and pulling it into solution as a complex ion.

So, at the end, I will say that one half a moles leftover or 71.5 grams of AgCl. So now we know how we can both increase and decrease the solubility of an insightful compound by either having some of the ions already present solution or using a complex ion to pull out some of the product. We can also use pH to change the solubility under very similar principles. If we'd have an equilibrium for PbF2 becoming Pb2+ + 2F- the ksp for this is pretty small, so its an insoluble compound.

How could we increase or decrease the solubility based on pH and not complex ion formation? Well we would have to have something bout pH. We would have to have either H- or Somehow interact with one of these options. Turns out that F- is gonna be a really good thing to work with. Because its a conjugate weak acids and so weak acids their conjugates are relatively small basis And so it's going to act F- like a base and take up H+.

So, if we were to increase the H+ concentration. AKA, lower the pH, we are going to get the formation of HF which means that the F- is gonna get pulled into this. Just like in the complex ion situation. Now that said remember, F- is a quote unquote strong base, but it's still relatively weak and so it won't be as strong of a pole.

But we will still be able to dissolve much more lead fluoride than under the other situation. So what would happen though, if we had a different type of conjugate base. What might happen if we work with silver chloride again and we try to add acid to pull the chloride ion out right. Because Cl- shares and equilibrium situation with HCl.

The answer is this would not work, there'd be no change. Right? So why is that? Because remember, HCl is a strong acid, which means the equilibrium lies almost entirely this way. We say that strong acids dissociate completely, which means that at this equilibrium there's zero this.

Which means we have no desire, the hydrogen ion and the cord ion are not going to go to the right at all. So when you have the conjugate of a strong acid or strong base that will not affect the solubility of an insightful compound. It's only when you have a conjugate of something that's relatively weak. Enzymes often require buffered solutions when they perform reactions in vitro.

So that they are not to nature and optimal pH can be kept. Which of the following salts would be the most soluble at an acidic pH. A, Barium sulphate becoming barium plus sulphate. B, Potassium nitrate becoming pottasium ion plus nitrate. C, silver sulphide becoming two silver ions plus sulphide. D, sodium iodide becoming sodium ion plus iodide.

Go ahead and pause the video, give it a shot and I will see you back here in a minute for the answer. So looking at these, we again have the idea of pH coming in and affecting equilibrium of an insoluble salt. And we want to say that the one that is going to be able to react with H+ is going to be the thing that will become the most valuable.

So H with S O four 2- is no good because this is the conjugate that they strong acid. H2SO4 as a strong acid, HNO3 string acid, HI strong acid. H2S is not a strong acid. There fore, this will have a good reaction or have some appreciable reaction I should say with hydrogen ion.

And we will get more pulled to the right and again more soluble. If you haven't yet memorized your strong acids, don't worry, you can go to acid base lessons and take some time with flashcards and figure them out. But you do need to memorize all of your strong assets. Anything that's not a strong acid is a weak acid. So you don't really have to memorize weak acids just be able to recognize them from their general form.

And one more question to help us finish practicing our common eye on effect. Fluoride in toothpaste is useful in strengthening enamel of teeth by combining with hydroxide appetite to form Fluorapatie. The compounds with a much lower Ksp. However, fluoride if ingested, can combine the calcium ions, which are found in all cell types to form an insoluble compound which the body may struggle to remove.

Which of the following compounds would be the best choice to precipitate fluoride ions so they cannot be absorbed into the body? A, Ca(OH)2, because CaF2 has a Ksp of 5.3x10^ -9. B, Ba(NO3)2, because BaF2 has a Ksp of 1X10^-6. C, Pb(NO3)2, because PbF2 has a Ksp of 2.7x10^-8. D, strontium nitrate, because strontium fluoride has a Ksp of 2.5x10^-9.

Go ahead and think about it. Pause the video while you're doing that, and I'll see you back here for the answer. In order to precipitate the Florida ions, we want to make sure that we have the least soluble compound. We wanna have the thing that when we add it, it will give us the most precipitate so that there's the least amount of fluoride left.

So we want the smallest Kcp the smallest Ksp. Remember Ksp is on equilibrium constant. So, small values of K have less product. The smallest one here will be the one with the largest negative number, so not this or this. It's gotta be one of the -9.

And so what's smaller is D. This is a smaller number than this one. So strontium flouride will be the best one. Now, I kind of ignored the idea of metals here. I mean, you might have thought, hey, I shouldn't give somebody lead ion and you'd be right about that.

But the most important concept was that the Ksp, being very small, will give you the most precipitate and this is an applied option of Ksp to healthcare aspect. Here's our summary of the major ideas within the common ion effect. Keep in mind that we can go both forward and backwards in terms of solubility. If we're dealing with the common ion solution or if we're taking out an ion that's common to a complex ion.

pH is also able to affect it and it depends on the type of compound you're working with. So today we only looked at acids affecting it, but keep in mind. We can also have hydroxide change the amount of H plus as well. So there's different ways that pH can interact. So make sure you practice those, and are ready for any type of question involving ions and solution.