So take a look at lead to nitrate up here in our molecular equation. Well, as you go from a molecular equation down to the ionic equation, all right, what we're doing is we're splitting that cation up away from its anion, all right? So again, it's lead to nitrate.
So we get lead to right there, and it's nitrate. Remember, nitrate's NO3 minus, okay? NO3 minus.
And again, remember what this subscript... that's telling us. It's telling us there are two of those groups for every lead to cation.
So we get this coefficient out in front, the stoichiometric coefficient of two out in front. The other thing you'll notice is that we've carried the phase along with us. All right.
So it's lead two nitrate in its aqueous phase. It's two nitrates, all right, in their aqueous phase. All right. And then as you see with our ammonium sulfide, okay, again, Remember, sulfur likes to form a two minus charge. So that's our sulfide, carries its phase as well.
So it's still aqueous, all right? And our ammonium, all right, that's one of our two polyatomic cations we want you guys to know. So NH4 plus, all right? Again, our subscript, there's two of them. So we need a two out in front, okay?
Now, when we come over here, let's look at the ammonium nitrate first, all right? It's still aqueous, all right? So really inside of a aqueous environment, all right, or an environment that's kind of chocked full with water, all right, these guys are still gonna be falling apart, okay?
So we're going back to the solubility rules, all right? Our ammonium, no matter what it's with, all right, it's always soluble, okay? So it's gonna look like this.
It's gonna look like itself, all right? Again, there's a stoichiometric coefficient of two out in front. which means there are two of those ammonium and there's also two of these nitrates. Okay. And again, they're aqueous.
All right. Now the one thing that you'll notice that we did not split apart was that lead sulfide that led to sulfide. Okay.
And that's because it's a solid, right? It doesn't want to interact with the water. All right. So therefore we do not break up our solids.
Okay. And so So lead sulfide or lead 2-sulfide up in our reactant ends up, or in our product, excuse me, in our molecular equation, stays as lead 2-sulfide in our ionic equation. We do not split that up. It is still a solid. Now, what you can see here is that there's some things that are similar from one side of the equation to the other side of the equation.
All right. So you can see that our nitrates. And our ammonium cations, all right, are the same on either side of the equation, okay?
These are called spectator ions, all right? Spectator ions are these ions that are in solution, and they are not participating directly in any sort of chemistry, all right? Again, that's where the name comes from.
They're just kind of wallflowers on the wall, okay? They are spectators. They're just there and merely observing, all right? So they're not taking part in the chemistry, okay? So what we can do with those guys is we can think about eliminating them.
And then what results is what we call our net ionic equation, all right? Our net ionic equation is essentially describing the chemistry that is happening inside of that test tube, all right? Remember, our nitrates, there's two nitrates on our reactant side, two nitrates on our product side, all right? So essentially, we can cancel these guys out and get rid of them.
We can eliminate them, all right? Because again, they're not doing anything. They don't change from one side of the equation to the other side of this ionic equation, all right?
So we can get rid of them. Same thing with our ammoniums, all right? And what you see this left are the species that are actually taking part in the chemistry that we see.
And the chemistry that we see, right, is taking two aqueous solutions that are clear. And when we mix them together, we get a solid that falls out. All right, we get a precipitate that falls out.
And so everything that's left over after we get rid of our spectator ions, this is called our net ionic equation. Again, this is essentially describing the chemistry that is happening inside of that test tube and only those species that are taking part in that chemistry. All right, so let's do a sample problem. All right, so right...
the molecular equation, all right, the complete ionic equation, and the net ionic equation for the reaction between chromium three chloride. and sodium hydroxide. Okay. So remember the first thing we have to be able to do is we have to be able to take the names of our compounds. All right.
And actually write their chemical formulas. And so chrome three. All right. That means that we have chromium three plus. And remember chloride looks like this.
All right. And when we make our compounds, what do we want to do? We want to make a neutral compound. All right. So we get chromium three chloride.
OK, it's going to take three of these guys to balance out that charge to give us an overall compound that is neutral. All right. And it's going to react with sodium hydroxide. So we got sodium. and our hydroxide, okay?
So they come together in a one-to-one fashion, one sodium per one hydroxide, all right? And they will give us sodium hydroxide. Now we're talking about these double displacement reactions, these precipitation reactions.
And so again, and you can also go back to your solubility table, okay? And what we'd be able to see is that these guys should both with the aqueous. Because again, what do we want to do first?
We want to write out the molecular equation, like what's going on in this tube. Okay. And so then what do we want to do?
We want to do our double displacement. We want to be able to predict, all right, what products we're going to get from this reaction. Again, we're starting with two soluble ionic salts.
All right. We're going to mix them together and we want to see if some reaction happens. All right.
And so we got to figure out what the products are. And we'll also remember this is double displacement. So again, we want to swap the anions between the cations. And so the chromium, instead of interacting with chlorides, we can think about it interacting with the hydroxide.
Okay. And again, it's going to take three of those hydroxides. All right.
So essentially balance out that three plus on our chromium. And so we should get chromium hydroxide. And again, what's going to happen with the sodium? Well, the sodium no longer wants to interact with the hydroxide. All right.
Let's think about it that way. It's going to interact with the chlorides. All right. And these guys are going to come together in a one-to-one fashion.
Okay. So this is our essentially the start of our molecular equation. Okay.
There's a couple of things we have to do next, right? We have to predict like what the phases are of our products. Okay.
And so what we can do is we can go to our solubility table. We can focus on the sodium chloride first. And remember that our sodium is a group one metal and all group one metals, no matter what they're bound to are going to make something that is soluble. Now this should make some sense too, because we've all taken some salt and we've dumped it in water. And we know that those crystals disappear.
All right. So the salt will dissolve. All right.
So we can go ahead and say that this is aqueous. We know that that should be true. All right.
But how about the chromium hydroxide? All right. Well, we can go. I don't remember anything on this table about chromium.
And hopefully, except for down here when we're dealing with chromate. But chromate is not chromium. All right.
But we do. All right. We do have hydroxide.
on our table, all right? And remember, hydroxides tend to be insoluble. Now, there are some exceptions if those hydroxides are bound up with either a group 1A metal, all right? So these guys up here, or ammonium cation, all right?
Or some of our heavier group 2 metals, strontium 2 plus and barium 2 plus, all right? Well, chromium does not fit that bill, okay? So therefore, we should expect... that chromium hydroxide is going to in fact form a solid, okay? It's going to form a solid.
So this should be a reaction. In other words, we should see something if we would mix our chromium-3, all right, chloride and our sodium hydroxide together into a new vessel. We should see a precipitate form.
All right, so now the last thing we need to do is think about balancing this thing out, okay? Well... We have one hydroxide on our reactant side. We have three hydroxides on our product side.
Remember, it's easier to balance our polyatomic anions or our polyatomic cations as groups instead of breaking them up into their constituent atoms. All right. And so what we want to do is put a three.
out front. Okay. Now that gives us three sodiums that we have to deal with on our product side. So we can put a three out front here.
And now that gives us the three chlorides that we need to balance out the three chlorides that we have on the chromium three chloride on our reactant side. Okay. All right. So this is our molecular equation. All right.
We start with our reactants. We've predicted what our products are. We've predicted what the phases of our products are. And then we've balanced that guy out.
Now, the next thing we want to do is write out our net ionic, not our net ionic equation, our complete ionic equation. All right. And so remember what that does. That's taking all these guys in our molecular equation and essentially splitting up.
into their constituent ions, okay? All right, so we've already partially done this, all right? So our chromium three chloride will end up as chromium three plus.
Again, we gotta keep our phase on there, all right? And then again, we have our three chlorides. Again, let's keep our phases. And now we can break our sodium hydroxide up, all right?
And we have a stoichiometric coefficient of three out there, which means that there are three sodiums for the three hydroxides. So we got three sodiums. Again, these will be aqueous. And we get our three hydroxides, which are also aqueous.
And now we can come over to our products. Remember, when you have a solid, the solid would rather stay as a solid. chromium three plus in this case would rather interact with those hydroxides than interact with water all right so when we're talking about going from a molecular equation to a an ionic equation we keep our solids as a solid we do not split them up all right so we would get chromium hydroxide and again that's going to remain a solid all right and then of course we're going to split up our sodium chloride and we have three of them so we should end up with three sodiums Those are going to be aqueous. Let's keep our phases plus our three chlorides, which are also aqueous.
Okay. All right. And now what we want to do, we want to go from our molecular equation to our ionic equation.
All right. To our net ionic equation. And going from an ionic equation.
to a net ionic equation, we really only want to report on or show the species that are actually taking part in the chemistry, all right? So in other words, we need to get rid of our spectator ions, those guys that are just floating around in solution and they're not really contributing to any sort of chemistry, all right? And so what we want to do is we want to compare our reactant side to our product side and see if there's anything that shows up on either side of that reaction arrow that does not change. And what we can see is that we got three chlorides, okay, on our reactant side and three chlorides on our product side.
And we also have three sodiums on our reactant side and three sodiums on our product side. So we can cancel those guys out. And what we will see and what we're left with is essentially all the species that are taking part in the chemistry. Do not forget to bring along your stoichiometric coefficients.
Do not forget to bring along your phases. And so at the end of the day, this net ionic equation, all right, is showing us the chemistry that's happening. And again, it should be balanced. There's three hydroxides on our reactant side, three hydroxides on our product side.
There's one chromium on our reactant side, one chromium. on our product side, all right? And the other thing that can help you essentially give you comfort that you've actually done this correctly is that your charges should also balance, all right? So over here, we have a three plus plus three minus, right?
So those essentially come together to give us zero, all right, and on our product side, we have a neutral solid, which so also has a total charge of zero. So even our charges check out. Charges check out.
Okay, so this is how you go through this type of question, all right? You got to start with your reactants. You got to predict your products first thing, okay?
After you predict your products, all right, go to your solubility table or start to get comfortable with these solubility rules so you can know if they're going to remain aqueous or if they're going to form a precipitate or a solid, okay? Then what you have to do after that is you have to balance it. And when you've done all those things, that is going to be your molecular equation. Again, these are the types of equations we've been seeing a lot of in class thus far. All right.
Then what you can do, all right, is you can take your molecular equation. You can write essentially an ionic equation from it where we're taking all of these guys and splitting them up into their constituent ions. Of course, being certain that we keep solids as solids.
Okay. If it's a solid, it's not going to end up wanting to interact with water. Therefore, it should remain as a solid, not an aqueous species. All right, so we can break up all these guys that have these aqueous phases.
All right, then what we can do, all right, is essentially get rid of our spectator ions. And getting rid of our spectator ions takes us from an ionic equation now to a net ionic equation, where that net ionic equation is showing us the real chemistry that's happening inside of that test tube. Okay, and again, as a final check, all right, After you make sure you bring along your phases, you make sure you bring along the stoichiometric coefficients, you should double check that your charges actually check out as well. In other words, the charge on the reactant side, the total charge on the reactant side, should equal the total charge on the product side.
Now, if you get to this point, if you get up to that molecular equation, and you see after you've gone through those solubility rules, and you see that perhaps you don't have any solids or precipitates form, In that case, what you would label that equation as is a no reaction. Because ultimately, if you get everything that remains aqueous, you should be able to split everything apart when you're talking about the ionic equation. And therefore, when you're trying to make a net ionic equation, everything is going to cancel out. If you've balanced it correctly, everything should cancel out and you would get essentially no reaction. All right.
that. All right. Here's a participation question. All right. So write the molecular equation, the complete ionic equation, and the net ionic equation for the reaction between ammonium carbonate and calcium chloride.
What is the name and the formula of the precipitate that forms? Okay. So again, take these names of these two reactants, figure out what their formulas are, then predict what the products are, and then predict.
if any of those products are solids or not, all right? You got a hint, okay, that there's likely to be a precipitate that forms, okay? As you go from that molecular equation, all right, you have to make sure it's balanced, all right?
But then you can take all those species that are aqueous, okay? You can split those guys up into their constituent ions, all right? Make sure that a solid remains a solid.
You do not split your solids up, okay? And then what you can do is you can get rid of spectator ions. and take your net, your complete ionic equation, and essentially construct your net ionic equation.
All right, so that's the first one. The next one, write the molecular equation, complete ionic equation, and the net ionic equation for the reaction between sodium sulfate and silver acetate. Again, identify the spectator ions that remain in solution after the reaction has ended, okay? So in the first question, we want you to tell us what the precipitate that forms, all right? And this question, all right?
we want you to report on what those spectator ions are. Okay. All right. If you guys have questions, don't hesitate to reach out and ask. I hope you all have a great rest of your afternoon and I'll be seeing you soon.