 All right, there is one last application in this solutions unit And that is to write in that ionic equations. There is this procedure we follow to do this first thing we do is Either write an equation balance an equation we have to get a balanced equation and that balance equation has to have state in it So liquid gas and aqueous and I said get a balanced equation because it depends on how we're going to be asked to Do these questions it could be where they give you the reactants and you have to figure out the products And then you've got to balance the whole thing and then you got to use your solubility tables to figure out what's solid what's aqueous and so on going back to the first Semester and using a lot of those skills that we learned there Along with stuff that came from third quarter very cumulative kind of thing Which I think a lot of you are figuring out is very difficult if you don't know the parts Well at any rate you need a balanced equation that has all the states in it aqueous solid liquid and so on the next thing that we're going to do in the process is Go through and break down all the aqueous Things into their component ions. We're going to dissociate all the aqueous solutions and dissociate Just again brings break the ionic compound down into its component positive negative ions then We're going to replace all the aqueous solutions with the ions So we'll go back through and cross out all the aqueous solutions and rewrite it with their component positive negative ions instead We will cross out We call spectator ions These are ions that we have on both the reactant and product side. They're just hanging out there There's really nothing anything happening to them. That's why we cross them out And then rewrite the equation With whatever's left simple Right All right, so let me do a couple of these as a demonstration to show you how it's done Maybe assist you with being able to figure out this homework. That's all gonna fit on the screen or not. It does All right, so here's our reaction. We have aqueous sodium chloride reacting with aqueous led to nitrate to produce solid lead chloride lead to chloride and Aqueous sodium nitrate. This is a precipitation reaction Precipitation reaction because we have two aqueous solutions combined to make a solid that solid is called precipitate Now step one get a balanced equation with the states in it Well, this one already has states in it, but this one's not balanced So that's the first thing we'd have to do is balance this out And I'm just going to visually balance it scanning through adding coefficients where I find they're needed So I start with sodium. There's one sodium on this side. I go over to this side of the equation There's only one sodium there. So that's good One chlorine on the reactant side two on the product side that's out of balance So I'll throw a two out in front of the sodium chloride to fix that I'd have two chlorines there two chlorines there Now whenever I add a coefficient in this method, I have to go back and I have to start the whole thing over Two times one is two sodiums only one sodium over here So I've got a two there to balance the sodiums out and again start over Two times one two sodiums two times one two sodiums Two times one two chlorines two times one two chlorines One times one one lead one times one one lead One times two two nitrates two times one two nitrates now. It's a balanced equation So again that first step in the process is a little bit vague because it all depends on whatever you're given to do If you're given the equation and there's no states in it You would have to use ye old solubility table to figure out what the states are If it's not balanced, you'll have to balance it if they only give you the reactant side You'll have to figure out the product side. There's all kinds of ways this can be given to you Once you have the balanced equation the next thing to do is to associate all the aqueous solutions. So Anything with an aq on it. You got to break it down into ions So the two Nacl needs to be broken down into ions Splits right there breaks down into positive sodium and negative chlorine There are two sodiums and there will be two chlorines Whenever you have a coefficient it applies to both the ions that you're going to break it down into we have two Nacls Because of that we'll get two NAs and we'll get two Cls Think of it as the number of atoms have to be the same on both sides two times one means two sodiums two sodiums two times one Two times one means two chlorines two times one is two chlorines The charges have to cancel out over here two times plus one is plus two Two times negative one is negative two, so the charges cancel out Moving on to the next aq the lead to nitrate PB N03 to breaking it down splits right there a positive lead ion and Two nitrates When you have a subscript it's going to end up changing the number of moles of that particular ion So we have two on that nitrate, which means we have two moles of nitrate now This is on your list of polyatomic ions as a negative one Two times negative one is negative two this side has to cancel it out So lead is a plus two thus the name lead to Nitrate lead to chloride is a solid we don't break those down. We only break down the aqs Sodium nitrate is an aq so we have to break it down Splits right there Again whenever you have a coefficient it carries through to both So we have two NA is each one with a charge of plus one and we have two nitrates And again each one of those are negative one. So there we have dissociated all The aqueous solutions Now we're going to replace These three things in the equation with what they break down into so we're getting rid of All the aqueous stuff So the NACL gets replaced with this The lead to nitrate gets replaced with this We didn't do anything with a lead to chloride because again, it's solid and we don't do anything with the solids and The sodium nitrate gets replaced with that So that is the replacement part. We have gone through we dissociated all the aqueous solutions Then we go back in and we replace all those aqueous solutions with their component ions We dissociated them for a reason next step Cross out the spectator ion. The spectator ion is one that you find on both sides of the arrow So two NA on this side two NA on that side that is a spectator ion We cross it out to cl minus one over here. No cl minus one over there We leave that one PB plus two no PB plus two leave that one to n o three negative one to n o three negative one It's on both sides cross it out cross out the spectator ions Now we rewrite the equation with whatever's left that's left to cl minus one That's left PB plus two and that's left PBCL two What we've just done is we have distilled this entire Chemical equation down until what's really important What's really happening here is that the chloride ions and the lead ions are getting together to make lead to Chloride that is a net ionic equation Let's do to get one more in here without making this video too too long Let's do one We're got to use a solubility table Let's do One off the back of your practice sheet. I don't know what is not going to take us forever. I'm gonna do number 10 zirconium for hydroxide which it tells us is a solid Hydrogen nitrate what you're gonna learn to call nitric acid in the next unit That's aqueous zirconium for nitrate and it doesn't tell us to say to that and what that's the equation That is what we're gonna do. All right, this equation needs to be balanced So we need to take care of that We'll do the long way this time one zirconium on this side Hydroxides over here, but it's not appearing to be over here Really is but it doesn't look like it is so we're gonna break it down four oxygens For hydrogens, what's on the inside times? What's on the outside one times four one times four and Then we have another hydrogen there so we can change that to five and Then nitrate and a three and one of those Same list on the side ones are conium one oxygen Two hydrogens For nitrates, I got a balance it Really doesn't matter where we start So we have four hydrogens over here. We want to get four hydrogens over there We can stick a four out in front of the water to do that Four times two gives us eight hydrogens four times one gives us four high oxygens hooray now We can either fix the hydrogens or we can fix the nitrates That doesn't really matter. Let's go ahead and fix the nitrates first because it's gonna be easier just looks easier, right? Five and eight one and four. Let's fix the nitrates next We'll stick a four there so follow me four times one gives us four hydrogens here four times one gives us four hydrogens there for eight happy accident as good old Good old. What's his name the painter? Oh? Ross would say Happy accident. We fixed that and then four times one gives us four nitrates now that is a balanced equation Bob Ross that's his name. All right, so we got a balanced equation Congratulations next thing I do is figure out the states of these well water is easy. We know that's a liquid good enough This however zirconium nitrate is going to be trickier and our Solubility table isn't really going to address this one So we need a basic set of solubility rules to look at I didn't give you one of those so what you can do is you can get on the old internet and search solubility rules and Take a look at them. So that's what I'm doing right now I'm just pulling up some solubility rules and I need to take a look and see that I can find one with zirconium in it and As I look over these solubility rules, I'm gonna look at nitrates and oh three and see what it says about nitrates in here And I'll see anything about zirconium Specifically would be an odd one to see anyways in the set of solubility rules This is salt baits this is hydroxides Salt's containing nitrate ions and oh three negative one are generally soluble So this is gonna be an aq Most likely and again, that's just what I did I just went on the internet search solubility rules and I looked it up and I found nitrates in it and oh three is generally So that's how that's done. Hopefully the rest of them are on there Or I just give you a really hard assignment to do on your senior trip, which would be fun. Anyways, so hey Enjoy that the rest of you. You'll be practicing in class with me. So be easier now balanced equation we have to go through and break down the aq's so we have that as an aq for h and oh three and We have that as an aq Z are and Oh three for This one breaks down there and gives me four h pluses and four and oh three minus ones and Then this breaks down right after the zirconium giving a z r with some positive charge and Again, the subscript belongs to the nitrate. So it's four and oh threes and each was negative one Four times negative one means that's negative four This one has to be plus four to compensate Which is why I was calling it zirconium for nitrate now We got to get rid of the aq's and replace them with what we just did that was a solid so that just stays the way it was The hydrogen nitrate or as I said, you're gonna call nitric acid before too long Is this that one we broke down into that and that's just liquid water that doesn't break down Whoo fun stuff Look for spectator ions that one's not on the other side h pluses. I don't have h plus. So that's important four and oh three four and oh three Boom and boom. That's the only things we get the cancel out. So that ionic equation rewrite it with what's left z are oh h four plus four h plus ones produces z are plus fours and Water again distilling it down into what really matters Yeah, that's all I gotta say is yeah