hello and welcome to chapter 4 where we will talk about aqueous reactions and solutions stoichiometry these are the learning objectives that you need to be familiar with for your exam so those exchange reactions that's that precipitation reaction that we're talking about this is a little picture little gift of a double displacement reaction so you can see how one thing just they literally change dancing partners we're going to talk a lot about exchange reactions in this chapter so an exchange reaction the word just means the metathesis word just means to transpose so we're changing dancing partners and the steps to complete and balance these reactions the first step is going to be determine which ions are present this makes it so much easier students usually don't want to do this because it's a little bit of extra work however when they do it they're like oh my gosh this makes it so much easier so I'm just giving you that word of advice go ahead and find out which ions are present it will make your life easier the second step is to write the chemical formulas for the products and this is going to be based off of that exchange the next step is to check solubility and that's the table that we just worked with again please print off the solubility rules that are given to you on the exam and use those to practice and then the last step is going to be balance the equation so here is a basic [Music] formula that you will see as an example of what an exchange reaction is and you've seen this before in a previous chapter so we have our ax plus B Y yields a Y so that a is kicking B out of its spot and combining with Y and then you'll have B combining with X so an example of that would be the silver nitrate agno3 and the KCl so the AG since it's a Cateye it's positive it's going to combine with the CL the positive is and the negatives go together you will not have the AG going with the K that breaks the rules that's not what we're doing here in Kim one so you'll have AG going with Co and then the K plus or the K will go with the no.3 they are going to balance accordingly on the other side and we'll go through a couple of examples of this so let's practice exchange reactions so we're going to identify the precipitate if there is one that forms and the following solutions are mixed and we're going to write the balanced equation for each reaction so those four steps that I just gave you this is what we're going to do for each one if you can find the pattern it makes it easier to learn so let's do these so we have ni no.3 2 and NaOH our first step is going to be to identify the ions so if we do the reverse of the crossover method you can see that nickel is a 2 plus because there's a 2 outside of the parenthesis on the no.3 and hopefully you have installed all of the polyatomic ions by now and you know that nitrate is a negative 1 then we have our NaOH and you can look at the periodic table to determine the charge of the NA because of where it's in column 1 you know where it is you know that column 1 has a charge of 1 plus so what you're seeing is how chemistry builds on all of the information you've previously learned and then we know because we've memorized all of our polyatomics that hydroxide ions a negative 1 all right it's a step 2 we're going to write the chemical formulas for the products so we know that we have our ni 2 plus the cation from the first molecule is going to combine with the hydroxide the anion the negatively charged part on the other molecule so we are going to have our ni 2 plus and Oh H 1 negative when we combine those we're going to do our crossover method we get ni parentheses o h2 so we're just doing that crossover method again and then the other one that we're going to create the other product we're creating is the cation for the second molecule that's our sodium plus and then the no.3 so we have the NA plus going with the no.3 minus it's a one-to-one relationship so that equals that neutralizes that's nano3 is going to be our product the third step is going to be check solubility so we're going to write our chemical equation out we have what we started with ni no.3 two that's what was given to us plus NaOH and when the question gave us that they said Solutions of when we say solutions of that means they're aqueous solutions so we will put aq as our phase for those two the products and we can also check our solubility rules our products we're gonna check our solubility rules we're going to find our hydroxide on our solubility chart and then we're gonna see if nickel is an exception so this ends up being a solid so please practice this you have the answers right here to check your work and then we have nano3 so all nitrate compounds are aqueous so in a no.3 is an aqueous compound all right so we have steps one through three done now we have step four we're balancing the chemical equation so this should be familiar from Chapter three hopefully you've practiced this because you're going to be doing this a lot so we balance that chemical equation it's going to be a huge skill to make sure you're using the crossover method because if your chemical equations not getting balanced for you and it looks it's just real clunky and it's awkward what you'll want to do is go back and see hey is my molecule balanced did I get the correct ions so these four steps well usually if you're pretty concise and focused on your first on your first steps then that final step four will be so much easier so though our balance equation you can see it right here so we have our nickel nitrate plus two NaOH yields in IOH two and two in a and O three so this is our balanced chemical equation and what I always suggest with these is to go back and practice them just start out with the the two reactants and practice writing these all these steps out and getting a final balanced chemical equation okay next one NaOH and k2so4 first step is going to be identify the ions so we have an na plus and os- a k plus and an so4 2 minus we are going to write our chemical formulas for the products so our na is going to go with our s o4 and you can see the charges there so we end up with in a 2 s o4 because that charge on that the sulfate and then we have a k+ and o h- giving us the product of Koh step three is check solubility we're going to use that table and we get and two aqueous compounds as products and that's absolutely fine remember we said if there is a precipitate great if not we just write our pride our complete chemical formula with our solubility and we'll still be able to balance that chemical equation so we get a 2 in front of the NaOH and a 2 in front of the Koh we have na 2 s and copper 2 acetate step one identify the ion so we have na plus s 2 minus again we can find that s on the periodic table and see that it's a two negative charge because of where it is on the periodic table or we can do the reverse of the crossover method here and see that that two subscript on the na tells us that charge is a to negative the copper is a 2 plus and then the acetate is a negative one and we know that from our knowing our polyatomic ions so we're going to write our chemical formulas for the products we have our copper two plus going with the sulfur it's a two to two so that two plus and a two negative equals zero gives us a neutral compound it's just going to be C us remember ionic compounds are always lowest common multiple then we have an na plus and a acetate the C two H three O 2 is a negative one it's going to be a one-to-one relationship so those are going to be our two products step three is going to be check solubility so when we look at our solubility we get an aqueous compound for our sodium acetate all acetate compounds are aqueous and then our copper sulfide is going to be a solid so double check that don't take my word for it practice using this you're going to need to identify solubility on your exam so please practice this and then finally step four we're balancing the chemical equation so we write our chemical equation out and we see that we need two of the sodium acetate on the product side and that balances out the equation please feel free to work these again I would highly suggest taking these same exact examples that I'm given giving and to write down those reactants figure out the ions figure out the products checklist solubility and balance the equation so we have sodium carbonate and magnesium sulfate when we identify the ions we see we have a sodium plus and a co 3 2 minus magnesium 2 plus and a sulfate 2 - we write our chemical formulas for the products and we see we have magnesium combining with carbonate each one they have it one has a two plus one has a two negative so that's gonna just be a one-to-one relationship mg Co 3 and then our sodium is a positive one so4 is a two negative so it'll give us na2so4 we'll check our solubility and we see that our magnesium carbonate is going to be a solid and our sodium sulfate is aqueous and then finally we're going to balance our chemical equation so it's actually already balanced so there was nothing for us to do we just always need to double check that and make sure that that equation is balanced we have PB no3 2 and na 2's when we identify our ions we're going to have PB 2 plus and since PB is one of those transition metals the only way that we're going to be able to identify the charge on that is by doing the reverse of the crossover method so we see the two outside of the no.3 and we know that that makes that a PB 2 plus when we write our chemical formulas for our products you'll see that you'll have PB s so the 2 plus and the 2 negative on the sulfur those will cancel out that equals 0 remember the whole point is to get a neutral compound so we have PB s as one of our products and then nano3 as our second product when we check the solubility we see that our PB star lead to sulfide is a solid and our sodium nitrate is an aqueous compound all nitrate compounds are aqueous so that's there and then when we balance our chemical equation we have a 2 in front of the sodium nitrate and that balances that whole chemical equation for us we have an ammonium phosphate and a calcium chloride when we identify the ions we notice we have an ammonium ion that's a positive charge a phosphate ion that's a three negative charge calcium is a 2 plus and chloride is a negative one when we combine the these ions to form the products we get our NH 4 goes with their CL and so we have a positive one and a negative one nh4cl is that product the next one is calcium 2 plus is going to combine with a phosphate 3 minus so we end up with ca3 po4 2 then we'll check the solubility and we see that our ammonium chloride is aqueous all ammonium compounds are aqueous and then we'll have a calcium phosphate and that's going to be our solid or our precipitate and then when we balanced our chemical equation we get a 2 in front of the ammonium phosphate a 3 in front of the cacl2 a 6 in front of the ammonium chloride and just a 1 it's understood to be 1 there in front of the calcium phosphate so that's our balanced chemical equation this skill is integral to continue so if you're not sure if you understand it or you just kind of been you know working along with me you haven't really been practicing these I would suggest you stop here go back and practice these because what we're going to do is we're going to take this molecular formula the things that we've been creating so far and we are going to level up we are going to apply the learning that we've just received onto something a little more expanded we're going to use that knowledge so go ahead and feel free to pause the video where it is and go back and practice every one of those and you have all of the answers so you can go back start with those reactants break them down into ions create the products check the solubility and then balance and this needs to be pretty second nature and what we know is practice makes permanent and just be willing to make those mistakes now so that when you get to the exam time it's going to be super easy for you ok so if you're ready let's continue on so we're still doing metathesis or exchange reactions so the first thing what we've just learned and we've spent a good amount of time on hopefully is the Molecular equation the level up the next step is going to be doing the complete ionic equation and a net ionic equation if you can do the Molecular equation easily these next two steps will be so much easier for you so we have our molecular equation and that tells us the reactants in the products in the molecular form so molecular just means the two ions are combined to be neutral right and this is what we've just learned to do and this equation is one of the ones that we've already done so this is an example of a molecular equation our complete ionic equation this is that next step so we take our molecular equation and then we're able to break down that molecular equation into what we call a complete ionic equation that means all the ions are present and the way that we get from the molecular to the complete ionic is we're dissociating all of our strong electrolytes our strong acids our strong bases and the soluble ionic salts we're breaking those down into ions so all of our aqueous compounds are going to be broken down into ions and I'm going to show you an example of this and what this does for us as chemists is it actually accurately reflects the species that are found in the reaction mixture this tells us what's actually going on with it within that reaction so what does a complete ionic equation look like well we started with our Molecular equation here we have sodium carbonate plus magnesium sulfate and these are aqueous and we have magnesium carbonate and sodium sulfate here now we move on to our next step our net ionic equation this is where we cross out anything that does not change from the left side of the equation to the right side of the equation the only things that are going to be left are going to be those things that change ie react during the course of the reaction so let me show you what that looks like so this is the same equation that we just did we have our molecular equation complete ionic that I just went through and explained now we're going to talk about our how we get to our net ionic equation so what we're going to look at is we're going to look at the overall and we're going to look for our ions so I just start at the beginning I have to na plus aqueous what I'm gonna do is I'm gonna look at the product side and I'm gonna find two in a plus aqueous if I find it I'm going to notice it and then I'm going to cross it out if they're exactly the same on both sides of the reaction that means they didn't really do anything and I have a term for that in a minute so I'm just gonna cross them out just trust me for a minute then I go to the Co 3 - - and I noticed that ok Co 3 2 - aqueous on the reactant side but on the product side that's not the same it's mg Co 3 solid so I keep it that means something happened with that one so I don't need to cross it out I'm gonna keep it so I put a box around it the next one is the mg two-plus aqueous again I go to the right side of the and the reactants or the products and I look for mg 2 plus aqueous okay the mg is in a completely different form on the right of the product side so I don't do anything with it I put a box around it and I keep it then I go to the sulfate so4 2 minus aqueous I look to the product side and what I see is the exact same polyatomic this exact same molecule so4 2 minus aqueous so since I find it the same exact way I cross it out okay so now we're ready for our net ionic equation all of the things that I crossed out go away and we're going to talk about what those are what happens - and all of that the things that I put a box around those I'm just copying down to the next level these are the things that are actually reacting to create something so I would just copy down Co 3 2 - aqueous exactly the way it shows up mg 2 plus aqueous + mg Co 3 solid so those things that we crossed out what those are called our spectator ions so I always think of maybe this soccer game or a football game or a baseball game you go in you pay for your ticket and you sit in the stands well did you affect the game at all no not really you didn't get in there and throw the ball or kick the ball or anything like that you're a spectator in that sport so you're you know you you feel important in everything however you didn't affect the game itself so what you are as a spectator and it's the same way with these ions these things these spectators ions do not change in the chemical reaction they're still important we still need to mention them in the molecular and complete ionic equation because they're you know they're paying their their ticket and they're you know supporting the players however there's still spectator ions they didn't affect the game itself the actual what happened in the game and so when we get to the net ionic equation those are going to be deleted and so if we're back to our example this is the same exact thing we just went through we have our Molecular equation our complete ionic equation with our spectator ions crossed out and then our net ionic equation written and now I've added the last part is the spectator ions themselves I don't care how many spectator ions I have I just want to know their identity so I want to know that my spectator ion was an na plus and I want to know that it was an so4 to - I don't care that there's two na plus I just want to know the identity of the spectator ion so I don't need to carry the coefficient with that so we have steps to writing net ionic equations guess what the first step is it's to write the balanced Molecular equation if you know how to do the balanced Molecular equation everything else and you follow the four steps of that everything else becomes so much easier the second step is to dissociate all so electrolytes and aqueous compounds you will find that all strong electrolytes are aqueous we're going to cross out anything that remains unchanged from the left side to the right side of the equation and then step four our final step is write the net ionic equation with the species that remain so let's take a look at these from the previous written balanced Molecular equations we've already done that work so hopefully it'll make it a little easier right complete ionic and net ionic equations for the reactions and that that occur in each of the following cases identify the spectator ion or ions in each reaction so I have the balanced chemical equations here so let's work through these so my Molecular equation hopefully in your notes you have identified the ions because that will make this a lot easier so I have the nickel nitrate plus the two NaOH yields ni o h2 plus two in a and O three I look at my complete ionic equation I break down anything that's aqueous so I have ni 2 plus plus 2 no.3 negative ok so I told you when you have a subscript on a polyatomic that you don't put that as a coefficient unless you have two of those polyatomics so if I look at ni no.3 parenthesis 2 that 2 does come out as a coefficient because it's not part of the polyatomic it's telling me that I have two of those to balance that molecule out hopefully that makes sense the three stays exactly where it is because that is part of the identity of the polyatomic and then I have to na plus aqueous remember the face does not change plus two O H so that two that's in front of the NaOH it applies to the sodium and to the hydroxide so that two coefficient will be for both of those ions and then on the product side I have ni o h2 solid so that solid does not break down into ions so we keep that we just write that down and then we have two in a plus and two no.3 again that coefficient means that I have two molecules of it so I have two of those ions now we look at our what's the same on both sides and what's different so we have our ni 2 plus and I just start from the from the left and go to the right I've ni 2 plus aqueous if I look on the product side I do not have ni 2 plus aqueous I have ni o H to solid so that stays the same so I just put a box around that I find my no.3 negative aqueous same phase then that means that and it's a 2 in front it needs to be the same coefficient everything if it's not the same coefficient you probably made a mistake in the molecular equation so we see that the 2 no.3 negative aqueous is both is the same on both sides so we are able to cross that one out then we go to na and we find to na plus aqueous on both sides and since there are the exactly the same we can cross those out and then we find our 2 O H negative aqueous it does not have a twin over on the product side so we keep it we just put a box around it or underline it or however you want to do that so when we write our net ionic equation we're going to write the things that are boxed ni 2 plus aqueous plus 2 O H negative aqueous yields ni o H 2 solid and the spectator ions are na plus and no.3 minus again I don't care about the coefficients I just want the identity of the spectator ion alright let's do this one so here's our molecular equation I'm going to break down everything that's aqueous so I end up with 2 na plus plus s2 minus keeps the same phase plus Cu 2 plus plus my acetate so I need to make sure to be able to identify polyatomic ions the two that's outside of the parenthesis on that acetate is going to go as a coefficient in front of the acetate ion and the phase stays the same same thing on the product side I have to na plus plus two of the acetate C 2 H 3 o 2 minus aqueous however the C ooh the copper to sulfide is going to stay the same because it's a solid is not going to break down into ions then I go through and I look at what has changed and what stayed the same so I have to na plus aqueous I find that on the product side exactly the same I Circle it and then I cross it out my sulfa s to minus aqueous it's not the same on both sides so I just draw a box around it or you can underline it or highlight it however you'd like to do that the copper two-plus again it did change there's nothing that there there's not a twin to it on the product side however the acetate is exactly the same on both sides so I Circle it and then I cross it out so now I'm ready to write my net ionic equation so I'm just going to copy it down what's in the boxes I wrote Cu two plus first plus s two minus aqueous it's fine as long as you have both species in your reaction it's absolutely fine whatever order that you write it in and then my product a see us solid my spectator ions na+ and my acetate C two H three O 2 minus R are the identities of my spectator ions hopefully this is getting a little more easier for you because you're practicing it over and over again okay last one we have lead nitrate sodium sulfide lead sulfide and sodium nitrate we break those aqueous compounds down and you can see we have a PB two plus two no.3 minus 2 na plus aqueous plus an s2 minus aqueous yields there's my PB s that's a solid so I don't break that apart plus my two in a positive aqueous plus two no.3 minus aqueous I look at side to side and I see that PB 2 plus is not the exact same on the product side so I just put a box around that one and then I look at my next one over is 2 no.3 negative aqueous I have an identical one on the product side so I circled that one and I'm able to cross that one out I do the same thing with na they are identical so I just can cross those out my sulphide it is not the same on both sides so I just draw a box around that then I go to my net ionic equation and I just copy down what's in the boxes PB two plus aqueous plus s2 minus aqueous gives me PB s solid and I identify my spectator ion so I have a sodium and a nitrate as spectator ions for this reaction so this is the end of this section thank you for listening