in this video we're going to focus on precipitation reactions and also writing that ionic equations so let's say if you have the reaction between silver nitrate and calcium chloride now the first thing you need to do is you need to be able to determine the products of this reaction so precipitation reactions are double replacement reactions the first and the last ions will pair up together and the two in the middle will pair up as well silver has a plus one charge and you can tell based on the formula because nitrate has a minus one charge and because it's only one nitrate silver has to have a plus one charge to neutralize the negative one charge of the nitrate ion chloride has a negative one charge all of the halogens like fluoride chloride bromide iodide they all contain negative one charges as an ion because the magnitude of the silver and chloride ions because the magnitude of the charges are the same you could simply write them in a one to one ratio or if you prefer you can use the crisscross method which is going to be a g1 cl1 so you really don't need to write the one so we're just going to write it as agcl now when calcium pairs up with nitrate let's see the formula that's going to be produced as a result typically the positively charged ion is written before the negatively charged ion so using the crisscross method to write the formula it's going to be ca1 and o32 now whenever you have multiple polyatomic ions you need to enclose it inside a parenthesis so the other product is ca no3 2. now the next thing we need to do is balance the equation so feel free to pause the video and balance it and then unpause it to see if you have the right answer notice that we have two nitrates on the right side so we've got to put a 2 in front of agno3 and we have 2 chloride ions on the right i mean on the left so we need to put a 2 in front of agcl notice that the reaction is now balanced our next step is to write the phases of every substance silver nitrate is it in the aqueous phase or in the solid phase nitrates are always soluble so anytime you see no3 it's going to be in the aqueous phase now what about calcium chloride and silver chloride chlorides are generally soluble with the exception of silver mercury and lead so silver chloride is insoluble but calcium chloride is soluble now that we have our complete molecular equation the next thing we need to do is write the total ionic equation so every substance that is in the aqueous phase we need to separate it into ions anything else like the silver chloride which is in a solid phase we're going to leave it the way it is so silver nitrate contains the ag plus ion and the no3 ion and it has two of each so you need to distribute the two calcium chloride has the ca 2 plus ion all of these ions are in the aqueous phase by the way now on the right side we're going to leave the agcl the way it is and this is going to remain in the solid face and for calcium nitrate it's going to be ca2 plus and two no3 ions what are the spectator ions in the total ionic equation these bacteria ions are those ions that look exactly the same on both sides so this includes the nitrate ions and the calcium atoms now what remains is the net ionic equation so notice that we have a coefficient of 2 for what remains if we divide everything by 2 the net ionic equation is going to be a g plus plus cl minus produces agcl so this is the net ionic equation and of course the ions are in the aqueous phase so that's how you can write the net ionic equation from precipitation reaction so let's go ahead and try another example so let's say if we have a solution of lead nitrate and sodium bromide typically the solutions on the left side these substances are dissolved in water so it's usually in the aqueous phase nitrates are always soluble and alkaline metals like lithium sodium potassium they're always soluble so make sure you know your solubility rules so go ahead pause the video predict the products of this reaction balance the equation write the total and ionic equation after that so let's begin by writing the products of the reaction so pb is going to pair up with br now what is the charge on the lead ion notice that it has two nitrates each nitrate ion has a negative one charge which means that pb has to have a plus two charge to neutralize it now bromide is a halide so we know it has a negative one charge so using the crisscross method it's going to be pb1 br2 so now let's pair up the other two ions sodium and nitrate sodium is an alkali metal in group one and alkali metals usually form plus one charges nitrate we know has a minus one charge so because the charges are the same these two ions will combine in a one to one ratio so we don't need to put the one for no3 so we could just write n a o three any time the charges are the same you could just write them together so now our next step is to balance the equation so notice that we have two nitrates on the left side so we got to put a 2 in front of nano 3 we have two bromine atoms on the right so we need to put a 2 in front of nabr and so now the reaction is balanced so now we need to determine the phases of the ions on the right side nitrates are always soluble so this is in the aqueous phase and lead bromide the halites like chloride bromide iodide they're generally soluble except with silver lead and mercury so this is an exception so lead to bromide is insoluble so now at this point we need to separate everything in the aqueous phase into ions so we have the pb2 plus ion two nitrate ions two n a ions don't forget to distribute the two and two bromide ions now on the right side we're going to leave pb br2 the way it is and we're going to separate this compound into two ions so we can see that the spected ions are the sodium ions and the nitrate ions so now what remains is the net ionic equation which is the reaction between the lead 2 ion and the bromide ion and this is going to produce led to bromide so that is it for this video thanks for watching and have a great day