The accompanying photo shows the reaction between a solution of Cd(NO₃)₂ and one of Na₂S. (b) What ions remain in solution?

Video transcript

Hey everyone in this example, we need to complete and balance the given molecular equation and provide a balanced net ionic equation including faces. So our first step is to write out the ions present. What we have is the PB two plus cat eye on, we have the C two, H 302 minus an eye on so that acetate an ion. And we have specifically two moles of this acetate, an ion based on the subscript given in the compound name. Next for ammonium bromide, we have the NH four plus catalon and the B R minus an ion. And so now in order to form our product, we're going to pair up the opposite charges. So we pair up the acetate and the ammonium and then we would pair up the lead two plus and the B R minus an ion. And what we would form as our first product, which is going to be lead bromide, So PB BR two. And based on our sustainability rules for bromides, this is going to be an insoluble solid precipitate because it's bonded to lead two plus, which is one of the exceptions for Celje bility for bromides. Our second product is going to be the formation of ammonium acetate, so NH four C two H 302. And this is going to have an AQ label because all acetate are soluble according to our sustainability rules. Our next step is to make sure this is balanced and in order to balance this out, we would place a coefficient of two in front of our ammonium acetate as well as a coefficient of two in front of our ammonium bromide. And so now everything is completely balanced and this is for our molecular equation. So our next step before we can get to the net ionic equation is to write out the complete ionic equation which includes spectator ions. And so what we would have is all of the ions present. So we would write out PB two plus on our reactant side, which gets the label because recall that all ions are soluble plus two moles of our acetate an ion. So two molds of the c two H 302 minus an ion added to two moles of our ammonium cat ion as well as two moles of our b r minus and ion. And sorry, these both get the label. So this completes our ions on our reaction side. And now for our product side we would keep our solid compound as is, we can't break that up into ions because it's not soluble. And so this would remain as lead bromide, solid. And next we would have our two moles of our NH four plus cat ion And then two moles of our acetate an ion. And so now we want to recall that when we have the complete ionic equation, we want to identify everything that is matching as far as the ions on both sides of our equation. So recognize that we have two moles of our acetate, an ion on the product side as well as our reactive side. So we're going to go ahead and cancel these out because these are spectator ions. And next we also have two moles of ammonium cat eye on on the product side as well as the reactive side. So we would cancel these out as well. So everything else left behind is what we would have for our net ionic equation. And so carrying everything down, we should have PB two plus with the accu label still plus two moles of R B r minus an ion producing our solid product which is going to be the lead bromide or the lead to bromide solid precipitate. And so this will complete this example as our balanced net ionic equation including our phases. So I hope that everything we went through is clear. If you have any questions, please leave them down below. So everything boxed in yellow first represents our balanced molecular equation as well as our balanced net ionic equation for our final answers. If you have any questions, I'll just leave them down below and I will see everyone in the next practice video