In this module we're going to look at molecular, ionic, and net ionic equations. Our objective in this unit is to get to the net ionic equation so that we can see exactly the change that's happening in a reaction between two ionic compounds. So most of the equations that we've looked at up until now are were actually molecular equations. Meaning they showed us everything that's actually involved in the reaction and showed them as compounds. Sometimes, it's more useful to understand what the actual chemical changes that are happening, and as a result it's better to have a net ionic equation when you're dealing with the reaction between two ionic compounds. Before we can get to the net ionic equation, we first need to look at the complete ionic equation. And from that we can then get to the net ionic equation. So we're going to look at the process of how we go from one to two to three. From molecular to the complete ionic to the net ionic. So when we look at a molecular equation what we see is that the formulas and the compounds are written as though all species existed as molecules or whole units. Remember that when we have ionic compounds in solution, if they are aqueous, meaning they are soluble in water, those ionic compounds are going to dissociate into their ions. So if I look at my example here, I really don't have ammonium sulfide and copper nitrate in solution. What I have are ammonium ions, sulfide ions, copper ions, and nitrate ions. But this is the molecular equation that shows these as molecules. It's not necessarily showing us the chemical change that's happening. Then we can go do a complete ionic equation. To do this, we simply show anything that's dissolved. So that anything that's labeled as aqueous in the ionic form. Because that's how it actually exists in water. And so here we have our ammonium sulfide, has been broken down into ammonium ions and sulfide ions. Copper nitrate becomes copper ions and nitrate ions. Our ammonium nitrate is also broken up into ions, but notice that our copper sulfide remains as CuS because it's solid, it's insoluble in water. And so that's the precipitate that forms from this reaction. So now that we have our complete ionic equation, now what we're going to do is look for those ions that are actually not involved in the reaction. Basically looking for things that do not change from the reactant side to the product side. And when we say something doesn't change we have to look both at the formula, in this case our ammonium ion, and the phase is aqueous. I notice that on the other side of the equation of ammonium ion, aqueous, I notice these are exactly the same. So I know that NH4 plus is a spectator ion because it doesn't change as I go from left to right. When I look at sulfide I see S2 minus aqueous on the left side, but on the right sulfur is now in a compound. So I know that's not going to be an aqueous ion because I go from the aqueous ion of sulfide to having sulfur within a solid compound. Same thing for copper. It is not a spectator ion because I go from copper 2 plus aqueous to copper sulfide in the solid state. And then I get to nitrate, NO3 minus in the aqueous phase, and I see here that in the products, I also have nitrate with a minus 1 charge in the aqueous phase. So this will be a spectator ion. So in this case I have two spectator ions, NH4 plus and NO3 minus. Neither of these have any changes as I go from the left to the right side, therefore they're considered spectators. Now, what I can do when I write my net ionic equation is I basically eliminate those spectator ions. I'm showing only those species that are actually involved in the reaction. And so what I'm left with is sulfide and copper ion reacting to form copper sulfide. So when we need to write a net ionic equation first we want to write a balanced molecular equation for the reaction. It does need to be balanced. We need to make sure we're balanced at each step along the way. We're going to rewrite the equation to show dissociated ions in solution. So anything that's labeled aqueous will be broken up into its ions. And then we need to identify and cancel out spectator ions, so those things that do not change from the left to the right. So let's look at an example. When aqueous solutions of copper(II) nitrate and potassium carbonate are mixed, a precipitate forms. Write the net ionic equation for this reaction. Our correct answer is number two. Notice that when we look at our options, we have copper(II) nitrate and potassium carbonate. So what we have present in solution are Cu2 plus and O3 minus, K plus, and CO3 2 minus. We know that copper nitrate is soluble, because it was an aqueous solution, we were given that information in the problem, as was potassium carbonate. Therefore, we know that neither one nor four can be our answers, simply because we're looking at the same product, and already know that those two substances are soluble. Now, when I look at two and three, the remaining two answers, I only have to worry about these two options. So, I can look at copper carbonate and potassium nitrate, because those are the only two possible compounds that conform because copper was already paired with nitrate. The only other thing it can be paired with is carbonate, because I can't form a compound between copper and potassium or between nitrate and carbonate, because I can't have two cations or two anions. So when I look at my options of copper carbonate or a potassium nitrate, what I see is that potassium nitrate is soluble, it would not be solid, it would be aqueous. Therefore it's not going to form a precipitate and it's not going to have any change between the left and the right side of the reaction. And so what I would find is that my K plus and my NO3 minus are actually my spectator ions. And the only possible product I have here is the copper carbonate. And I see I go from copper to plus, carbonate to minus, to copper carbonate. In the next module we're going to look at acid-based reactions.
