When aqueous solutions of copper two nitrate and potassium carbonate are mixed, a precipitate forms. Write the net ionic equation for this reaction. So, the first thing I want to do, is actually write the formulas for my copper two nitrate, and the potassium carbonate. So, I know that I have copper and nitrate. I know that nitrate has a plu, a minus one charge. So, that means that my subscript on copper is going to be one. And that for nitrate, it's going to have a subscript of two, because the charge on copper is two, as given in the name of the compound. So, we have copper nitrate. And we're told that this is an aqueous solution. So, we mark it as aqueous. Plus potassium carbonate. And we get the two as a subscript on carbonate, from the charge on the carbonate ion, which is minus 2. And we know that this is also aqueous. So, now we have our two reactants. Now we have to figure out our products. Now, notice that there's only one pair of products that could form, and potentially form a precipitate. Because copper is already paired with the nitrate, there's no other way that we can combine these two and possibly get a precipitate. So, we have to look at copper being paired with either potassium or carbonate. Well, copper and potassium are both cations, so they are not going to form a compound. So, the only other possible compound we can form is between copper and carbonate. So, we can say CuCO3, and we actually won't have any additional subscripts, because the charge on copper is two, the charge in carbonate is minus 2, so we have a one to one ratio between those two ions. We're going to put the parenthesis in, but we're going to leave them blank for right now. And we're going to look at our other product, which is going to form between the potassium and the nitrate. So, potassium has a plus 1 charge, nitrate has a minus 1 charge. So, we actually won't get any other subscripts in there, because we have a one to one ratio between those. I do see that I'm going to need a two, in front of the potassium nitrate, in order to get a balanced chemical equation. I'm also going to leave the parentheses blank there, until we determine what the phase is of that substance. So, we know we started with two aqueous solutions, now we have to evaluate the solubility of both copper carbonate and potassium nitrate. And what we know about carbonates, is that they're generally insoluble compounds, so we're going to label this with an S, because copper is not one of the exceptions for carbonate. When I look at potassium and nitrate, I know that either of those ions will make something soluble, so I can label this as aqueous, so what I see is that the only precipitate that forms is copper carbonate, and so that's going to be the precipitate that forms. But what I want to actually write is my net ionic equation, so what I have written here is a complete molecular equation. Now I'm going to write my ionic, or my complete ionic equation. And when I do that, what I'm going to do, is write anything that is labelled as aqueous, as an ion. So, I have copper two plus aqueous, plus nitrate aqueous, plus potassium aqueous, plus carbonate aqueous. Because if I were to look at the solution before, I combine these two solutions, what I would see are these ions present in the solution. I would not see units of copper nitrate or potassium carbonate, but I would actually see the ions. So, this is what's present in the solutions, when I combine them, before any reaction occurs. So, now, I have all the ions on the reactant side, now I need to look at the product side. And I'm not actually going to break up the copper carbonate, because remember, it's a solid, it's insoluble in the solution, in the water, so it's not going to be broken into its ions, it's actually going to precipitate out of solution as that solid compound, not as those ions, but I still have my 2k plus, and my 2 nitrates. Because those are an aqueous solution. So, now I have my molecular equation here. I have my complete ionic equation here. Now what I want to do is look for things that are exactly the same on both sides, those spectator ions, and cancel them out. And I see that copper goes from copper two plus aqueous, to copper in the copper carbonate compound. There is a, there is a change there, so that's not a spectator ion. Next I look at nitrate, and I see aqueous nitrate ions, aqueous nitrate ions, so I can mark them out. Likewise with potassium, I can mark those out, and I notice that with my carbonate, it goes from aqueous ions into a solid compound. So, what I'm going to end up with for my net ionic equation, is Cu2 plus aqueous, plus CO3 2 minus aqueous, yields copper carbonate solid. So, what I've done is, I've paired this down to the stuff that is actually changing in this chemical reaction. Because these are in aqueous solution, all of these species in the reactants that are present as ions, as are the potassium nitrate in the products, this, what I've shown down here at the bottom, is the only thing that's actually changing. So, I'm going from copper and carbonate ions in the aqueous phase, to the copper carbonate solid product. So, this bottom reaction is my net ionic equation.
