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Net Ionic Reactions 2,804 views
Let's talk about how to write Net ionic equations. Net ionic equations come from double replacement reactions. So everything in this, everything that is below, is a double replacement reaction. Which means that, two compounds come together and switch partners.
So we have the first one. Notice this little 'aq' here, that means it's an aqueous solution. Well that is definitely something you need to note, so let's keep that in mind.
When we have double replacement reactions, we basically switch partners. So I'm going to make my double rainbow, and that helps me identify what bonds with what. Now don't forget, I'm going to ignore these subscripts because they do not carry over into my products.
So I have sodium, and don't forget sodium is first, because it's a metal. It was first in my reactants, so it must be first in my products. By first I mean the first one in the compound. I'm going to write above the charge and so it's going to bond with sulphate, SO4. I'm going to write above the charge for that.
So this is now when it's a balanced compound; Na2SO4. Then my next one is Ba and CO3. Ba is a +2, CO3 is a -2, so therefore they're actually electronically balanced. So notice I didn't carry up the subscripts, except for this 3 and this 4, because it's part of the polyatomic ion. You'll get used to that as you do more of them.
So let's balance the equation. It looks like it's balanced, great. So now we have to figure out the state of matter for the products. The state of matter for the products, with that look in my solubility chart, Na2SO4 is soluble. So we're going to say aqueous. This is our soluble, which means it dissolves in water. It does not form a precipitate. BaCO3 is a solid. It is insoluble. It does form a precipitate, I'm going to write 'ppt' to show it's a short-hand version of precipitate. The solid within an aqueous solution.
So now if I want to write the complete ionic equation, I'm going to write complete ionic equation. I'm going to break up everything that says aqueous. When you think of something, a solution aqueous and compound in aqueous solution, an ionic compound breaks up into its ions. So I'm going to break up Barium and Sulphate and Sodium; I have two of them so I'm going to write 2Na, plus Carbonate. When it breaks up into its ions, make sure you include the charges. It produces soluble, so it breaks up into its ions; 2Na + SO4 -2 and the solid. Solid stays together. I mean it's boxed it off. So we notice stay it together. Keep it together. So it's BaCO3 and then write the solid there.
We're going to do a net ionic equation. By net I mean, after I take out the final ionic equation, after I take out the spectator ions. So notice, I have 2Na's on this side, 2Na+ and 2Na+ is on that side. So these are what we call spectator ions. I'm going to write this down here, spectator ions. Spectator ions are exactly as what you think. So let's think of the word 'spectator'.
When you are a spectator at a basketball game, you are sitting in the crowd, and you're watching a basketball game happen. Does your presence matter to the outcome of the game? No. It does not have any influence over which team is going to win. You are just merely watching the game take place. Sometimes we like to think it matters but, in reality, our presence does not matter.
So, the same thing with spectator ions. Their presence does not matter in the overall equation. They don't influence what is actually happening, so I'm going to take them out. I'm going to take out this one. Here is one also. We're left with our overall net ionic equation which is Ba+2 + CO3 -2 yields BaCO3, this is a solid. So this is our net ionic equation. Our spectators were sulphate which is SO4 -2 and Na+.
Let's look at another one. We have ammonium phosphate is reacting with solution of Sodium Sulphate. So we can again do a double rainbow and our products are going to be ammonium Sulphate, which is NH4+1SO4 -2, so we need 2 of these. And Na is going to bond with Sulphate so it's Na3. I'll tell you why it's a 3. This is +1, this is -3, so we need 3Na's. I need 3PO4. That was just not natural for me to do that.
So these are with aqueous solutions. Our reactants are always going to be aqueous. That's just something we know. Our products however, we have to figure out again which one is soluble, which is insoluble. So look at my solubility chart and I notice that ammonium sulphate, it is soluble. And sodium phosphate is also soluble, what does that mean?
It means they're both aqueous, so I'm going to put 'aq'. That also means everything breaks up. So my complete ionic equation, I'm going to put CIE, is everything is going to break up. Actually I didn't balance this first. So this is a 3, this is a 2 and this is a 2 this is a 3. So we're now going to balance everything.
So we have 6 ammoniums, this is 6NH4+ + 2PO4-3 + 6Na+. I'm getting this from this 2 goes to this. Then 3SO4, yields 6ammoniums + 3sulphates + 6sodiums + 2phosphate, I think I'm done here. Now I need a cross on my spectators. So I here cross out with this. I have 2 phosphates, I have 6 sodiums and I have 3 sulphates. I have no reaction left. So overall, I'm going to say this is no reaction.
A hint; if you have two soluble compounds, no reaction. There's nothing happened. So we have the solution that end up with everything saving this ionic form. Nothing came out, no solid formation. So this is in overall no reaction. Say no precipitate, write ppt, no reaction.