0:00:00.060,0:00:00.893
- [Instructor] Check this out.

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I have two clear colorless[br]solutions over here.

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Let's pour them into each other.

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We pour the first one and[br]we pour the second one.

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And boom, we now get a[br]white colored solution.

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Here's another example.

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Again, two colorless solutions.

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We pour one into another, and boom!

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We again get a beautiful[br]yellow colored solution.

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What's going on over here?

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To find out, let's dig a little deeper.

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Here's a more curious[br]question for us, okay?

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so in the first case, what did we do?

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We poured sodium chloride[br]and silver nitrate together,

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and that gave us a white[br]colored solution, right?

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But now if I were to[br]change just one element,

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instead of silver, if I[br]had potassium over here,

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everything else is the same,

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so if I had poured potassium nitrate

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and sodium chloride into each other,

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I wouldn't have gotten anything.

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I would've just gotten[br]a colorless solution.

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It's not that interesting.

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And therefore, I couldn't[br]find any footage online.

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So this is just an edited[br]image, but you get the point.

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We wouldn't get anything[br]interesting over here.

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But the big question is why?

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Why silver nitrate and sodium chloride

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gives us a white colored solution,

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whereas potassium[br]nitrate, just one change,

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and sodium chloride does[br]not give us anything?

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Let's look at it. Let's[br]look at it one by one.

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So in the first case,

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we are reacting sodium[br]chloride aqueous solution

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with silver nitrate aqueous solution.

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What do we get?

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Well, remember that in aqueous solutions,

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ions usually dissociate.

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So over here, we'll get basically NA+ ions

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and CL- ions.

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And over here, we have[br]Ag+ ions and NO3- ions.

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So when I pour them together,

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we just get all those ions together, okay?

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Now, because these are together,

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they can form new combination.

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Of course, cations should always[br]combine with anions, okay?

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So NA can now combine with NO3,

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but there's nothing special because again,

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they will dissociate.

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But Ag can also combine with Cl.

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When Ag combines with Cl,[br]something interesting happens.

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What? Well, guess what?

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AgCl is insoluble,

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and therefore, it will precipitate out.

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And that's the reason why[br]this whole thing looks white

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because of the AgCl precipitation.

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So what do we end up with?

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We'll end up with AgCl,[br]which is insoluble.

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So that's why it's written[br]as solid over here.

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It precipitates out.

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So let me just share it over here

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to show that it's precipitating, okay?

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And what remains in the aqueous solution?

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Well, sodium ions and nitrate ions,

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so we get sodium nitrate aqueous solution.

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The white color is due to[br]the AgCl precipitating out.

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Now, if you zoom out and[br]look at the reaction,

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see what has happened,

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sodium and silver cations[br]have switched places.

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Sodium has replaced silver over here

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to get sodium nitrate,

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and silver has replaced sodium over here

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to give us silver chloride.

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So since there are two[br]cations replacing each other,

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there's a double replacement happening.

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This is called, no surprise,[br]a double replacement reaction.

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We can also call this double[br]displacement reaction.

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So what we witnessed was a[br]double replacement reaction

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and one of the products precipitated

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giving us that white color.

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Now, before we look at the other one,

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a quick question for you is,

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can you identify which of the[br]elements underwent oxidation

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and which ones underwent reduction?

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Pause and think about this.

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Okay, whenever I want to think about that,

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I just look at the[br]charges on the elements.

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Well, over here, sodium[br]has a positive charge.

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It's a cation.

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On the other side, well,[br]it's still a positive cation.

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So no change happened to[br]the charge on the sodium.

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Nothing happened to it, okay?

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What about Ag?

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No change.

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The same is the case[br]with the anions as well.

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No change, no change,

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which means look, nothing[br]is undergoing an oxidation,

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nothing is undergoing a reduction,

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so double replacement reactions[br]are not redox reactions.

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And you may be wondering,

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why are you excited about the fact

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that it's not a redox reaction?

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I'm excited because I used to think

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that all chemical reactions

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must involve electron[br]transfers, and therefore,

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all chemical reactions should[br]have something oxidizing

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and something else reducing.

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But I was wrong. Look,[br]right in front of our eyes.

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We can see examples of chemical reactions

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where there are no electron transfers,

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where there is no oxidation or reduction,

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so that's pretty cool.

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But anyways, now let's[br]look at the other one.

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What happens when I[br]pour these two together?

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Well, let's look at the reactants.

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This time, the reactants[br]are NaCl and KNO3.

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Both are aqueous solutions.

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I pour them together.

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So just like before, I will now have

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all the four different[br]kinds of ions over here.

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Na can combine with NO3.

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Remember, cations can only[br]combine with anions, okay?

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Those are the only new[br]combinations you can form.

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So Na can combine with NO3-,

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but again, it'll dissociate.

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K+ can also combine with Cl-.

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But what's important over here

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is that KCl, potassium[br]chloride, is soluble.

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Therefore, when K and Cl combine,

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again, they will dissociate.

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So nothing happens over here.[br]There's no precipitation.

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I'll just end up with a solution

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where all the four different kinds of ions

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are just floating around together.

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So no chemical changes happened.

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And that's the reason why I[br]don't get any colorations.

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I don't get anything over here.

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So over here, I get[br]essentially no reaction.

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So you notice the key difference?

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The key difference was AgCl was insoluble.

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That's why it precipitated out.

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And that's why in order for us

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to get a double replacement reaction,

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we need one of the[br]products to precipitate.

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If both are soluble and[br]they form aqueous solution,

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then nothing will happen.

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We'll just get a solution

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with all the four different kinds of ions.

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No chemical change at all.

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So in general, we can now write down

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what a displacement reaction looks like.

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We can say that if you have[br]an aqueous solution of AB

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reacting with an aqueous solution of CD,

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then a double replacement reaction,

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the two cations replace each other.

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So A will now combine with[br]D and C will combine with B.

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But that'll only be the case[br]if one of them is insoluble

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and precipitates out.

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Precipitation, sorry, is the key

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to having double replacement reaction.

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So look, if you pour any[br]two aqueous ionic solutions,

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do not expect to get a[br]double replacement reaction.

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You'll only get them if one[br]of the products is insoluble.

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But now we'll be wondering,

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how do we know whether a particular salt

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is soluble or insoluble?

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I'm glad you asked that question

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because that brings us[br]to the solubility chart.

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A solubility chart is basically that,

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it tells us whether a[br]salt is soluble or not.

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So here's how we can read it.

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If you wanna look at potassium chloride,

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here's potassium cation,

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here's the chloride anion, sorry.

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And now we can just say,[br]hey, this is where they meet

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and so this is the solubility[br]of potassium chloride

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and you can see it is soluble.

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But what about silver chloride?

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Silver is here.

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Chloride is here.

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Again, try to make them[br]meet. And what do you notice?

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Silver chloride is insoluble.

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And what about this[br]yellow slightly soluble?

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Well, don't worry too much about that.

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We'll only work with the soluble

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and the insoluble ones, okay?

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And just by looking at this chart,

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you can see some trends.

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For example, you can see salts of lithium,

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sodium, potassium, and even ammonium.

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Almost all are soluble.

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Of course, there are some exceptions,

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but they're all soluble.

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In contrast, salts of[br]lead are almost insoluble.

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You can also see salts[br]which have nitrate ions

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and acetate ions, pretty much soluble.

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Anyways, now equipped with[br]this solubility chart,

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we can predict whether certain[br]double replacement reactions

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are gonna happen or not, okay?

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So let's check that. Here's the first one.

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We're gonna pour lead two[br]nitrate aqueous solution

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and potassium iodide[br]aqueous solution together.

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What will we get?

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Pause the video and try[br]to do this yourself.

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First, think about what[br]the potential products are

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by swapping the cations

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and then check whether[br]one of them is insoluble.

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If it is, then it'll precipitate it out.

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We'll get the reaction.

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If both are soluble, we'll get nothing.

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So pause and try.

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All right, here it goes.

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So one of the potential products is

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lead cation combines with iodide ion.

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So before writing, let[br]me just check over here.

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Where is lead?

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Lead is over here

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and iodide lead cation, okay.

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Iodide is over here.

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So if you look at that,[br]there you go. It's insoluble.

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So I know immediately,

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lead

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iodide,

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and I need to be careful,

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lead has a +2 charge and iodine[br]over here has a -1 charge.

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So to compensate, I have[br]to put two over here.

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So I get lead two iodide.[br]That is insoluble.

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So that will precipitate out.

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And what else will I get?

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Well, potassium can combine with nitrate.

0:08:49.950,0:08:52.050
And again, we can check[br]for it. Where is potassium?

0:08:52.050,0:08:54.060
Potassium is here.

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Nitrate is over here.

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So if I go down, go over[br]here, look, it's soluble.

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So I'll get potassium[br]nitrate, which is soluble,

0:09:02.880,0:09:06.780
charge is +1, -1, okay,[br]so I'll just get this.

0:09:06.780,0:09:08.790
So I'll get an aqueous solution.

0:09:08.790,0:09:10.500
And, of course, I'll[br]have to balance it out.

0:09:10.500,0:09:11.340
Let's quickly do that.

0:09:11.340,0:09:13.050
So I have two iodine over here,

0:09:13.050,0:09:14.850
so I'll put a two here.

0:09:14.850,0:09:17.550
So two potassium, so I'll put a two here.

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And that balances everything out.

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And this is the experiment[br]that we saw earlier.

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We are pouring potassium[br]iodide into lead two nitrate.

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What is that yellow color?

0:09:28.440,0:09:32.370
That's basically the lead two[br]iodide being precipitated.

0:09:32.370,0:09:33.780
And now the aqueous solution

0:09:33.780,0:09:36.840
contains potassium and nitrate ions.

0:09:36.840,0:09:38.400
All right, why don't we try another one?

0:09:38.400,0:09:41.400
This one looks a little bit intimidating,

0:09:41.400,0:09:42.600
but the idea is the same.

0:09:42.600,0:09:45.330
So why don't you pause the[br]video and try this again.

0:09:45.330,0:09:47.040
All right, we start by thinking about

0:09:47.040,0:09:48.450
what the potential products are.

0:09:48.450,0:09:51.300
How do we do that? We[br]swap the cations, okay?

0:09:51.300,0:09:56.300
So ammonium cation, let's combine[br]them with the acetate ion.

0:09:56.400,0:09:57.540
Again, before writing it,

0:09:57.540,0:09:58.980
let's just look over here.

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So where is ammonium?

0:10:00.240,0:10:04.140
Here's ammonium, and acetate is over here.

0:10:04.140,0:10:06.810
So let's look at that.[br]Oh yeah, that is soluble.

0:10:06.810,0:10:08.730
So this one is soluble.

0:10:08.730,0:10:13.050
The other one would be[br]sodium and sulfate ions.

0:10:13.050,0:10:17.250
So sodium is here, sulfate is here.

0:10:17.250,0:10:20.073
What do we get? Oh, that's also soluble.

0:10:21.330,0:10:22.680
Nothing is insoluble over here.

0:10:22.680,0:10:25.500
What we'll get is soluble,[br]so nothing precipitates out,

0:10:25.500,0:10:27.210
which means we'll just[br]end up with a solution

0:10:27.210,0:10:30.150
where you have all these[br]four kinds of ions.

0:10:30.150,0:10:33.843
So that means we will get no reaction.

0:10:35.040,0:10:36.090
All right, so the final thing

0:10:36.090,0:10:37.260
is that there's a special kind

0:10:37.260,0:10:39.000
of double replacement reaction,

0:10:39.000,0:10:41.940
which we call acid-base neutralization.

0:10:41.940,0:10:44.010
Now, we'll talk about[br]what acids and bases are

0:10:44.010,0:10:46.020
in detail in future videos,

0:10:46.020,0:10:47.850
we'll look at all the cool[br]properties and everything,

0:10:47.850,0:10:49.860
but for now, think about acid

0:10:49.860,0:10:51.510
as basically an ionic solution,

0:10:51.510,0:10:55.410
which has hydrogen cation[br]and some other anion,

0:10:55.410,0:10:58.080
and base as an ionic solution,

0:10:58.080,0:11:01.590
which contains a hydroxide anion.

0:11:01.590,0:11:04.080
And, of course, some metal cation.

0:11:04.080,0:11:06.240
For example, consider HCl,

0:11:06.240,0:11:09.420
which is an acid because[br]it has a hydrogen cation,

0:11:09.420,0:11:11.040
reacting with sodium hydroxide,

0:11:11.040,0:11:14.160
which is a base because[br]it has a hydroxide anion.

0:11:14.160,0:11:15.273
What will happen?

0:11:16.170,0:11:18.540
Well, we just swap the cations.

0:11:18.540,0:11:21.750
So sodium will combine with chlorine

0:11:21.750,0:11:25.830
to give me sodium chloride,[br]and that is soluble,

0:11:25.830,0:11:27.810
so I'll get an aqueous solution.

0:11:27.810,0:11:29.670
But the interesting part over here is

0:11:29.670,0:11:31.920
what happens when hydrogen[br]combines with OH-?

0:11:34.020,0:11:37.290
What do we get? This is[br]no longer an ionic salt.

0:11:37.290,0:11:40.320
This is H2O. This is water.

0:11:40.320,0:11:42.270
Water is covalently bonded.

0:11:42.270,0:11:46.530
So we now end up with a[br]covalently bonded molecule.

0:11:46.530,0:11:49.500
So we will get water, H2O.

0:11:49.500,0:11:51.510
And since it's no longer[br]an ionic solution,

0:11:51.510,0:11:52.920
we just write as liquid.

0:11:52.920,0:11:54.690
So look what we get in general.

0:11:54.690,0:11:57.090
When you combine acid with a base,

0:11:57.090,0:12:01.260
they neutralize each other[br]to give us a salt and water.

0:12:01.260,0:12:03.210
So this is a special kind of[br]double replacement reaction

0:12:03.210,0:12:05.070
because there are no[br]precipitates over here,

0:12:05.070,0:12:06.060
but the reaction happens

0:12:06.060,0:12:09.693
because we get a covalently[br]bonded liquid water.