4.04 Precipitation Reactions

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Del curso dictado por University of Kentucky

Chemistry

588 calificaciones

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University of Kentucky

Chemistry

588 calificaciones

This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.

De la lección

Week 4

We will explore how compounds react with one another to form new substances and then write balanced chemical equations to represent what is happening in a reaction. We will explore several different types of reactions including precipitation, acid-base, oxidation-reduction, and combustion reaction.

4.01 Writing Balanced Chemical Equations 12:10

4.01a Balance the following equation 1 2:57

4.01b Balance the following equation 2 3:32

4.02 Aqueous solutions 12:52

4.03 Solubility of ionic compounds 6:23

4.04 Precipitation Reactions6:34

4.05 Molecular, Ionic and Net ionic Equations 6:14

4.05a Net ionic equation 5:46

4.06 Acid-base reactions 15:16

4.07 Oxidation-reduction reactions 16:49

4.07a Oxidation Numbers 2:25

4.07b Oxidized and Reduced species identification 2:37

4.08 Combustion Reactions 5:35

4.08a Balance the following equation 3 4:12

Conoce a los instructores

Dr. Allison Soult
Lecturer
Chemistry
Dr. Kim Woodrum
Sr. Lecturer
Chemistry

In this module, we're going to look at precipitation reactions.

By the end of this module, you should be able to write a precipitation reacting,

given two ionic compounds, be able to identify the product, and

balance the resulting equation.

Now that we've looked at the solubility rules, and

can identify what compounds are soluble and what ones are insoluble.

We can actually look at our first type of reaction,

which is a precipitation reaction.

So a precipitation reaction, is one that results in the formation of a precipitate.

And a precipitate is just a solid that separates out from the solution.

Frequently, when you hear meteorologists on the news,

they talk about precipitation.

They talk about water or snow coming out of the sky.

Well, what we're looking at here, is solid coming out of solution.

And we're going to use our solubility rules that we've covered,

to determine the precipitate, or the potential precipitate, in our reaction.

Let's look at an example, we have two compounds here, ammonium sulfide, and

copper nitrate.

These are both soluble compounds.

Note that we have our ammonium ion, and a nitrate ion, so

we have two soluble compounds.

So, these are listed as being in the aqueous phase,

because they're dissolved in water.

Now what I need to do is figure out what the potential products are.

Remember, that if I have a cation such as ammonium, what I know is that it cannot be

paired with another cation, it can only be paired with an anion.

And I'll see that when it's paired with a sulfide, it's aqueous, it's soluble.

However, I have to look and

see what the other compound is, and in this case, it's going to be, nitrate.

So I could form ammonium nitrate.

Now notice that I did not bring the subscripts with me, because I

know that the charge on ammonium is plus 1, the charge on nitrate is minus 1,

and so I go back to what I learned in dealing with ionic compounds.

And see that the formula for this compound will be NH4NO3.

So this is one of my potential products, because it pairs a cation, and an anion.

Now I can look at my other potential product, and I have one cation and

one anion remaining, so it has to be copper.

It can't be with nitrate, because it came in with nitrate, and

we already know that compound is soluble.

So the only other option is to pair the copper with the sulfide.

And so, what I see, is I get CuS.

Now remember, copper in this case had a plus 2 charge,

sulfur had a minus 2 charge, so when I criss-cross down,

I wind up with Cu2S2, which I can then simplify this to CuS.

Now what I have to look at, is determine whether or

not these two potential products are soluble or not.

When I look at ammonium nitrate, both of these ions fall on my always soluble list.

So, I'm actually going to write this as aqueous,

meaning this compound will dissolve in water.

For the CuS, I know that my sulfides are insoluble, and so I'm going to

write this as S for solid, because this is going to be my precipitate.

When I put ammonium sulfide and

copper nitrate in water, these compounds are going to dissociate completely,

because they're ionic compounds, therefore they're strong electrolytes.

And basicall,y they're going to try and

find another ion, that they can pair up with.

And as a result, precipitate out of solution.

And the only pairing of ions that can do that are the copper and the sulfide.

And so, we do get a precipitate out of this reaction, and it is copper sulfide.

I still need to go back and balance my equation, because I notice that I

have two ammonium ions on the left, and only have one on the right.

So I add a two in front of the NH4NO3, and

that also resolves the issue with the nitrate, that I have two nitrate ions, and

now I have two nitrate ions on the product side.

For copper and for sulfide, I have one copper and

one sulfide on the left, I have one copper and one sulfide on the right.

So I know have a balanced chemical equation,

which shows me the precipitate from my reaction.

Let's look at another example, and see what our potential products are.

The first thing we need to look at are our original compounds.

You have sodium sulfate and magnesium chloride.

These are both aqueous substances.

Our sodium is always soluble, sulfates are soluble,

This is not one of the exceptions, and our chlorides are soluble, and

this is not one of the exceptions.

So, I have two soluble compounds,

which means these are going to completely dissociate in water.

So, I'm not going to have units of sodium sulfate in water,

I'm going to have sodium ions, and sulfate ions.

Likewise, I'm going to have magnesium ions, and chloride ions.

So, now I need to find if there's any other potential pairs of ions that

could form a compound, and if that resulting compound will be soluble or not.

I notice that sodium is my cation here in the sodium sulfate,

and it's paired with sulfate.

We see that, that's an ins, that, that's a soluble compound.

Our only other option for sodium is to pair it with chlorine, and so

we could potentially form NaCl.

And we look at our other potential compound, here we have magnesium,

and we have sulfate, and so I can write the formula down for that, as MgSO4.

Now, for sodium chloride, I see that it is stable in water,

and so I'll mark that as aqueous.

I also see that for magnesium sulfate, sulfates are generally soluble, but

magnesium is not one of the exceptions.

And so, as a result, I actually have no precipitate forming, and so

I have no reaction occurring.

Because there's no change in the reactants and products.

Because everything, all of my reactants, and

all of my products are going to be dissolved in water.

They will be dissociated completely into their ions.

In this reaction, we're going to look at the formation of a precipitate.

When we look at two aqueous solutions, in this case we have calcium hydroxide which

is an aqueous solution, and cobalt nitrate which is an aqueous solution, they

are going to react to form solid cobalt hydroxide, and aqueous calcium nitrate.

So, we're actually going to see the solid forming from these two aqueous solutions.

So, I'm going to to pour, the calcium hydroxide into our beaker.

[SOUND] And then, I'm going to add drops of

our cobalt nitrate, which is a purple solution.

And we should see the formation of a precipitate.

And wee see that it is a blue precipitate.

What you see going to the bottom,

is actually a solid forming, from the cobalt nitrate, and the calcium hydroxide.

In the next module, we'll continue on with this idea of equations, and

we'll talk about molecular, ionic, and net ionic equations.

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