Predicting Precipitation Reactions

Okay, so now we're going to talk
about predicting

the double
replacement reaction products.

So we're going to predict
these precipitation products.

So what happens if we mix

the following solutions?

So we have silver nitrate.

It's aqueous;
we know that from our solubility rules.

And we have sodium chloride.

It's aqueous;
we know that from our solubility rules.

So let's think
about what this might look like.

So we have a beaker...

here,
and we are going to put in--

we're going to pour
in some silver nitrate.

So we would have
some silver plus ions floating around,

and we would have some

NO3- ions floating around,

and then we would put
in some sodium chloride.

This would be our NaCl aqueous.

Over here we put

in our silver nitrate

aqueous.

Okay, so when we put
in our NaCl aqueous,

we have some Na+ ions floating
around,

Cl- ions floating around.

Okay, so what happens

when we mix those two together?

What kind of products do we get?

How do we predict what happens?

[Undeciphered] [Muttering to self]

And we can use the idea

of double displacement reactions
to predict our products.

I'm going to erase this
so it's not in the way,

but we can still keep our beaker
with the stuff in it.

And so we know
in double displacement reactions,

everybody switches partners.

So the cation from one goes

with the anion from the other,
right?

A+ goes with D-.

The anion from the first one goes
with the cation from the other.

So everybody switches places.

So how would that look
in the reaction

that we have up here?

Well, it would look like silver

would bond with chlorine,

and...

nitrate would bond

with sodium.

So if we go back to our beaker,

that would look like...

these two guys will bond together,

they would run into each other.

And then these two guys would run
into each other.

So everybody switches partners.

So then we have to write--
well, what could those products be?

And in order to keep track
of what the products could be,

because when I write
ionic compounds,

the charges switch and drop,

I am going to write the charge.

So Na plu-- Na is plus one.

I'm going to write the charge
for NA down here as a one.

I know NO3 is minus one,

so I'm going to write
a one down here.

And this will become
more obvious when we get--

why I'm doing this when we get

to reactions that don't have ones.

Silver is plus one,
so I'm going to put a one here.

Chlorine is minus one,
so I'm going to put a one down here.

Okay, so then I can read
off what my products are.

I can just say, Okay,

my products are going to be...

I'm going to go silver--

one silver; chlorine, one.

So I'm going to go AgCl,

one of each.

I don't yet know
if it's aqueous or solid.

I'm gonna leave it blank.

And then I'm going to go

nitrate, one;

sodium, oops--

yes, sodium, one.

And I always write
the metal first.

So even though I looked
at nitrate first,

I'm going to write sodium first.

So I have sodium nitrate,

one of each,
because that's what it told me.

I don't yet know
if that is going to be

aqueous or solid.

Okay, now I'm going to go
to my solubility charts.

Is-- I'm going to go
to the solubility charts.

Go to...

solubility charts.

Is silver chloride solid

or aqueous?

Is it soluble?

Hopefully, you say, No, it's insoluble!

It's not soluble,
according to our solubility chart,

so it is a solid.

What about sodium nitrate?

Well, all salts
of sodium are soluble,

all salts
of nitrates are soluble,

so it's aqueous.

Okay, so now we have
what the products are.

The last step is

to balance the equation.

It is easiest
to balance the equation if I move--

remove all of this

so that I can
just see my equation.

Okay, so now I have....

I'm going to balance
this equation.

I can keep my polyatomics
together,

as long as those
aren't breaking apart.

So I'm going to balance the polyatomic
as one unit

instead of as individual atoms,

because it will make
my balancing a little bit easier.

So on the reactant side,

I have silver and nitrate,

and sodium and chlorine.

And on the product side,

I have silver

and nitrate,

and sodium and chlorine.

And I have one silver
on the reactant side,

one nitrate, one sodium,
and one chlorine.

And I have one silver,
one nitrate, one sodium,

and one chlorine.

So this equation
is already balanced.

So this is what happens

when I mix those two solutions.

What does that look like
in this beaker

that we have down at the bottom?

Well, it looks like we--
let's see,

sodium nitrate's going to keep floating
around in solution,

but what is the silver chloride
going to do?

So this silver chloride
in our beaker

is going to bond together,

and instead of floating
around in solution,

it is going to form a solid
at the bottom.

So we are going to get

sodium
and nitrate floating around,

and then a solid forming,

and that solid
is going to be silver chloride.

And that's what we would see.

We would see a clear solution
of sodium nitrate

and a solid forming
at the bottom,

that would be silver chloride.

Okay, let's try another one.

We're going to predict the--

what happens when we--

same reaction, sorry about that.

We're going to predict
what happens when we combine

lead nitrate
and potassium iodide.

What happens
when we combine those two?

And so we know
it's a double displacement reaction.

So the first thing
that we're going to do

is combine the anion--

sorry, the cation from one

with the anion from the other...

the anion with the cation.

So that's what's going to bond.

That's how we're going to form
our products.

Now we're going to put the charges
where they need to be.

Remember,
the charges switch and drop.

So I know that lead
in this compound,

because there are two nitrates
that are one minus,

this lead must be plus two.

So I'm going to put a two
down here.

This iodine--
iodine is always minus one,

so I'm going to put a one
right there,

because those charges
are going to switch and drop.

Nitrate is minus one.

I'm going to put a one
under potassium.

And potassium is plus one,

so I'm going to put a one
under nitrate.

Okay, now I'm going to use those numbers
to write my formulas.

So I have lead with a one,

Pb, one lead,

and two iodides, PbI2.

I don't know
if it's solid or liquid--

or, sorry, solid or aqueous,

so I'm going to leave that blank.

I then, remember,
I need to write the next one,

but I have to write
the metal first.

Even though NO3 comes first
in the equation,

I have to write the metal first,
which is K,

and there is one potassium

and one nitrate.

Because there are ones here,

that tells me I'm going to write
one potassium, one nitrate.

I don't yet know

if it is aqueous or solid.

Now I'm going to go
to my solubility chart.

Look and see
in my solubility chart

is lead iodide

soluble or insoluble?

It's insoluble,

so this will be a solid.

Potassium nitrate.

Potassium nitrate is soluble,

so it will be aqueous.

Okay, I got to that step.

Now I want to balance
my reaction.

I'm going to erase this part

so that I can have more space
for balancing.

Okay, so remember,
if the polyatomic stays together,

I can balance it as a unit.

So on the reactant side,

I have lead,

I have nitrate,

I have potassium, and iodide.

Lead,

nitrate, potassium, and iodide.

Okay.
And so on the reactant side,

I have one lead, two nitrates,

two there,
one potassium, one iodine.

On the product side,
I have one lead,

one nitrate, one potassium,
and two iodine.

Okay,
so lead is already balanced.

We're going to leave that
that way for now.

I have two nitrates
on the reactant side,

one on the product side,
so I'm going to put a two here.

That will give me...

a two in front of nitrate

and potassium.

Now I need to balance the number
of potassium.

So I have two
on the product side,

but only one
on the reactant side,

so I'm going to put a two here,

which will...

give me two potassium
and two iodine,

and completely balances
my reaction.

So, this is what happens

when I combine a solution
of lead nitrate

with the solution
of potassium iodide,

I get...

a precipitate of lead iodide.

So we can think
about what's happening here.

So we have a beaker.

Okay, and inside that beaker,

we would put...

lead nitrate.

Okay, so we would have...

water...,

and we would have
lead two plus ions.

And then for every lead two plus,

we would have
two NO3 ions floating around.

And then we would also have...

a solution of KI.

Oops.

And so we will have a K+

and an I-, okay.

And for every K+,

there will be an I-.

And then
what's going to happen is that...

every time

the K+ and the NO3,
they're going to keep floating around

in solution
because they're going to be aqueous,

but this lead two plus

is going to find the I-,
and that--

they're going to collide together

and form solid
at the bottom of the solution.

And so they're going to come
out of solution,

and we would see a solid forming

at the bottom of this solution.

And that solid would be PbI2.

And that's what we would physically see
with our eyes

when we did this reaction.

Okay.

So now I want you
to try one by yourself.

I will not leave us too big
of a space in the video.

So you want to stop it,
try it yourself,

and then check your answer.

So what happens
when you mix barium chloride

and sodium sulfate?
[Pause for work.]

Okay, so the first thing we do
when we're predicting this,

it is a double displacement
reaction,

so we determine
what is going to react and--

with each other?

How are we going to form
these products?

Double displacement reaction.

So Barium

is going to connect
with sulfate.

Chlorine will connect
with sodium.

Now let's look at the charges
on these things.

Barium is plus two,
so a two will go down here.

If we look up sulfate
on our polyatomic,

that is a minus two,
or two minus,

so we put a two down here.

Chlorine on the periodic table,
it's in the halogens,

so that's a one minus,
so we put a one down here.

Sodium has a one charge,
a one plus charge,

so we put a one over here.

We use those numbers
to write our formulas,

so we don't get
the wrong formulas.

Okay. So then if we went
to write

barium sulfate,

we would look--
we would go to write it,

and we would go,

Oh, it's

Ba2(SO4)2.

Well, what's wrong with that?

The twos, those twos reduce.

We can divide them both by two.

Remember when we
were doing our formulas?

So if we divide them both by two,

we end up with BaSO4.

And then we don't know
if it's aqueous or soluble yet.

And we have

Na1Cl1.

Even though they're twos here,

we don't look at these subscripts
to write our formulas.

We look at the numbers down
below here

that we got from the charges.

So it's NaCl...

not Na2Cl2.

We don't-- those subscripts

go away when we're writing
our products.

We get the new subscripts
from the charges.

Okay.

So, barium sulfate.

Is barium sulfate soluble
or insoluble?

It is insoluble.

This is a solid.

Is sodium chloride soluble
or insoluble?

It is soluble,
so this is aqueous.

And so now

we just need
to balance this reaction.

Okay, so we have--

on this side we have barium
and chlorine,

and sodium and sulfate.

And we have barium
and chlorine,

and sodium and sulfate.

So we have one barium,
two chlorine,

two sodium, and one sulfate.

One barium, one chlorine,

one sodium, one sulfate.

And so we need two chlorines
on the product side;

we have two on the reactant side.

So I'm going to put a two there.

And that gives me...

two chlorines and two sodiums,

and balances the rest
of the reaction.

And I'm already done,
that is my answer.

Okay, I want you to try one last one
before we finish.

Calcium nitrate and potassium phosphate.
[Pause for work.]

Okay,
so double displacement reaction.

First we've got to predict
what's going to connect.

So we have calcium connecting
with phosphate,

nitrate with potassium.

Now we're going to look
at the charges

so that we can write our formulas.

We're not looking at the subscripts
that are on them right now.

So calcium is plus 2.

And you can go through
and write all the charges up there,

all together, if you want to.

Phosphate is minus three. Okay.

Now we are going to put the number
on calcium over here

on phosphate because charges
are switching and dropping,

and those are the two
that are connected.

So calcium, two, go there.

The three from phosphate

will go over under calcium.

The one on nitrate will go
under...

potassium.

And the one on potassium will go
under nitrate.

And we're going to look at those numbers
to write our formulas.

Okay, so calcium

will be calcium.

And we're going to look
at that three, three,

phosphate, two.

Okay, and then we have...

we don't know if it's aqueous
or solid.

And then we have the metal first,
potassium.

There's a one here,

so one potassium and one nitrate,

because there's a one here.

Don't know if it's solid
or aqueous.

Okay, we look
at our solubility charts.

Calcium phosphate is insoluble,

so it is a solid.

Potassium nitrate is soluble,

so it's aqueous.

Now we have to balance.

We can balance our--

we can keep our polyatomics
together

as long as they
aren't being separated.

So we can keep NO3 together
and PO4 together.

We don't have to balance oxygen
by itself.

It'll make our lives
a little bit easier.

So if we have
on the reactant side calcium,

we have nitrate units,

we have potassium units,

and we have phosphate units.

And over here we have calcium units,
and nitrate units,

and potassium units,
and phosphate units.

Then we count them.

We have one calcium
in the reactants,

two nitrates, three potassium,

and one phosphate.

Don't get tricked by that four,
it's part of the phosphate.

And then we have three calcium,

one nitrate, one potassium,
and two phosphate.

Okay, now we need to balance.

We have three calciums
on the products.

We need three on the reactants,

so we'll put a three here.

That will give me
that three...

here,

and it will give me six nitrates.

Three times two is six.

Okay, that tells me
that I need six nitrates,

so I need a six in front
of potassium nitrate

in the products.

That will give me

six nitrates and six potassiums.

Now my nitrates are balanced,
but potassium isn't.

So I need...

a two...

in front of potassium phosphate
in the reactants,

so that I have...

six potassiums for the reactants

and two phosphate groups.

And that balances everything,

and I am finished.

Okay.

So that is

predicting
precipitation reactions,

predicting double
displacement reactions,

and which are also
precipitation reactions.

So now you should be able
to do this on your own.

There is some homework practice
and some activity practice.