4.10 Molar Solubility and the Solubility Product

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Do curso por University of Kentucky

Química Avançada

335 classificações

University of Kentucky

Química Avançada

335 classificações

A chemistry course to cover selected topics covered in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. Prerequisites: Students should have a background in basic chemistry including nomenclature, reactions, stoichiometry, molarity and thermochemistry.

Na lição

Aqueous Equilibria

This unit continues and expands on the theme of equlibria. You will examine buffers, acid/base titrations and the equilibria of insoluble salts.

4.01 Buffers and the Common Ion Effect10:01

4.02 pH of Buffer Solutions14:04

4.02a Aqueous Equilibria Worked Example 13:58

4.03 Buffer Action25:21

4.03a Aqueous Equilibria Worked Example 25:22

4.03b Aqueous Equilibria Worked Example 35:11

4.04 Buffer: Preparation and Capacity8:45

4.05 Strong Acid - Strong Base Titration17:20

4.06 Titrations Involving Either a Weak Acid or a Weak Base36:45

4.06 Part 1 Aqueous Equilibria Worked Example3:46

4.06 Part 1.a - Aqueous Equilibria Worked Example4:06

4.06 Part 1.b - Aqueous Equilibria Worked Example4:55

4.06 Part 1.c - Aqueous Equilibria Worked Example4:21

4.06 Part 1.f Aqueous Equilibria Worked Example8:01

4.06 Part 1.h Aqueous Equilibria Worked Example3:57

4.06 Part 2.a Aqueous Equilibria Worked Example8:25

4.06 Part 2.b Aqueous Equilibria Worked Example5:53

4.07 Polyprotic Acid Titrations5:53

4.08 Indicators10:19

4.09 Solubility Equilibria4:50

4.10 Molar Solubility and the Solubility Product10:37

4.10.a Aqueous Equilibria Worked Example2:20

4.11 Molar Solubility and the Common Ion Effect7:20

4.11 2.b - Aqueous Equilibria Worked Example4:42

4.11 2.c Aqueous Equilibria Worked Example3:45

4.12 The Effect of pH on Solubility7:07

4.12.b Aqueous Equilibria Worked Example3:42

4.12c Aqueous Equilibria Worked Example3:30

4.13 Precipitation Reaction and Selective Precipitation15:03

4.13a Aqueous Equilibria Worked Example9:05

Conheça os instrutores

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

Within this learning objective we are going to learn how to utilize

what we learned in the previous one

which was to write the reactions for these

insoluble salts and write the equilibrium constant expressions for the salts.

We going to be able to use that and convert

between what is called molar solubility

and the solubility product we learned about there.

And we going to learn how to rank salts and the solubility product we learned about there.

And we going to learn how to rank salts

according to their solubility. Which one is more soluble, which one is less soluble.

By looking at either molar solubility or

K_sp. So as you look at these boxes down below in this

slide here. We see that we can convert between K_sp. So as you look at these boxes down below in this

slide here. We see that we can convert between

what is called K_sp and molar solubility.

So the K_sp is one-way

to express how soluble salt is and in general

the smaller the K_sp the less soluble the salt.

Now that comparison really only works well

when you are comparing salts that have the same general formula. Now that comparison really only works well

when you are comparing salts that have the same general formula.

So you might compare a salt

with a formula like MX to MY.

They have a 1 to 1 ratio

but it might not be good for comparing

MX to another substance of MX_2. Where there is

two anions for every one cation.

So you have to be a little careful when comparing

K_sp's for

these insolubles salts and say

who is more soluble and who is less soluble.

But the other way of expressing solubility is what is

called molar solubility and here is the definition.

It is abbreviated with a lower case 's'

and it is the number of moles of solute in one liter

of saturated solution.

It is in unit of moles per liter, in other words it is

molarity. And it is basically

the molarity of the amount

of the salt that dissolved.

So if you are given one of these

two. Either the K_sp or the molar solubility

you can calculate the other.

By way of determining the concentration you can calculate the other.

By way of determining the concentration

of the ions in solution.

In our first example problem

we are given the K_sp

of this lead chloride and we are asked to determine the molar solubility.

So as with any

equilibrium problem, you want to start with a reaction

we are not giving you the reaction, we are expecting you to know the reaction.

And you put the solids on the left, and the ions on the right.

I will leave room for an ICE table.

I am going to write the K_sp expression.

We have the lead.

We have the chloride and it is squared.

So what is happening here? We have a beaker

of water and we are going to dump into that water

some lead chloride.

Lead chloride is not very soluble so

most of the lead chloride will sit at the bottom.

We are going to dump and excess

of lead chloride in there.

And some people do that the say

excess. But that looks too much to me like

X - 5, and writing out

excess I just don't do.

So what I general write is I put some

of the lead chloride into solution. I put a bunch.

And I will put zero because none of it has dissolved yet.

I will allow

instead of using X's, I am going to use S's.

in this change line because

S and I always use a script s because my

s looks too much like a 5.

Is moral solubility.

And where I have written this s here

that is how much lead chloride dissolves.

So by definition that s

or what sits right here in my change line

by definition is molar solubility.

And so that is why in any text book you look at they use S's here.

Well we will follow that S change

across to the product side.

A S and 2S.

And only a little bit of that lead chloride dissolves so there is still some

in the E line. So we will have S and 2S.

We come to our expression and plug in what we know.

We know the K_sp, that was given to us that was 2.4 x 10 ^-4.

For lead there was an S, so we will put an S here.

For Chloride there was a 2S. For lead there was an S, so we will put an S here.

For Chloride there was a 2S.

So we will put a 2S here and we will square it.

That will give me, 2.4 x 10 ^ -4

Is equal to 4 do you see where the 4 came from?

The two square, and then S cubed. Is equal to 4 do you see where the 4 came from?

The two square, and then S cubed.

See why it is cubed?

You have squared this S and you have multiplied by another S.

So now we can divide by 4 on both sides.

And give me and S cubed So now we can divide by 4 on both sides.

And give me and S cubed

equal to 6.0 x 10 ^-4

Then we can take the cube root of both sides

and that will give me S equal to

3.9 x 10 ^-2.

Now it has a unit, it is how much has dissolved in mole

per liter so it is in molarity. Now it has a unit, it is how much has dissolved in mole

per liter so it is in molarity.

And that is the value

for the lower solubility of lead chloride.

OK in this problem we have molar solubility

given to us and we are trying to determine K_sp so it is

the reverse of the previous problem.

We begin by writing the reaction.

Same as we did before

we have the solid on the left

the ions it breaks apart into

on the right.

I will do an ICE table.

Before I do that I write my K_sp expression.

Manganese 2 plus Before I do that I write my K_sp expression.

Manganese 2 plus

carbonate so far nothing is different.

We are going to dumb some

of the solid into our water

before anything happens, we don't any of those ions present.

Now here is where it varies.

We could put minus s here, but we know what

the solubility is. Remember S stands for the molar solubility.

And that was given to me. the solubility is. Remember S stands for the molar solubility.

And that was given to me.

So i know how much of this is dissolving.

Not very much, but some is dissolving. That quantity.

We carry the change line across so we are going to produce

4.2 x 10 ^ -6 of this substance

and produce 4.2 x 10 ^ -6

of the carbonate.

So there is a large excess of the solid, so there is still some sitting on the bottom.

But we now know the quantity at equilibrium

of these two substances.

So K_sp's value would simply be

4.2 x 10 ^-6

which is the manganese's concentration

times 4.2 x 10 ^-6

which is the carbonate's concentration.

When I multiply those two value I will get 1.8 x 10 ^-11. which is the carbonate's concentration.

When I multiply those two value I will get 1.8 x 10 ^-11.

and that is the K_sp of Manganese 2 Carbonate.

We are going to work this one together

but before we work it together I want you to figure out We are going to work this one together

but before we work it together I want you to figure out

the molar solubility

of this salt, so lets describe what is happening here.

You were given the solubility not is molar per liter

but in grams per liter. We cannot use the K_sp

expression unless we know things in molarity.

So you job is to take this information

and determine which one of those

answer is the molar solubility

of this compound.

Well did you choose B?

If so that is the correct answer.

To two decimal places.

Or two significant figures.

Now we are going to complete this problem together.

New we have space to work the problem

and we are going to start by writing the reaction

the solid is M_2X_3.

And it is turning into the ions

in aqueous solution, there are two of the metal ion.

It probably has a 3 plus charge.

And 3 of th anion, it probably have a 2 minus charge.

Its charge is not important

but that is probably the charges that they are

I will leave room for an ICE table

and I will write the expression I will leave room for an ICE table

and I will write the expression

K_sp expression down here. K_sp

is the metal ion's concentration squared

times the X concentration cubed.

So we will fill in our table we have some

solid that we put into the beaker.

There is none of the ions to begin with.

We figured out with your previous There is none of the ions to begin with.

We figured out with your previous

question that you answered what the molar We figured out with your previous

question that you answered what the molar

solubility of this salt is.

I carrying just a little bit extra sig. fig here

just to carry extra until the end.

It was actually 1.25 just to carry the extra.

This is the moles of solubility.

We carry the change line across, so it has to be twice

1.25 x 10 ^-19 here.

And 3 x 1.25 x 10 ^-19 here. 1.25 x 10 ^-19 here.

And 3 x 1.25 x 10 ^-19 here.

So there is still going to be some of the solid on the bottom of my flask.

But now we are going to have 2.5 x 10 ^ -19

and 3.75 x 10 ^ -19.

Now we are ready to obtain the K_sp value.

For the metal we put 2.5 x 10 ^ -19

and that is squared.

For the none metal we put

3.75 x 10 ^-19

and that is cubed.

We you plug these into your calculator

and obtain a value, you will get 3.3 x 10 ^ -93.

So that is an absurdly small K_sp value

for this hypothetical compound.

But I wanted to show you that

these coefficients come into play twice.

They come into place in the change line

and they come into play in a power.

So that is the end of this learning objective where we have

learned to covert between K_sp value and molar solubility.

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