4.11 2.b - Aqueous Equilibria Worked Example

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Do curso por University of Kentucky

Química Avançada

354 classificações

University of Kentucky

Química Avançada

354 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

Here in part B we are going to see the effects of adding sodium chromate to

the solution of silver chromate.

Now very commonly what students will try to do when they come across a problem like

this is to react these two as reactants and try to come up with a product, but

that is not the way you work these problems.

What you will want to do is write the reaction as you did in part A.

The silver chromate dissolving in solution

to give you two silver ions plus chromate.

And the change or the difference in part B versus part A comes in this I line.

You will put some of the solid into the beaker where there's already

some solution present of sodium, or some ions present of sodium and chromate.

So there's not any silver ions present but

there is 2.6 times 10 to the minus 3 molar of the chromate present.

And so I'll put that number here saying it's this number.

Now word of caution.

The reason it's the same number is because there is

one of the chromate in sodium chromate.

So the concentration of chromate is the same as this.

But if there were two chromates in the formula, which there isn't, but

if there were, then I would need to double that value that they're giving me here

before I put it into the table,

because that would be the concentration of the ions in solution.

So we will use minuses here because that's how much

of the sodium chromate will dissolve.

That's the molar solubility.

I will add 2s here.

I will add s over here.

That will still give me some solid in solution, I mean in the container,

I have 2s for that, and I have 2.6 x 10 to the minus 3 plus s for this.

Now, I will always want to write the Ksp expression it's

the silver ion concentration squared.

I'm looking at the reaction, I'm doing the powers a lot of times students try to skip

this step to save time, but

then they invariably forget to work with those coefficients.

Then I will look at the silver concentration, and that was 2s, and

it's squared, and then I will look at the chromate concentration, which I see here,

2.6 times 10 to the minus 3 plus s, so all of this is multiplied.

Now we're going to assume that S is very small compared to the other values,

and ignore this term for simplicity sake.

We'll make sure we go back and check that it was a valid assumption.

So all of this is going to be equal to the number

of ksp which is given in the problem.

1.2 times 10 to the minus 12.

And so I will multiply the 2 times 2 that will be 2 squared and

then I'm going to multiply that to the 2.6 x 10 to the -3.

And that is going to give me 1.04 times 10 to the minus 2 times s squared.

So then I will divide both sides by the 1.04 times 10 to the minus 2,

And that's going to give me s squared equal

to 1.1 times 10 to the minus 10.

So then I can take the square root of both sides, and that will give me s, and

the value for s is 1.0 times 10 to the minus 5.

So then we check our assumption, and if we take this value and

subtract it from the 2.6 times 10 to the minus 3,

we will still get 2.6 times 10 to the minus 3 or add it to it, add it,

it won't change the values, so that's a good assumption for us.

And so, this is the molar solubility of silver chromate

In a solution that contains sodium chromate.

Now let's see if it makes sense.

The solubility of the silver chromate in pure water,

that was determined in part A,

was 6.54 times 10 to the minus 5.

Okay, and this is 1 times 10 to the minus 5, so the molar solubility

has decreased by the addition of this small amount of the sodium chromate,

so it is supporting what we know about a common ion situation.

Common ions always decrease the solubility of an insoluble salt.

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