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4.11 2.c Aqueous Equilibria Worked Example

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Química Avançada

Universidade de Kentucky

4.7 (422 classificações) | 51K alunos inscritos

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.

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Avaliações

4.7 (422 classificações)

5 stars
315 ratings
4 stars
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3 stars
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2 stars
4 ratings
1 star
2 ratings

KO

Jul 17, 2017

Excellent content and delivery!\n\nThis course has many well-explained examples and practice questions to reinforce the understanding of the taught concepts. Thank you!

CW

Sep 17, 2015

I used this course as a significant portion of my preparation for the CLEP Chemistry exam, and passed today with an A! Thank you!

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

Ministrado por

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

Transcrição

We are ready to see how the silver

chromate dissolves in the solution that contains silver nitrate.

It's a very similar problem as part b, just a few subtle differences.

We'll write the same reaction that we wrote for part a and part b, and

that is that the silver chromate.

Solid is on the left side of the equation and

the ions it breaks apart into are on the right side of the equation.

So nothing different here.

In the I C E table,

we start out by putting some of the solid into the solution.

The solution already contains silver ions.

We look up here, we want to figure out how much silver ions are present.

Well there's one silver in this silver nitrate, so

this is the concentration of the silver ions that goes here.

Common mistake for students is for them to see the 2 and want to double this number.

This is how much silver's presence before the Silver Chromate has ever had a chance

to dissolve.

So it has nothing to do with this coefficient.

And there's no chromate in part c.

So, once again, molar solubility.

This is how much the Silver Chromate dissolves.

We will produce two S here, and + S here.

If you dissolved a little bit, you will still have some left as a solid,

you have 0.15 + 2s here, and you have s here.

Once again we will do our Ksp expression.

Ksp is a silver concentration squared times a chromate concentration.

We will plug in the values that we have Ksp is 1.12 x 10 to the -12.

Silver's concentration is 0.15 + 2s and that is squared.

And chromate is just s.

Now here we're going to assume that just a very small amount dissolves.

So we can ignore this term, or at least we think we can.

We'll check that before we're finished, and we are going to square the 0.15.

And then we'll have times s.

So we divide both sides by 0.225.

And this will give me s.

And s = 4.98 times 10 to the -11.

So the silver makes a big impact.

On the Ksp, on the solubility, sorry, it doesn't change the Ksp at all,

has a big impact on the molar solubility, much more so than in part b.

Part b the molar solubility just dropped down ever so slightly.

It went from what it was in pure water, which was 6.5 times 10 to the -5,

and this became 1.0 times 10 to the -5, so it dropped a little bit.

Because of this square term, having some silver concentration

already present in solution really effected the molar solubility,

and this is way lower than what it is in pure water.

Now let's check our assumption, was it a valid assumption?

Let's look right here.

If we double s, okay, so that's going to be about,

let's say, 1 times 10 to the-10.

That's when you double this number.

We double it, it's going to be way too small to even think about when

you add it there to change this number of 0.15, it's still definitely 0.15.

So that was a very good assumption for that part of the problem.

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