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4.06 Part 1.b - Aqueous Equilibria Worked Example

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Advanced Chemistry

Universidad de Kentucky

4.7 (415 calificaciones) | 50K estudiantes 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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4.7 (415 calificaciones)

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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!

De la lección

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

Impartido por:

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

In this section we're going to begin adding some sodium hydroxide to the flask.

We're going to start with 10 milliliters of sodium hydroxide and

calculate the pH after the addition of that.

If we look at our graph here, we're at the point where we have

10 milliliters of sodium hydroxide being added.

And so, it's an acidic solution.

I'm not sure exactly what the pH is, but it's acidic.

And we're going to see why it should be acidic as we proceed with

the problem itself.

Now as a weak acid, we know we can write HA for the acid.

I want to write And not NAOH for the base,

because the NA is a spectator and is really not part of this reaction.

It makes it much easier if you leave it out.

It's a one way reaction because a strong base forces it to completion.

The acid donates it's H+ to the base.

That will leave the acid as A- and the base becomes water.

This is a one way reaction, so we know an ICF table will apply.

And you always put moles into an ICF table.

Now we're given the molarity and the volume of each substance, and we know that

molarity times volume will give you the moles, as long as the moles is in liters.

So let's start with the moles of the acid.

The molarity of the acid 0.249.

And the volume of the acid was 20 milliliters, or 0.2 liters.

This will give me the number of moles of the acid of 4.98 times 10 to the minus 3.

For the moles of that would be the same as the moles of sodium hydroxide,

because there's one hydroxide in every sodium hydroxide.

So we'll take the molarity of sodium hydroxide,

which is the same molarity, times volume.

Now, it's 10 milliliters or 0.01 liters, and

that will give me 0.00249 or, 2.49 times 10 to the minus 3.

And that would be moles.

I will plug those numbers in here.

4.98 times 10 to the minus 3,

2.49 times 10 to the minus 3.

For the reaction, I have not produced any of the A-, and

I don't care about the water.

I will consume the smaller quantity.

That's the 2.49 times 10 to the minus 3.

And I will produce that on this side.

2.49 times 10 to the minus 3.

That will leave me, when I subtract these values, at 2.49 times 10 to the minus 3.

I will have consumed all of the strong base, and

I will have some of the conjugate base of that weak acid sitting here.

Now we see that we have converted half of our HA over to A-.

And we end up with the same amount of both of those.

And if you have a weak acid and

its conjugate base in solution, you have a buffer.

So we can use the Henderson-Hasselbalch equation.

pH equals pKa plus the log of the number of moles of base over

the number of moles of acid.

Well, the pKa is the negative log of the Ka,

1.54 times 10 to the minus 5.

Plus the log of the number of moles of base, which was right here, okay?

2.49 times 10 to the minus 3, over the number of moles of acid,

2.49 times 10 to the minus 3.

The log of 1is 0, so this term goes away.

And so pH equals pKa, or the negative log of 1.54 times 10 to the minus 5.

This is equal to 4.812.

So we went through a lot of work there to do this problem.

But one of the things that you should know, is handy to know anyway, is that

when you're halfway to the equivalence point, and that's where we are here.

Equivalence point would be at 20 milliliters, and we're halfway there.

Whenever you're halfway to the equivalence point,

pH is equal to pKa, at one half the distance

To the equivalence point.

Okay?

So we have got a solution that is a mixture of a weak acid and

its conjugate base.

We have equal amounts of both.

That term goes away.

So if that term goes away, then pH is equal to pKa.

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