4.03a Aqueous Equilibria Worked Example 2

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Из курса от партнера University of Kentucky

Advanced Chemistry

405 оценки

University of Kentucky

Advanced Chemistry

405 оценки

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.

Из урока

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

Познакомьтесь с преподавателями

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

We've previously calculated the pH of the buffer that is described in this

problem here.

And we determined the pH to be 9.427.

This was the concentration of ammonia that's actually given in the problem.

But with that previous work, we determined that the ammonium,

the NH4+ concentration, was 0.748.

Now we're asked to calculate the pH

when we take a small amount of a strong acid and add it to the buffer.

The buffer will neutralize that strong acid.

And take it out of the system, and that's why it resists change to pH.

So to work this problem we have to first write the reaction between

that strong acid, which I always let H3O+ represent my strong acid.

And the portion of the buffer that will neutralize it,

which would be the weak base.

Ammonia is a weak base.

And I will do a proton swap.

The H+ from the acid is donated to the base and

that will turn the base into ammonium and the H3O+ will be converted to water.

So that is the reaction.

Because the acid is strong, it is a one-way reaction and

we will an ICF table, and we will plug moles into that table.

Now we're ready to go determine the moles of each of these substances.

Let's begin with the moles of H3O+.

The moles of H3O+ would be the same as the moles of HCL.

HCL.

So find out the molarity of the HCL, which I do, it's given to me in the problem.

And I know the volume of the HCL which I do,

I will have the most being 0.0002 moles.

And I can plug that in to the table.

To get the moles of ammonia,

we're going to take the molarity times the volume.

The volume that we'll use is a volume of the buffer

of this portion of the titration.

So we're using a 75 milliliters, not the 500 milliliters,

because we're not titrating with all of that.

This is going to give to me 0.05625 moles, now that's an extra

significant figure that I'm just going to carry along for now.

And that's the moles of ammonia.

Now before this reaction takes place, there's still some ammonium present.

Because it's a buffer and it contains ammonium in a buffer.

So I'm going to go ahead and calculate how much ammonia we're starting with.

I'll take the molarity of the this molarity we determined in the previous

example problem that we did in the last learning objective.

And we'll multiply by the 75 milliliters, or

the 0.075 liters here, and this is going to give me 0.0561 moles.

And the water we can leave out.

Now we're going to consume this smaller quantity, and

that would be the strong acid.

And I will subtract it from both sides.

I mean from this side, and add it to this side.

This is going to leave me with none of the H3O+.

It's going to bring my ammonium down to 0.05605.

It's going to increase my ammonium up to 0.0563.

Now, because I have present a weak base,

ammonia, and it's conjugate acid, I have a buffer.

And so the Henderson–Hasselbalch equation will apply.

pH equals pKa.

Well pKa is the negative log of the Ka.

The Ka is Kw divided by Kb, I did not write Kb in this problem.

But it was in the previous problem.

It's 1.76 x 10 to the -5.

That's the Kb of ammonia.

So I want to divide by 1.76 x 10 to the -5, and

I'm going to do the moles of the base over the moles of the acid.

The moles of the base are sitting right here, 0.05605.

The moles of the acid is sitting right here, so at 0.0563.

And that will give me a pH = 9 0.243.

And as we look at this pH, 9.243,

we see that it is slightly lower than what the pH was of just the buffer.

It's resisting change.

It's not completely not changing at all, but it is resisting it even so slightly.

It is going down, and

that is to be expected because you are after all adding an acid to the solution.

So you expect the pH to be a little smaller or a little bit lower.

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