4.02a Aqueous Equilibria Worked Example 1

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En provenance du cours de University of Kentucky

Advanced Chemistry

354 notes

University of Kentucky

Advanced Chemistry

354 notes

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.

À partir de la leçon

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

Rencontrer les enseignants

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

Let's begin this problem by examining what kind of solution we have.

And it tells us right here at the beginning that we have a buffer, so

let's underline that word.

And when we have a buffer, we know that the Henderson–Hasselbalch

equation is a good way to calculate the pH of a buffer.

So let's write the Henderson–Hasselbalch equation,

pH = pka + log [B]/[A].

Now you can also use instead of concentrations,

you can put moles of base over moles of acid.

And this problem, one way is about as easy as the other.

We are trying to calculate the PH so we need to know the PKA.

Well the PKA is the negative log of the KA value but

I don't have the KA value, I have the KB of ammonia.

To get the KA we will take KW divided by

KB which is 1.76 times 10 to -5.

Okay, then plus the log of the concentration of the base

over the concentration of the acid.

NH3, nitrogen with three things attached, that's my base.

The acid is this portion of the salt.

So the acid is NH4 plus.

So I need a concentration of those two items.

They gave me the concentration of the base where I can put 0.750 molar right there.

Now I need to know the concentration of the acid.

To get the concentration of the acid, I need to know the moles

of the acid over the volume in liters.

Well it's 500 millilitres,so that bottom number was easy enough to determine.

500 millilitres or let's write it this way,

0.5 liters, 500 milliliters is 0.5 liters.

What I need to know is the moles of that acid.

And they gave me grams of it.

And they gave me the formula of the salt which is ammonium chloride.

So I'll get the molar mass of ammonium chloride, one nitrogen,

look at the formula there.

One nitrogen, four hydrogens, and a chlorine, because it's the mass of

the whole salt, so we need to use the molar mass of the whole salt.

When you add those values together, you get 53.50.

I would go with the grams and the moles.

So 20 grams divided by the molar mass will give me moles of this acid,

divided by the volume will give me moles per liter, or molarity of this acid.

And when you divide those values out,

you obtain a number of 0.748.

It's almost as concentrated as the base, but not quite.

So when I take the negative log of this ratio, I have 9.246.

Since there is, I'll go ahead and write this out,

log of 0.750 / 0.748.

Since there is a slight more base than there is acid,

then this may be a little bit higher in pH.

And when you actually work the numbers out it does raise it up to

that third decimal point to 9.247.

So that is the pH of this buffer solution.

And it's very common when you've got a buffer based upon a base and

its conjugate acid salt that you would have a pH greater than 7.

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