4.05 Strong Acid - Strong Base Titration

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From the course by University of Kentucky

Advanced Chemistry

333 ratings

University of Kentucky

Advanced Chemistry

333 ratings

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.

From the lesson

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

Meet the Instructors

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

We are ready to begin our next round of learning objectives

dealing with acid and base titration.

within these acid/base titrations

we are going to learn to calculate the pH

At various points

along the titration.

Now if you will think back to along the titration.

Now if you will think back to

what you have learning in the past

you have learned about acids and bases

titrations, most likely.

You might want to go back and review

what you have learned.

But one of the things you learned to do with

acid and base titrations is you are given

an acid

where you know the concentration and you are

trying to determine the concentration of

an unknown base.

You know the concentration

of a base and you are trying to determine the

pH or the concentration of an unknown acid.

And its a stoichiometry problem where you are

covering from the molarity and the volume of

the known or the standard solution

getting into moles of that

substance and determining

the moles of the other substance.

Within these units

that we are talking about here Within these units

that we are talking about here

and the next couple of learning objectives

we are going to look at how pH

changing during the course of this.

We are going to look at three types of titration.

We are going to begin

in this lesson with strong acids

and strong base titrations.

This will be learning objective number 5.

In learning objective number 6

we are going to have one of those

be weak. Either the acid will be weak

or the base will be weak. be weak. Either the acid will be weak

or the base will be weak.

We will never do calculations in which

both the acid and the base are weak.

There are very many equilibria involved

and it gets very complicated.

So we are going to do make sure that at least and it gets very complicated.

So we are going to do make sure that at least

one of those two will be strong.

So lets being, our learning objective number

5 will be

calculating the pH of a strong acid

and strong base titration

and any point along

the titration, from the beginning

to the, what we call the equivalence point

and beyond.

What happens when a strong acid and a strong base

react? We we can write the reaction

as follows.

Here is an example of a strong acid

and a strong base. The

cation of the base and the anion of the

acid will form the salt.

And the H of the acid

and the OH of the base

will produce the water.

We are going to consider just the net

ionic equation.

It would be H+ of the acid

and OH- of the base

combining to form water.

Or we could and I am going to

back this up, we could write this

typically will write this

with H_3O+

for the strong acid with H_3O+

for the strong acid

and OH- for the strong base.

And then the reaction is just a little bit different

we have two waters.

Because remember when a strong acid

dissolves, you could either say Because remember when a strong acid

dissolves, you could either say

it 100%

ionizes

to produce those ions.

Or you could say, it is to produce those ions.

Or you could say, it is

reacting with water

to 100%

give you the anion

NaH_3O+

so either way, and I typically choose

writing strong acids as H+.

Now this is a one-way reaction.

And with a one-way reaction we are going to use

what is called an ICF table

and we saw those tables

we when were dealing with adding and acid

to a buffer or a base to a buffer.

We know that it is a one-way

push it to completion type of reaction.

So lets look at how pH is changing

during the course of a titration

prior to actually getting into the calculations.

Here is a schematic of the reaction taking place.

We have in the flask a strong acid. Here is a schematic of the reaction taking place.

We have in the flask a strong acid.

That is our .1 molar HCl.

And lets say we have 25 ml of it.

We are going to start adding base

to this and I have got a sodium hydroxide in the

portion I am interested in is the OH- of that base.

And here is our reaction H+ and OH- forming water.

I am going to plot, and kind of determine

pH at every point along this titration

as the base is being added.

At the very beginning

before I have added any base what so ever

We have got the pH of this strong

acid. Well, if it is .1 molar We have got the pH of this strong

acid. Well, if it is .1 molar

and we take the negative log

of .1

we will have a pH of 1.

So we have a red dot there, if you look at the graph

at one, the beginning.

Now we are add a little bit

the base, and we are going start adding that base Now we are add a little bit

the base, and we are going start adding that base

into the flask.

When a little bit, look at the base up here into the flask.

When a little bit, look at the base up here

when a little bit of the base is added

it will react with some

of the acid.

And the base will go away, the base will be turned into water.

But there will be a lot of acid left.

But not as much as we started with.

So the concentration

of that acid is

going to be dropping a little bit. of that acid is

going to be dropping a little bit.

during the course of this titration.

Well, if that concentration is

dropping a little bit the pH

is going to come up a little bit so we see

a slight rise of

the pH. Well, we keep adding and

we keep adding and it going to rise and rise and rise.

But it is not changing drastically just

slowly increasing as the

H+ concentration

is dropping a little at a time

as we continue to add a little bit

base and add a little bit of base.

But there will come a time where

the amount of acid and the amount of

base are exact equal the amount of acid and the amount of

base are exact equal

and we have added just enough OH-

to react with all of the H+.

At that point we have reached what is called the equivalence point.

There number of moles of base is going to be stoichiometrically

equal the number of moles of acid.

So there are a certain number of moles of acid in here

right, we had a certain number of moles and we

kept on adding until all of those right, we had a certain number of moles and we

kept on adding until all of those

we finally converted to acid

and we had the picture

and all the acid had been converted

to water, and we have the right on the right.

When the titration is between

a strong acid and a strong base When the titration is between

a strong acid and a strong base

you will have converted all the H+

and OH- over to water.

And the pH of water is 7. and OH- over to water.

And the pH of water is 7.

So always the pH is going to be 7

at the equivalence point when they are both strong.

But we can keep on adding

we can go beyond

the equivalence point

and what happens when we continue to add

base? If we were take this system

and add some more

base to there, suddenly we got

base present.

We already have consumed all of the acid. base present.

We already have consumed all of the acid.

There is no acid left to neutralize.

so the pH is going to shoot up.

It is going to shoot up very rapidly and then just

gradually increase as we add more and more

base to the flask.

So we are going to be able to calculate the pH

at any point. So lets go back to this procedure.

This is a similar procedure at any point. So lets go back to this procedure.

This is a similar procedure

that we saw when we added

an acid or base to a buffer.

The first thing we will have is a one way reaction

in between the acid and the base.

We will use an ICF table

and put moles into the table which is a way

of doing a limiting reactant problem.

We will examine that final line

and determine our next course of action.

At this point there could be three

possibilities of what could happen.

It there remains a conjugate base pair

then we would use the Henderson-Hasselbalch equation.

This would never happen

when we are in this portion of our

lesson, because we are dealing with

strong acid and strong base. There will never be a weak

thing present and its conjugate.

But if there is a titration and it is weak

then this would occur prior to the equivalence point and we will

come back to that later.

If there remains

only a weak acid or a weak base present

this occurs when we are at the equivalence point

if one of them is weak. Again,

in this portion of our lesson this will not happen.

What will happen in this portion of the lesson

is there may remains

some strong acid, or some

strong base, and we put the word strong here as well. some strong acid, or some

strong base, and we put the word strong here as well.

And when that happens you would

calculate the pH directly

from the molarity of that

strong acid or strong base that is left over.

With this procedure

we are going to hold and ready

these two portions

there are not going to happen when they are both strong.

But we will see them later

when we are in learning objective

number 6.

OK we are going to determine the pH

beginning with a small amount

of sodium hydroxide being added to the acid.

So we are only adding 5 ml.

Begin by writing the reaction.

The strong acid I

use H_3O+ to represent the strong acid.

I let OH- represent the strong base.

I will form 2H_2O.

I will do an ICF table according to my instructions. I will form 2H_2O.

I will do an ICF table according to my instructions.

One way go to completion.

Now I need the moles of H_3O+

and the moles of OH-. Now I need the moles of H_3O+

and the moles of OH-.

Lets begin with the moles of H_3O+

coming from the acid.

The molarity of HCl coming from the acid.

The molarity of HCl

is .1.

If I multiple by the volume which is

.025 liters

that will produce from me .0025

moles of H_3O+.

and I can put that into the table.

For the moles of base

I see that the sodium hydroxide has one OH in it.

So the moles of OH would be the molarity I see that the sodium hydroxide has one OH in it.

So the moles of OH would be the molarity

of the sodium hydroxide

times its volume, again in liters

0.0050 liters.

That is going to give me 0.005

moles of base. And I can plug that into my table.

I don't care about the water, I am just going to put a line through that.

I am going to consume the smaller of the too.

That is the base. I am going to consume the smaller of the too.

That is the base.

I would have, of course, produce

some of it. I had a lot

to begin with a few more moles

that is still going to be a lot there

and I will be left with that is still going to be a lot there

and I will be left with

none of the base.

So at this point, I have

H_3O+

in solution, acidic solution, and it is a

strong acid. So I can obtain the pH

by taking the negative log

of the H_3O+ concentration. by taking the negative log

of the H_3O+ concentration.

Well, this is not

the H_3O+ concentration.

I plugged everything in this table using moles.

So this is the moles of H_3O+.

So if I want to know the concentration of H_3O+

I have to divide by the volume. So if I want to know the concentration of H_3O+

I have to divide by the volume.

And I want to divide by the total volume.

So I need to take the negative log

of the 0.002 So I need to take the negative log

of the 0.002

moles and I have

25 ml of the acid

5 ml of the base

so that is .030 5 ml of the base

so that is .030

liter, 30 ml, or .030 liters.

When I

divide those numbers out

it is equal to 0.067.

That is the concentration

when I take the negative log of that That is the concentration

when I take the negative log of that

value is 1.18.

So that is the pH of the solution.

So it should be acidic.

We are before the equivalence point

we neutralize all the base We are before the equivalence point

we neutralize all the base

and we have converted the base

to water but there is still a lot of extra acid in there.

Now lets get to the equivalence point.

We are going to add 25 ml

of the base to the 25 ml of

the acid. Since there are both at the same

molarity, we know we are at the equivalence point.

But lets imagine that you didn't realize that.

And you start redoing the problem But lets imagine that you didn't realize that.

And you start redoing the problem

blindly as we have been before.

Eventually you are going to recognize you are at the equivalence point.

We would take the molarity and the volume

of the acid. It is the same molarity

of the same volume of acid we had before.

So we are going to place that number

.0025 here. So we are going to place that number

.0025 here.

And if wanted to figure out the moles of

hydroxide we do the same thing.

We take the molarity

times the volume in liters

and we see that is it .0025 moles.

So at this point you might recognize

that hey we are at the equivalence point.

But lets say you didn't at this point.

This is what I see some students doing.

They just keep on working through the problem and they have a zero This is what I see some students doing.

They just keep on working through the problem and they have a zero

here and a zero here. They just keep on working through the problem and they have a zero

here and a zero here.

And then they scratch their heads and say

I have nothing to base my calculation on.

Keep in mind that if you neutralize all your acid and base

you will only have water and what is the pH of water? It is 7.

So at some point you are going to recognize you are at the equivalence point.

And whenever you are at the equivalence point between a

strong acid and a strong base

the pH is 7. strong acid and a strong base

the pH is 7.

OK now I am going to have you

stop the video after a bit

and try to work through this problem. stop the video after a bit

and try to work through this problem.

We have got a titration between

a strong acid and a strong base. The same one. We have got a titration between

a strong acid and a strong base. The same one.

But now we have gone beyond the equivalence point

and added excess sodium hydroxide.

So I want to see if you can calculate the pH of this solution.

Well, if you came up with 11.96, very well done.

That is the pH of that solution.

If you struggled with that

I am going to go through how to work

and get that 11.96.

You might not need to see this

in which case you can of course skip ahead.

So we have our reaction that we have been working with.

H_3O+ plus OH-

producing 2 waters.

We have our ICF table

the amount of acid hasn't changed

in any one of these problems. the amount of acid hasn't changed

in any one of these problems.

But now the amount of in any one of these problems.

But now the amount of

base has certainly changed.

30 ml of .1 molar base has certainly changed.

30 ml of .1 molar

is 0.0030 .

So we have more of the

base then we have of the acid.

So know when we consume

smaller amount it is the So know when we consume

smaller amount it is the

acid that going away.

We are left with 0.005 moles

of the base.

Since it is a base that is left over we can

first calculate

the pOH.

That would be equal to the negative log

of the hydroxide concentration.

Well this is just moles so we have of the hydroxide concentration.

Well this is just moles so we have

to divide by the total volume.

The total volume would be

25 ml of the acid The total volume would be

25 ml of the acid

and 30 ml of the base thats

55 ml

or .055 liters.

That would be the negative log

of 0.009.

That gives me a pOH

of 2.04.

Of course the pH

is simply 14 minus that 2.04.

That is how we get the 11.96. is simply 14 minus that 2.04.

That is how we get the 11.96.

Now hopefully when you saw this

problem, you realized it had to be basic

and you eliminated these two answers right off the bat.

But you could not distinguish between the other two.

Without first doing this process.

So that is the end of

learning objective 5 in which

we determined the pH

when the acid and the base are strong. we determined the pH

when the acid and the base are strong.

We have

a procedure to follow. We do an ICF

table there will always be either at the a procedure to follow. We do an ICF

table there will always be either at the

equivalence point in which case the pH is 7

no calculation is needed.

Or you have got more acid

strong acid in there, or more strong base in there

and you can calculate of either the acid or the base

using the negative log of that concentration.

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