3.04 Finding pH

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Química Avançada

353 classificações

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University of Kentucky

Química Avançada

353 classificações

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.

Na lição

Acid-Base Equilibria

The concept of equilibrium is applied to acid and base solutions. To begin, the idea of weak acids and bases is explored along with the equilibrium constants associated with their ionization in water and how the value of the equilibrium constant is associated with the strength of the acid or base. The autoionization of water is discussed and how temperature affects this process. A variety of problem types are covered including calculations of pH, pOH, [OH-], and [H+] for both strong and weak acids and bases. Aqueous salt solutions are classified as acids and bases and the multi-step ionization of polyprotic acids is discussed. Finally, the concept of Lewis acids and bases is discussed and demonstrated through examples.

3.01 Acids and Bases9:15

3.02 pH and Kw13:32

3.02a Finding Kw2:10

3.03 Acid Strength6:34

3.04 Finding pH10:25

3.05 Strong and Weak Bases12:51

3.05a Calculating equilibrium concentrations of a weak base, part 12:51

3.05b Calculating equilibrium concentrations of a weak base, part 21:36

3.06 Ions as Acids and Bases18:15

3.06a Acid-base properties of aqueous salts, part 11:43

3.06b Acid-base properties of aqueous salts, part 21:07

3.06c Acid-base properties of aqueous salts, part 34:13

3.07 Types of Acids12:06

3.08 Polyprotic Acids4:46

Conheça os instrutores

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

In this module we will look at how we found the pH of both strong and weak

acid solutions.

Our objective in this unit is to calculate the pH are both strong and

weak acid solutions

because we have different procedures for each. Finding the H+ concentration of a

strong acid is fairly easy.

So finding the pH of that same solution is a straightforward process.

When we look at a strong acid

we know that it is a strong electrolyte meaning it ionizes is completely.

HCl goes to H+ and Cl.

If we have a solution that is 0.050 molar

HCl, as in that is the concentration that we measured

then we can know that the complete dissociation, or complete ionization

will result in all other HCl being converted into H+ and Cl- ions

So if we have a complete dissociation and we know the initial concentration of HCl.

Then we can also find the final concentration of H+ because it

will be the same as the initial.

Therefore H+ is going to be equal to 0.050 molar.

Once we know the H+ concentration finding the pH

is easy. pH equals

negative log a the H+ concentration.

So in this case pH is equal to negative log

of 0.050 molar.

And out pH

will be equal to

0.30. Note that I have

too sig figs in my original concentration. So I have two decimal places in my answer.

Finding the H+ concentration of a weak acid is a little bit different.

Because we don't get a complete dissociation or ionization.

We have to figure out what fraction for acid dissociate into the ions.

If we look at our reaction we have HA

yields H+ plus A- and this is an equilibrium process.

If you're not sure if something is a weak or strong acid the presence of that

equilibrium arrow

indicates that it is in fact a week acid. If we look at our initial concentration of HA

we see that is not equal to the final H+ concentration as it was for the indicates that it is in fact a week acid. If we look at our initial concentration of HA

we see that is not equal to the final H+ concentration as it was for the

strong acid.

Because not all of our HA is divided or dissociated into the H+ an A-.

We have to look a K_a the acid dissociation

our ionization constant for the acid.

And remember this is just an equilibrium constant for the acid.

So our procedure is going to be much more like what we did in the

previous unit talking about equilibrium.

We are going to use an ICE table to solve for that unknown

concentration. Let's look at an example.

What is the pH about 0.200 molar solution of HF.

And were given that the K_a value is equal to 3.5 x 10^-4.

Here we have HF, aqueous phase,

plus F-. I set up my ICE table remember, Initial, Change, and Equilibrium.

I know my initial concentration of HF is

0.200 molar because that's what was given in the problem.

I know my initial concentrations of H+ an F- are zero.

Because we have not had any dissociation to occur initially.

Then I look at my change row I'm going to lose

some amount of HF and I'm going to gain the same amount of H+ plus an F- because if

the coefficient it's all a one to one to one ratio.

Then I see that for my equilibrium row

I have 0. 200 - X and X.

Now, I can set up my law of mass action

plug in my terms from the equilibrium row

and solve for my concentration.

So let's look at that how we would do that.

So here we have H+ and F- are both equal to x.

So I can say that equal to x^2 and the HF is going to be 0.200 - x. So here we have H+ and F- are both equal to x.

So I can say that equal to x^2 and the HF is going to be 0.200 - x.

On the bottom equal to 23.5 x 10 ^-4.

Just as we did in the earlier equilibrium problems. We want to make the

simplifying assumption

that X is much much less then 0.200.

Given the small value K_a this is probably a reasonable assumption, but we

would still need to do the calculation just to make sure.

So now we got out X's is less than 0.20. We simplified our expression

so let's rewrite what we have left.

Now I have lost the of the -X because we have simplified.

Now I can multiply both sides by 0.200.

And when I do that I end up with

7.0 times 10 ^ -5 times X^2.

I can take the square root of that number and found that

X equals 8.4 times 10 ^-3.

That will be equal to my H+ concentration.

From that, if we wanted, that we could find the pH value are we go on to

find the OH- concentration using what we know about K_w.

finding the pOH- value.

Another way of looking at the acid strength is to look at the percent ionization of weak acids.

What we can look at is the H_3O+ concentration at equilibrium

or remembering that the same thing we've been calling the H+ concentration.

those mean exactly the same thing. Divided by the initial HA concentration

and multiplied by 100. The bigger the value of the percent ionization

the stronger the acid

and vice versa. The smaller the percent ionization

the weaker the acid. Let's look at an example of how we can use this

information to figure out the pH of the solution.

Here we have HA in equilibrium with H+

and A-. It is 0.350 molar weak acid that is 13.2 percent

ionized and we want to find the pH of the solution.

We have percent ionization

is equal to H+ at equilibrium over HA initial times a hundred.

I could plug in the values I know is equal to H+ at equilibrium over HA initial times a hundred.

I could plug in the values I know

13.2 percent equals H+ it equilibrium

over 0.350 initially

times 100. Now I can rearrange

and solve for my H+ concentration at equilibrium

and find that is 0.462.

That gives me the concentration it does not yet give me

the pH value. I have to take the negative log of that number

to find the pH value which is 1.335.

We have three-digit in our decimal place and the mantissa of the log

because we have three significant figures in our concentration.

When I look at a mixture of strong and weak acid

I have a couple of things to consider. First of all the strong acid When I look at a mixture of strong and weak acid

I have a couple of things to consider. First of all the strong acid

completely dissociate or ionize.

Because I go from HCl to producing H+ and Cl- and all of my HCl will dissociate.

So I have a fairly large concentration of H+ present.

When I look at the HA, if I just had the weak acid by itself

I would see the formation of some H+ and some A-.

This number even by itself is going to be a small number.

So compared to this H+ that I get from the acid.

However because the week acid process is in equilibrium process

and we've got our equilibrium arrow there. It is an equilibrium process.

What we're going to see is that the present at the H+ from the strong

acid ionization

will actually drive the equilibrium towards the left

for the weak acid and as a result the small amount of H+ I actually had

present from dissociation the HA

will actually get smaller in this mixture with a strong acid.

So what I find is that the primary H+ source

will be the strong acid and the amount of H+ produced from that weak acid

is so small

that it becomes insignificant and I can ignore it. Now from looking at a mixture

of two or more weak acids.

I have multiple reactions going on and just like I did with the strong and weak acids

I'm gonna find the strongest have those acids. if there are guys are fairly

different from one another say we have K_a values of 10^-3

10 ^ -5 and 10 ^-9

those K_a values are far enough separated from one another

that I really only have to look at the one that has the K_a of 10^-3.

And I will do all my calculations as if that were the only one at present.

Now I will a look at an example which one of the following is associated with the acid?

Now note these all have K_a values so they are all weak acids.

So we're asking which one is the strongest week acid at these four.

Hopefully you answered the first 11.9 x 10 ^-4.

This is the largest the K values

for these four.

Therefore, it will be the strongest acid of these.

We talked a lot about acids both strong and weak. In the next module what we are

going to look at is bases.

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