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3.06 Ions as Acids and Bases

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Química Avançada

Universidade de Kentucky

4.7 (422 classificações) | 51K alunos 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 (422 classificações)

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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!

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

Ministrado por

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

Transcrição

In this module we will be looking at how I and behave as either an acid or a base.

Our objective is to be able to determine the acidic or basic characteristics

of ionic compounds. Going back to the value of K_w that we looked at for water Our objective is to be able to determine the acidic or basic characteristics

of ionic compounds. Going back to the value of K_w that we looked at for water

we can use this for other solutions as well to understand the relationship

between the K_a value of a compound and its K_b value.

If I look at HF what I know is that it has a K_a value of 3.5 x 10 ^-4.

I can also find the K_b for its conjugate F-

by using K_w equals K_a times K_b.

I plug in the values that I know and solve for my unknown K_b value.

We typically don't report K_b values in data tables

We simply calculators from the tabulated K_a values.

Now if I am looking at how F- behaves in water in a solution.

I can use the K_b balue because F-

is behaving as a base whereas HF is behaving as an acid or proton donor.

When I look HF in water

I know that a check dissociated H+ and F-.

I can look at the K_a value for that and given the concentration of

HF I could find the H+ concentration or that pH of that solution.

When I look at something like KF or an ionic compound in water

it is ionic, it completely dissociates into K+ and F-.

So what I actually have to look at is what is that ion do.

What is F- do in water and what this K+ do in water.

If we look at the F- and how it behaves with water

we look at the potential products so they could form in equilibrium

if you go to HF plus OH-.

In this case F- is behaving as a base it is accepting a proton

so I need a K_b value for that reaction not K_a value.

If I look at K+ so I need a K_b value for that reaction not K_a value.

If I look at K+

the cation in that ionic compound and look at how it behaves with water

than what I see is that there are no reaction because the only thing that it

could form would be KOH or potassium hydroxide

which is a strong base. And strong base is completely dissociate in water.

So we're not going to see any products in this process

but for F- because it goes to the forming a weak acid

HF, I will see some other HF stick around and therefore some my F- will be but for F- because it goes to the forming a weak acid

HF, I will see some other HF stick around and therefore some my F- will be

consumed by our reaction with water.

Just as we look at in the previous module these anion

are the conjugate bases of acids.

It doesn't matter whether we're talking about binary acids are oxyacids

and whether we're talking about strong or weak acids. I can still talk about

anions as being the conjugate base\of the acid.

Conjugate bases are not always bases.

Even though we can write the conjugate have a particular acid

it doesn't mean it always has basic properties. Let's look at a couple of example.

The anion that is the conjugate base have a weak acid

is also a weak base. So HF

has a conjugate base of F-. F-reacts with water to produce

HF and OH-. Because HF is a weak acid

not all of it will completely ionized again after its formed.

As a result we will have a residual amount of hydroxide president solution.

If I look at the anion

that is the conjugate base of a strong acid

I find that it is pH neutra. HCl going to H+

and Cl- is how my HCl will

ionize or dissociate.

Cl- when reacting with water will actually produce no reaction.

The only product that could be produced as for chlorine or chloride to behave as

a proton acceptor and became HCl

but HCl is a strong acid, and will completely dissociate into the ions

and as a result I do not get a residual concentration of hydroxide

present in the solution as I did with the HF.

The six-strong acids which we covered earlier are the only six that we have to

worry about remembering.

Because anything that is not a part of the six strong acids is in fact

a weak acid. So hydrochloric hydrobromic, hydroiodic

nitric, perchloric and sulfuric (the first step to only for sulfuric)

are all strong acids. Their an anions the conjugate bases of them

will not react with water. Because the only thing they would produce

is the reforming the strong acid which then completely ionizes them in water

leaving us with no residual hydroxide concentration and therefore no basic properties.

If we look at the cation we see that they behave

as a weak acid. So the conjugate acid have a weak base

is a weak acid.

Let's look at an example with ammonia. NH_3 is a weak base.

As a base to taxes a proton acceptor. When it does that

it forms the conjugate acid of NH_3 which is NH_4+.

NH_4+ can then react with water produce H_3O+

and NH_3. The generic form for writing this

is BH+ plus water going to H_3O+ plus B.

This allows us to simplify what we write instead of worrying about the complete formula

of the anmine compound.

Just like with strong acids the cations of strong bases

will not react with water. Any of our hydroxides: lithium, sodium, potassium, calcium and others

Will have ions present in solution because they completely ionized

to form say lithium-ion and hydroxide ion.

That lithium ion or whatever cation we are looking at

will not react with water because the only possible product would then be forming

the strong base. We know that strong base completely dissociate in water

therefore we're not gonna get any residual H_3O+ present in our solution.

Let's look at an example of how we apply these ideas.

Let's look at an NaCl. Sodium chloride is a strong electrolyte and will completely

dissociate in water.

When it does I'll get sodium ions and chloride ions.

If I were to attempt to write an equilibrium reaction with chloride and water.

What I would find is my possible products or HCl and OH-.

However HCl is a strong acid and strong acids completely dissociate.

Therefore this reaction will not happen, because I will not get any HCl that hangs around

and as a result will not have any hydroxide ion what will also

stick around and change the pH of the solution.

So this reaction does not happen. The same thing occurs when I look at sodium ion reacting with water.

The only possible products are sodium hydroxide and H+.

However sodium hydroxide is a strong base and will completely dissociate in water

and so I will not get a formation have any NaOH

and this reaction doesn't happen as well. So what we have here

are a cation whose parent base was a strong base

and anion whose parent acid is a strong acid

and therefore I get a salt

that is in fact neutral. This salt, NaCl, has no acidic or basic properties.

Let's look at how we classify salt as being acidic basic or neutral.

In order to do this we're going to look at the parent acid of the anion.

So even though we talk about the ionic compound

we can still say what if this ion came from an acid

what acid did it come from? And we can look at the properties at that acid to

help us determine the acidity

or basicity have a particular ion. help us determine the acidity

or basicity have a particular ion.

Chloride comes from HCl which is a strong acid

and we know this because it's on the list of six strong acids.

ClO- comes from HClO

which is a weak acid because it's not on the list of six strong acids.

When we look at our parent basis for sodium ion we see that comes from NaOH

for NH_4+ it must have come from NH_3 as its parent base.

By understanding the parent acid parent base it will help us figure out the

acidity or basicity

other resulting salt. So let's look at some possibilities. acidity or basicity

other resulting salt. So let's look at some possibilities.

Let's look at example of how we can classify salt as acidic, basic, or neutral.

We will start with a simple example of KCl.

K+ it has the parent basis KOH. Cl- has the parent acid at HCl.

Because both the parent acid and base our strong

we will see no reaction happening with water and as a result we will get no

production of H+ ions

or hydroxide ions. Therefore if a substance has a strong parent acid

and a strong parent base, we will end up with a neutral salt.

And in fact this is what we see. If we took the pH

a KCl solution we would find that the pH

is neutral. Now

look at an good example of KF. Here we have the same

parent base, our KOH because we have a K+ ion.

It will not react with water so we don't see the production of any H+

however we see F- whose parent acid is HF

a week parent acid, F- will react with water to produce

HF and OH-. Because there's some residual

OH- present in the mixture in the solution HF and OH-. Because there's some residual

OH- present in the mixture in the solution

we will see that we have a basic salt.

Let's look at example a NH_4Cl. We have NH_4+ which has a parents base of NH_3, a weak base.

So we will get the production of some H_3O+ in solution.

Cl-, as we've seen before, has a parent acid of HCl

and therefore it will not react with water and not produce and OH- in solution.

Therefore when we combine NH_4+ and Cl- in a compound

we get that we have an acidic salt.

Now we only have one more scenario to go where we have a week parent acid

and a weak parent base. So we'll look at the example

of NH_4F. So NH_4+

will produce some H_3O+ in solution because it will

donate a proton to form some NH_3, a weak base.

F- has a parent acid of HF, so it's a weak acid.

and we will get a OH- to form. However because we have these competing factors

of some H_3O+ plus forming

and some hydroxide forming we don't know whether the salt is going to be acidic or basic.

What we will have to look at are the individual K_a and K_b values

for the two substances. What we will have to look at now

is compare the K_a and K_b values of the parent acid and base.

I'll compare the K_a value of HF to the

K_b value of NH_3. The one that is larger will dominate and will determine I'll compare the K_a value of HF to the

K_b value of NH_3. The one that is larger will dominate and will determine

the pH of the salt.

So if NH_3 has a K_b value that is greater

than the K_a of HF I will find that this is a basic salt.

Conversely if I find that the K_a value of HF

is greater than the K_b of value of NH_3 I find that I have an acidic salt.

So until I know those values I cannot determine the acidity or basicity

of a NH_4F.

Now let's look at an example. Which have these will behave as an acidic salt?

For looking at this we need to compare the parent acids and bases at each of

the ions present

in the salt. For sodium our parent base

is NaOH which is strong

NHCl which is also strong.

For KClO we have KOH

and HClO so we have a strong base

and a weak acid so this will be a basic salt.

For MgBr_2 we have MgOH_2

as a parent base. HBr

as a parent acid MgOH_2 two is strong

as is HBr so we will have a neutral salt.

Just like we did for the sodium chloride.

Which is also a neutral salt.

When I look at NH_4I, my parent base is NH_3 which is weak

my parents acid is HI, and so I will have

an acidic salt.

Because win the NH_4 reacts with water

I will get the production of some H_3O+ ions.

We can also quantified the pH of salt solutions rather than just describing them as acidic

or basic. In this case we're looking at sodium acetate.

Now sodium acetate is an ionic compound it will completely dissociate in water.

So I will have sodium ions Na+

and acetate ions.

And when I do what I find is that the sodium ions will not react with water because

their apparent bases in NaOH.

However acetate ions will react with water

and we show that reaction here. So acetate plus water

gives us acidic acid and hydroxide ions.

Now I can look and how I can calculate

that pH value by setting up a ICE table. It's an equilibrium reaction and we can

use ICE Tables anytime we're dealing with the equilibrium reactions

So I in my initial concentration of my acetate ions is 0.500

because my sodium acetate completely dissociated

and for every mole of sodium acetate I get one mole of acetate ion. So my

concentration is .5.

I don't look at the concentration at my H_2O because it is a pure liquid and we

don't talk about the concentration for pure liquid.

Acetic acid is zero and hydroxide is 0.

Now I look at my change a row, I am losing acetate I and I am forming a scenic acid

and I am forming hydroxide ions.

And I get my equilibrium row of 0.500 - X,

X, and X. Now what I need to look at is how do I figure out what the hydroxide

concentration is.

I know the K_a of acetic acid

but I need to know the K_b value of acetate.

So I have to use my K_w a value again. I know that K_a

times K_b equals K_w.

Which is 1.0 x 10 ^-14.

Now this problem didn't give us a temperature but if none is given we can

assume it at 25 degrees Celsius so this is true.

My K_a value is given as 1.8 x 10 ^-5.

I am trying to solve for K_b and I know K_w.

So I can solve this problem

and when I do I find that my K_b value is 5.6 x 10 ^ -10.

Now what I need to do is use that information

along with my law of mass action for my equilibrium process

and be able to determine the OH- concentration

which will allow me to find the pOH and then the pH value.

So have might law of mass action here, my K_b equals HA

my acetic acid in this case. Rimes my OH-

divided by the A- concentration. I don't include water because it's a pure liquid.

Now I can start playing in the values I know so I have 5.6 x 10 ^-10.

Equals HA and OH- are both equal to X so we're gonna simplify that to X^2

divided by 0.500 -X.

I am gonna make that simplifying assumption that X is much much less than

0.500.

Given the value K_b this is a reasonable assumption.

That the x value will be met very small.

Now, I can have 5.6 times 10 ^ -10

equals X squared over 0.500.

So after I multiply 5.6 x 10 ^-10 and take the square root.

I find that X, which is my hydroxide concentration

is equal to 1.7 x 10 ^ -5.

Now that I know my hydroxide I concentration

I can find the pOH the solution because I know that

negative log of 1.7 x 10 ^ -5 equals the pOH.

And I have pOH

equal to 4.8.

I can subtract that from 14

and find a pH value of 9.2.

So now I know the pH

and the pOH this solution.

In the next module will look at polyprotic acids and the different K_a values

associated with each

dissociation or ionization step.

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