In this unit we will look at how the concept equilibrium

is applied to acids and bases. Our objective is to look at two definitions

of acids and bases.

The Bronsted-Lowry definition and the Arrhenius definition

and look at specific examples for each of them.

Here we see examples acid

found in foods. We have citric acid found in fruit.

Lipoic acid found in Broccoli and Lactic acid

found in yogurt. These are just a few of the examples

where acid is present and food. We also see bases in many substances,

as many cleaners contain a base. We also see bases such as sodium bicarbonate or

calcium carbonate president tums, Rolaids

or other antacid tablets. Acids and

bases have additional properties

beyond those that define them as an acid or base according to either definition.

Acids tend to have a sour taste such as that seen in lemons or vinegar.

They can dissolve many medals and they neutralize bases.

A base tend to be bitter,

slippery, and has the ability to neutralize acid.

In fact the reaction between an acid-base

is called a neutralization reaction because it produces a salt and water.

Now the salt may have acidic or basic properties

which we will explore and the next uni. But the

acidity and basicity of that resulting salt is much less

then that of the original acid or base solutions. Here we have four acids

pictured. We see we have two binary acids meaning they contain two

elements HF in HBr and we have to

oxyacids because they are formed from the polyatomic ions of

nitrite and nitrate. However, what we see in common in all four off of these acids

is that they all have a hydrogen present that can be removed

and it is this dissociation or ionization of this hydrogen ion

that lends to the acidity at this molecule.

Arrhenius acid

is a substance that produces H+

in aqueous solution. Here we show HBr

and we show that it completely dissociate or ionizes into the H+

and Br- ions.

We represent this in the chemical reaction

by showing a one direction arrow. This indicates that the reaction goes

completely from reactants

to products. We show that we have an H+ in solution

but remember we're in aqueous solution and we're not going to see free

H+ ions floating around.

What we will see

are H_3O+ ions, which means exactly the same as H+.

So this the H_3O+ is what's truly present the solution.

The H+ is just an abbreviation, but they mean exactly the same thing.

Remembered that we also know that hydrogen

has one proton and one electron.

When it loses one electron to form H+,

all that remains is a single proton; and so we also

refer to this H+ as a proton.

Here we have an Arrhenius Base that shows KOH dissociating into K+

and OH- ions. This also is a strong base and this is indicated by the

presences of the one direction arrow

rather than an equilibrium arrow.

Now, we can look at the Bronsted-Lowry

definitions of acids and bases.

We have conjugate acid/base pairs that differ by a single

H+. If I look at NH_3, which is acting as a base because it's acting as

a proton acceptor

I look at it partner the conjugate acid at that base and I see it is NH_4+

because it has accepted

that proton. Now NH_4+ can behave as an acid because they can behave as a proton

donor. Together the NH_3 and the NH_4+

behave as a conjugate acid base pair. When I look at the next of

water, in this case, water is acting as an acid.

It acting as a proton donor

because it giving up a proton and becomes OH-.

Water and OH- are now a conjugate acid-base pair.

Water is also amphoteric and behaves as a base in some cases.

It depends on who is partnered with because when we look at defining what an

acid and what is its conjugate base and vice versa

we're looking at it in the context of a chemical reaction.

If we look at a reaction we can go through and define everything is either an acid

or base. The first thing I want to do is look at my

HF and see what substance is most similar to that on the product side.

I see I go from HF to

F-. That's going to be one to conjugate acid/base pair.

Also I see that have H_2O going to H_3O+,

and that will be my other conjugate acid/base pair.

And so what I want to do is look at

HF and F- and see how it's changing.

As I that HF is donating a proton

to something else, I'm not worried about where the proton is going, I just know

that HF is donating a proton.

We form F-, so HF is acting as an acid.

F- is now willing to accept a proton.

So it's acting as a base because they differ from one another by a single

proton, a single H+ ion,

they make a conjugate acid-base pair.

I can look at the same thing when I look at water.

Here water is accepting a proton going for H_2O to

H_3O+ therefore water is acting as a base.

When I look at H_3O+ plus, it's now willing to donate a proton.

So it's behaving as an acid. And together

H_2O and H_3O+ make a conjugate acid-base pair

because they differ by a single proton. We can

look at another reaction such as

NH_3, ammonia, plus water

in equilibrium with ammonium ion plus hydroxide ion.

We can do the same thing that we did on the previous example.

We want to identify the pair's, so I have NH_3

and NH_4+ and I have H_2O

and OH-. When I look at water in this example I see that water is

donating a proton

and becoming OH-, therefore H_2O is acting as an acid in this case.

OH- is willing to accept a proton.

So it acting as a base and together H_2O

and OH- make a conjugate acid/base pair.

I can also look at NH_3 and NH_4+ plus.

NH_3 is willing to accept a proton

which makes it a base. NH_4+

is willing to donate a proton which makes it an acid

and because NH_3 and NH_4+ differ by a single proton

they make a conjugate acid/base pair.

Now let's look at an example.

What is the conjugate acid at HSO_4-.

The answer is H_2SO_4. Because it asking for the conjugate acid of HSO_4-

it implies that HSO_4

is acting as a base. If it's acting as a base

then it is going to be a proton acceptor.

If HSO_4- accepts a proton

then it become H2SO_4.

Now, we could also look at example were HSO_4-

is acting as an asset. And when it acts as an asset it's going to act as a proton donor.

When it donates a proton it will go to SO_4^2-.

So we still have a conjugate acid/base pair HSO_4-

and H2SO_4. As well as another conjugate acid/base pair

from HSO_4 to SO_4^2-.

The differences is, are we treating HSO_4-

as a base or are we treating HSO_4- as an acid.

Next we will look at pH and K_w. This K value is the same that we look at in

the previous unit.

3.01 Acids and Bases

Advanced Chemistry

Meet the Instructors