4.05 Places In The Community Writing Assignment For Middle School

We are ready to begin our second learning objective in this unit,

and that is to calculate the pH of buffered solutions.

We are going to

learn how to do this in two different ways.

First way is using an ICE table.

So this is not a new concept

there is nothing really new to learn expect

how to apply an ICE table

in this environment. how to apply an ICE table

in this environment.

It is a similar procedure that you learned in

the last unit when when you were dealing with acid solutions or base solutions.

Then we are going to learn a new equation called the Henderson-Hasselbalch equation.

We are are going to learn how to calculate the pH using it. Then we are going to learn a new equation called the Henderson-Hasselbalch equation.

We are are going to learn how to calculate the pH using it.

Either method is fine, they both have their advantages.

The advantages of taking the energy

to learn the Henderson-Hasselbalch equation is that

it is a whole lot simpler, shorter procedure,

to calculate the pH when you utilize this equation.

So the first thing we are going to do is see this

buffer here, and we are going to calculate the pH of this buffer

using an ICE table.

So again, this is with no new method.

Now what you have to be able to do is write the

correct reaction, and this is where a lot of students struggle.

You don't react HF

and NaF with each other.

There are conjugate acid/base pairs.

They do not react with each other.

You want to choice the, They do not react with each other.

You want to choice the,

in this case the,

not the salt, but either the acid of the

base [so in this case it is an acid]

and write the reaction of that acid in water.

Now here I do things a slight different and write the reaction of that acid in water.

Now here I do things a slight different

then what you saw in the acid/base

lesson. Whenever I write

a weak acid, I do not just break it apart.

It is just another way of doing the same thing

I add my water, because I want

to be consistent I add my water, because I want

to be consistent

always having the water

as a part of the reaction.

Now the acid will donate

to the water, to leave you with F-

and H_3O+. to the water, to leave you with F-

and H_3O+.

But like I said I like to have the water there

the acid donates to the water and we have this exchange.

I will explain why I like to include that

as we proceed on down the line.

So I want to do an ICE table.

I see that is the concentration

of HF is .10 molar.

I will not include the water of HF is .10 molar.

I will not include the water

I come over to the right hand side and I will look up

and note that the amount of

NaF is .3 molar.

We know that breaks apart NaF is .3 molar.

We know that breaks apart

100% into Na+

and F- 100% into Na+

and F-

there the concentration of F- in solution

before anything happens what so ever there the concentration of F- in solution

before anything happens what so ever

is already .3.

That is the common ion.

And that is the difference between

a regular acid/base problem, and a common ion problem.

What is the difference? You will always have

Present in solution

prior to anything happening with this reaction, already some of that ion.

I will put a zero under the H_3O+.

because this reaction has not occurred yet.

There is a teeny teeny tiny amount

of H_3O+ in water There is a teeny teeny tiny amount

of H_3O+ in water

but such a small amount will not have to consider it.

So what is going to happen? We use out change lines

some of this will react

and according to the coefficients we will have equal

quantities of those substances produced.

Now I will write the equilibrium

expression for K_a, the law of mass

action. And it is F-

time H_3O+

over HF, we do not include time H_3O+

over HF, we do not include

the water because it is a liquid. over HF, we do not include

the water because it is a liquid.

We plug in what we know

K_a is 3.5 times 10 ^ -4.

And HF- 0.10 minus X.

Now at this point we are going to assume

X is very very small compared to

the .30 or the .10 .

We see that X is being

added to .30 here.

It is being subtracted from .10 here.

And if X is much much smaller then .10

and of course it is much much smaller And if X is much much smaller then .10

and of course it is much much smaller

.30 as well, then we can ignore this term

right here. Now do not ignore that X .30 as well, then we can ignore this term

right here. Now do not ignore that X

that is sitting here, we cannot get rid of it.

But we can ignore the ones that are added

or subtracted from that value.

Then I can solve for X, it will be 3.5 times 10 ^-4

time .10 and divided .30 .

This gives me a X of 1.2 times 10 ^ -4.

Now what you solve for X, you look back up and say what did they want me to determine?

They wanted me to determine the pH of this solution.

So this is the

H_3O+ concentraion. So this is the

H_3O+ concentraion.

I know that because I look up here and I see that H_3O+ concentraion.

I know that because I look up here and I see that

X is sitting right underneath that H_3O+.

So I can take the negative log

of 1.2 times 10 ^ -4

and that will equal the pH.

And the value is 3.93.

Comment on this pH, it is an acid pH

and when you make a buffer with a weak acid

typicality it is in the acid range.

So that is nothing really new. We are seeing an ICE

table, we are writing our reaction, we are solving for X.

and this is very very similar to what we saw

in the previous unit on acid/base equilibrium. and this is very very similar to what we saw

in the previous unit on acid/base equilibrium.

The only difference once again

is that there was a value

under the F

that we are starting with, and that is the common ion scenario.

So now lets learn the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation

is only usual for this common ion, or buffered solution. The Henderson-Hasselbalch equation

is only usual for this common ion, or buffered solution.

So we have got to keep that in mind. is only usual for this common ion, or buffered solution.

So we have got to keep that in mind.

You have to know, that you are dealing with a buffer

before you can use this equation.

But if you are, then you can utilize it.

Now lets talk about the pieces. First of all pK_a,

pK_A is simply the negative log of the K_a value just like

pH is the negative log of H plus concentration.

And pOH is the negative log of the OH- concentration. pH is the negative log of H plus concentration.

And pOH is the negative log of the OH- concentration.

This p function

is take the negative log of what you see after it. This p function

is take the negative log of what you see after it.

When we use this equation, it incorporates

all the steps we just saw in the previous example. When we use this equation, it incorporates

all the steps we just saw in the previous example.

It also incorporates the assumption all the steps we just saw in the previous example.

It also incorporates the assumption

that your initial concentration

is big enough so that X is small compared to it. that your initial concentration

is big enough so that X is small compared to it.

It is taking into account this fact

that we utilized in that.

This assumption is a good assumption

generally when you have a concentration of that acid.

for your buffer, or a base we will see later.

That is in the .1 range or larger.

If it gets much smaller it might not be a

great assumption and it will just get you close to the

pH, and not exactly the pH.

So lets do that previous problem that we did

but utilizing the Henderson-Hasselbalch equation.

pH equals pK_a. but utilizing the Henderson-Hasselbalch equation.

pH equals pK_a.

Which would be the negative log of

3.5 x 10 ^ -4

plus the log of 3.5 x 10 ^ -4

plus the log of

the concentration of the base is on top

so we have to find the base, this is the base

Но если не считать его изрядно устаревших представлений о рыцарстве, Дэвид, по мнению Сьюзан, вполне соответствовал образцу идеального мужчины. Внимательный и заботливый, умный, с прекрасным чувством юмора и, самое главное, искренне интересующийся тем, что она делает.

Чем бы они ни занимались - посещали Смитсоновский институт, совершали велосипедную прогулку или готовили спагетти у нее на кухне, - Дэвид всегда вникал во все детали. Сьюзан отвечала на те вопросы, на которые могла ответить, и постепенно у Дэвида сложилось общее представление об Агентстве национальной безопасности - за исключением, разумеется, секретных сторон деятельности этого учреждения. Основанное президентом Трумэном в 12 часов 01 минуту 4 ноября 1952 года, АНБ на протяжении почти пятидесяти лет оставалось самым засекреченным разведывательным ведомством во всем мире.

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