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4.06 Part 1.c - Aqueous Equilibria Worked Example

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4.06 Part 1.c - Aqueous Equilibria Worked Example

This unit continues and expands on the theme of equlibria. You will examine buffers, acid/base titrations and the equilibria of insoluble salts.

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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Enseigné par

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

Transcription

In this problem we're going to be continuing to add sodium hydroxide

until we have added 15 millimeter of NaOH to the weak acid.

Let us look at our titration curve,

at 15 millimeters look down on the bottom of the graph.

We're still in the acidic range and we're in this area in which we

have a buffer zone that's resisting change to pH as we add sodium hydroxide.

We're going to see that Henderson–Hasselbalch equation will become

handy tool for us to calculate the pH.

Let's begin by writing the reaction.

The weak base acid is HA, the strong base I'll use and I do the proton swap.

The H+ is donated to the Leaving with A- + H2O.

Because it's a one way reaction due to the strong base,

I can use an ICF table and plug moles into the table.

I don't care about the water, but I need to know the moles of the acid and

the base that are added.

Now, the moles of the acid we obtain from part B, but

we'll go ahead and do it again here.

We have a molarity of 0.249 moles/L

x the volume of 20 mL or 0.02L.

This is going to give me a value of 4.98 times to the -3 moles.

To obtain the moles of the hydroxide,

that would be the same as the moles of sodium hydroxide.

I will have the molarity at 0.249 moles per liter times the volume.

It's 15 millilitres or 0.0150 liters.

When I multiply these numbers, I have 3.74 x 10 to minus 3 moles of a hydroxide.

Plant those values into my table 4.98 x 10 to the -3.

3.74 x 10 to the -3, and since nothing has happened yet

I have zero for the conjugate base.

Now in a one-way reaction we react until we consume the smaller quantity.

So that's the 3.74 times the -3.

I will have consumed all of that strong base and

generated 3.74 x 10 to the -3 of the conjugate base of the weak acid.

When I follow the change line through I have 1.24 x 10 to the -3,

out of 0 and a 3.74 x 10 to the -3.

Now at this point we can now recognize also that is a buffer.

We saw it on the graph, but we see it here.

We have some weak acid, and we have some present of a conjugate base,

and that is a buffer we get to this conjugate base.

So we know any time you have present at in the titration,

both the weak acid in its conjugate base.

Henderson–Hasselbalch equation would be a, I would apply.

So PH = pka+ of the log of the number moles of

the base over the number moles of the acid.

Well we'll take the -log of the ka to get the pka, that's 1.54 x 10 to the -5.

The base is the a-, so this is 3.74 x 10 to the -3.

The acid is the Ha, that's 1.24 x 10 to the -3.

And when you work through the problem with your calculator

you will obtain a value of 5.292 for the pH.

So it is acidic as we predicted it should be.

The pH has risen a little bit above where it was at 10.

What was it a 10, it was,

4.812 so it has risen some, but it's not risen a lot,

considering that we've added 5 mL of a strong base to the solution.

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