4.06 Part 1.h Aqueous Equilibria Worked Example

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From the course by University of Kentucky

Advanced Chemistry

345 ratings

University of Kentucky

Advanced Chemistry

345 ratings

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.

From the lesson

Aqueous Equilibria

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

4.01 Buffers and the Common Ion Effect10:01

4.02 pH of Buffer Solutions14:04

4.02a Aqueous Equilibria Worked Example 13:58

4.03 Buffer Action25:21

4.03a Aqueous Equilibria Worked Example 25:22

4.03b Aqueous Equilibria Worked Example 35:11

4.04 Buffer: Preparation and Capacity8:45

4.05 Strong Acid - Strong Base Titration17:20

4.06 Titrations Involving Either a Weak Acid or a Weak Base36:45

4.06 Part 1 Aqueous Equilibria Worked Example3:46

4.06 Part 1.a - Aqueous Equilibria Worked Example4:06

4.06 Part 1.b - Aqueous Equilibria Worked Example4:55

4.06 Part 1.c - Aqueous Equilibria Worked Example4:21

4.06 Part 1.f Aqueous Equilibria Worked Example8:01

4.06 Part 1.h Aqueous Equilibria Worked Example3:57

4.06 Part 2.a Aqueous Equilibria Worked Example8:25

4.06 Part 2.b Aqueous Equilibria Worked Example5:53

4.07 Polyprotic Acid Titrations5:53

4.08 Indicators10:19

4.09 Solubility Equilibria4:50

4.10 Molar Solubility and the Solubility Product10:37

4.10.a Aqueous Equilibria Worked Example2:20

4.11 Molar Solubility and the Common Ion Effect7:20

4.11 2.b - Aqueous Equilibria Worked Example4:42

4.11 2.c Aqueous Equilibria Worked Example3:45

4.12 The Effect of pH on Solubility7:07

4.12.b Aqueous Equilibria Worked Example3:42

4.12c Aqueous Equilibria Worked Example3:30

4.13 Precipitation Reaction and Selective Precipitation15:03

4.13a Aqueous Equilibria Worked Example9:05

Meet the Instructors

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

Here at the 25 milliliters of added sodium hydroxide,

we have gone beyond the equivalence point.

Let's look at a titration curve.

At 25 milliliters, we're here at this point.

And if you read up, you see that we have passed up the equivalence

point by 5 milliliters and the pH shoots up very rapidly.

A few extra drops of sodium hydroxide and now we have a strong base

in the solution with nothing for it to react with and pH will skyrocket.

So, let's go back to our reaction that we have been writing for

every point of this titration curve,

H- plus H2O is my reaction.

We will do an ICF table as has been the case in every step,

we're not changing the amount of the acid.

We've had 4.98 times 10 to the -3 moles every time.

For the number of moles of the we'll take the molarity of sodium hydroxide,

which is 0.249 moles per liter times the volume.

And now it's 0.025 liters and

this is going to give me 6.225 times

10 to the minus 3 moles of hydroxide.

225 times 10 to the minus 3, 0 don't care.

Smallest one now is the HA, so

we're subtracting 4.98 times 10 to the minus 3 of each of these.

And we're producing 4.98 times 10 to the -3 of this.

This is going to consume all of the week acid.

It is going to leave 1.25 times 10 to the -3 of the hydroxide.

And we will have generated 4.98 times 10 to the -3 of the the A- ion.

Now at this point,

I see a lot of students who want to use the Henderson Hozbog equation

because they're used to seeing if there's something on both sides to use that.

But the Henderson Hozbog equation is good for a buffer and this not a buffer.

It does not have a weak acid in its conjugate base.

What does it have?

It has a strong base, a strong base and a weak base in solution.

Now if you've got a mixture of both a the strong base and a weak base, you can most

likely without any trouble just base your problem entirely upon the strong base.

So what we need to know is,

what is the concentration of After these have reacted and have been added together?

We have moles of 1.25 times 10 to the -3 and

we have a volume of, if we use this 25

milliliters plus the 20 milliliters because it's the total volume in there.

So that 45 milliliters or 0.045 liters.

This is going to give me the moles of 0.0278.

I'm sorry not moles but molarity of the hydroxide.

So then we can get the POH and the POH will be pretty high.

It's negative log of this value and

you take the negative log of that you'll have 1.556 as the POH and

in the pH would certainly be 14 minus that 1.556 and

that would be a pH of 12.444.

So we add the extra hydroxide, the pH rises rapidly,

you can base the whole calculation on simply the strong base amount.

We didn't even considered this at all.

It contributes too few hydroxides to even worry about.

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