3.05a Calculating equilibrium concentrations of a weak base, part 1

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来自 University of Kentucky 的课程

高级化学

316 个评分

University of Kentucky

高级化学

316 个评分

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.

从本节课中

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

与讲师见面

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

Let's look at some examples involving equilibrium concentrations of a weak base

solution.

What is the hydroxide concentration in a 0.150 molar NH3 solution?

KB equals 1.8 times 10 the minus 5.

NH3 is a base and we know it's a weak base.

If we weren't sure, we'd see the Kb value and

that indicates that we're dealing with a weak base solution.

We're given the reaction of a aqueous ammonia plus water

is an equilibrium within NH4 plus and Minus.

Since we're dealing with a weak base,

we're going to have to set up an ICE table, where we have I for

initial, C for change and E For equilibrium.

We're given the initial concentration of NH3 is 0.150, water is a pure liquid,

so we don't have to worry about its concentration.

Initially we have no NH4+, no Present.

For the change row we have to look back at the stoichiometry of the balance chemical

equation, so we have -x for the NH3 Plus x for

NH4 plus, and plus x for Minus.

For my equilibrium row, I simply sum the initial and change rows.

So now I have my equilibrium concentrations with respect to x.

Now I can set up the Kb expression.

Kb, equal to the NH4 plus concentration

times the Minus concentration

divided by NH3 Concentration.

The water is omitted from our Kb expression because it's a pure liquid.

I now plug in the values I know, which are the Kb value is 1.8 x 10 to the -5.

And if that's not given, that's a value you can generally look up.

I have x for my NH4+ concentration, x for my Concentration and

0.150- X for the NH3 concentration.

Now I'm going to make the assumption that x is much, much less than 0.150.

This will allow me to simplify the expression

to 1.8 x 10 to the -5 equals x squared over 0.150.

Now I can solve for x, and

x=1.6 times 10 to the minus 3.

This will my hydroxide concentration.

I also see that the value of x is only about 1% of the value of 0.150.

So my assumption was valid.

And I can find my hydroxide concentration just by using the value of X.

It will also be the concentration of NH4+ since those two concentrations are equal.

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