3.06c Acid-base properties of aqueous salts, part 3

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Do curso por University of Kentucky

Química Avançada

333 classificações

University of Kentucky

Química Avançada

333 classificações

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.

Na lição

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

Conheça os instrutores

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

Now, we'll finish out this problem by finding the pH of a potassium

acetate solution.

What is the pH of a 0.325 molar solution of potassium acetate?

The first thing we have to do is write the reaction that represents what's going on

in this mixture.

This is the reaction that we looked at from the first slide,

all we are worried about is the reaction of acetate with water.

So, CH3COO- +

H2O in equilibrium

with CH3COOH +

Now that we have our equilibrium reaction, we can set up our ICE table.

So, we have I for initial, C for change, E for equilibrium.

We have our initial concentration is given as 0.325 molar.

Water we don't worry about.

Our initial concentration of acetic acid is 0.

Our initial concentration of hydroxide ion is also 0.

Now we look at the stoichiometry of the balanced chemical equation to determine

our change row.

So I have -x +x and +x.

And for my equilibrium row, I simply sum up the initial and change rows.

So, 0.325- x, x, and x.

Now, I have my equilibrium concentrations with respect to x.

I can use the information that I determined on the previous problem

from our Kb value.

And I can set up the Kb expression which will be HA,

and I'm using that as an abbreviation for my acetic acid, times Minus concentration

over the CH3COO- concentration.

I know that Kb is equal to 5.6 times 10 to the -10th.

And that HA and Are both x, so I have x squared over my

acetate concentration, which is 0.325 minus x.

I'm going to make the simplifying assumption that x is much much less

than 0.325.

And given the small size of my Kb value,

this is going to be a reasonable assumption.

So, I can simplify my Kb expression.

So, I end up with 5.6 times 10 to

the -10th = x squared over 0.325.

And when I solve for x, which is going to be my hydroxide concentration,

I end up with 1.3 times 10 to the -5th.

Now we can take the information that we know about the hydroxide

concentration and we can find the pH of the solution.

But first, we have to find the pOH of the solution.

So, I know that pOH = the -log of my hydroxide concentration.

So, pOH = -log of 1.3

times 10 to the -5th, and

I end up with a pOH = 4.89.

And because no temperature is given, we're going to assume 25 degree celsius.

And when I do that I know

the pH + pOH = 14, so

pH = 14- pOH, or 14,

4.89, and

we get a pH = 9.11.

Given that this is a weak base, this acetate ion is behaving as a weak base,

we expect a pH greater than seven, which is exactly what we see.

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