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The Chemistry of Acids and Bases
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K gives the ratio of ions (split up) to molecules (dont split up)

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Slide 44

Ionization Constants for Acids/Bases

Ionization Constants for Acids/Bases

Acids

Conjugate

Bases

Increase strength

Increase strength

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Slide 45

Equilibrium Constants for Weak Acids

Equilibrium Constants for Weak Acids

Weak acid has Ka < 1

Leads to small [H3O+] and a pH of 2 - 7

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Slide 46

Equilibrium Constants for Weak Bases

Equilibrium Constants for Weak Bases

Weak base has Kb < 1

Leads to small [OH-] and a pH of 12 - 7

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Slide 47

Relation of Ka, Kb, [H3O+] and pH

Relation of Ka, Kb, [H3O+] and pH

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Slide 48

Equilibria Involving A Weak Acid

Equilibria Involving A Weak Acid

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

Step 1. Define equilibrium concs. in ICE table.

[HOAc] [H3O+] [OAc-]

initial

change

equilib

1.00 0 0

-x +x +x

1.00-x x x

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Slide 49

Equilibria Involving A Weak Acid

Equilibria Involving A Weak Acid

Step 2. Write Ka expression

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

This is a quadratic. Solve using quadratic formula.

or you can make an approximation if x is very small! (Rule of thumb: 10-5 or smaller is ok)

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Slide 50

Equilibria Involving A Weak Acid

Equilibria Involving A Weak Acid

Step 3. Solve Ka expression

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

First assume x is very small because Ka is so small.

Now we can more easily solve this approximate expression.

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Slide 51

Approximating

Approximating

If K is really small, the equilibrium concentrations will be nearly the same as the initial concentrations.

Example: 0.20 x is just about 0.20 if x is really small.

If the K is 10-5 or smaller (10-6, 10-7, etc.), you should approximate. Otherwise, you have to use the quadratic.

Slide 52

Equilibria Involving A Weak Acid

Equilibria Involving A Weak Acid

Step 3. Solve Ka approximate expression

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

x = [H3O+] = [OAc-] = 4.2 x 10-3 M

pH = - log [H3O+] = -log (4.2 x 10-3) = 2.37

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Slide 53

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