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Acids and Bases. Arrhenius Acids and Bases
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1. What is the pH of the solution?

2. What percent of the parahydroxybenzoic acid has ionized in this solution?

Slide 29

D. Comment on Polyprotic Acids

D. Comment on Polyprotic Acids

Example: H3PO4

Ka’s = 7.5 x 10-3, 6.2 x 10-8, 3.6 x 10-13

1. Show the three acid ionization equations with their accompanying Ka values.

2. Which is the strongest acid?

3. How would you go about calculating the pH of a given H3PO4 solution?

Slide 30

Weak Base Equilibrium Rxns

Weak Base Equilibrium Rxns

Base Ionization Constant

(base will accept proton from H2O, forming OH-)

1. Definition

Give the base ionization constant expression for the base: B(aq) + H2O(l) ⇌ BH+(aq) + OH-(aq)

Kb = ?

Slide 31

Weak Base Equilibrium Rxns

Slide 32

2. Relative Strength of Weak Bases

2. Relative Strength of Weak Bases

a. The Kb for ammonia is 1.8 x 10-5.

The Kb for phosphate ion is 2.8 x 10-2.

Which is the strongest base?

b. Which will have a higher pH:

a 1.0 M ammonia solution, or

a 1.0 M phosphate ion solution

Slide 33

B. Kb From Equilibrium Concentrations

B. Kb From Equilibrium Concentrations

The pain killer, morphine, is a weak base. A 0.01 M morphine solution has a pH of 10.1. Calculate the Kb for morphine.

Slide 34

C. pH Calculations from Kb of Weak Base

C. pH Calculations from Kb of Weak Base

The weak base methylamine (CH3NH2) has a Kb of 5.0 x 10-4. Answer the following questions for a 0.080 M aqueous solution of methylamine.

a. Write the chemical equation.

b. Write the base ionization equilibrium expression.

c. Calculate the [OH-], pOH, and pH for solution.

Slide 35

Relationship of Ka and Kb

Relationship of Ka and Kb

A. Conjugate Acid and Base Reaction May Be Written in Both Directions. For Example:

HA + H2O ⇌ H3O+ + A- Ka =

A- + H2O ⇌ HA + OH- Kb =

** Reaction will go in direction of stronger acid and base to the weaker acid and base.

** Compare Ka and Kb to decide direction.

Slide 36

B. Mathematical Relationship of Ka and Kb

B. Mathematical Relationship of Ka and Kb

(for previous equations)

Thus:

Ka x Kb = [H+] [OH-] = Kw

Ka x Kb = 1 x 10-14

and pKa + pKb = 14

Slide 37

C. Example Problems

C. Example Problems

Consider the dihydrogen phosphate ion, H2PO41-. Ka = 6.2 x 10-8.

1. What is its conjugate base?

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