CHM 2046C Module 11 Chapter 14 Sample Exam Answers

Part E: Equilibria of Acids and Bases Calculations 10 points

Write the appropriate equilibrium constant expression for the following reaction::

 

E-1: A solution prepared from 0.055 mol of butanoic acid dissolved in sufficient water to give 1.0 L of solution has a pH of 2.72. Determine Ka for butanoic acid. The acid ionizes at according to the balanced equation.

CH3CH2CH2CO2H(aq) + H2O H3O+1 + CH3CH2 CH2CO2-1(aq)

 

pH = 2.72 [H3O+1]= 10-2.72 = .00191 M

 

 

[CH3CH2 CH2CO2H]

[H3O+1]

[CH3CH2 CH2CO2-1]

Initial

0.055 M

0

0

Change

- 0.00191

+ 0.00191

+ 0.00191

Equilibrium

0.053

0.00191

0.00191

 

[CH3CH2 CH2CO2-1]1 [H3O+1]1

Ka = -------------------------------------

[CH3CH2 CH2CO2H]1

 


[0.00191]1 [0.00191]1 (0.00191)2

Ka = -------------------------------- = ----------- = 6.8 x 10-6

[0.053]1 0.053

 

E-2: What are the equilibrium concentrations of acetic acid, the acetate ion, and hydronium ion for a 0.10 M solution of acetic acid (Ka = 1.8 x 10-5)? What is the pH of the solution? Ionization Reaction(you write):

CH3CO2H(aq) + H2O (l) H3O+1 (aq) + CH3CO2-1 (aq)

 

 

[CH3CO2H]

[H3O+1]

[CH3CO2-1]

Initial

0.10 M

0

0

Change

- x

x

x

Equilibrium

0.10 - x

x

x

[CH3CO2-1]1 [H3O+1]1

Ka = -----------------------------

[CH3CO2H]1

 

 

Assume x in 0.10 x is so small you can neglect it. Therefore:

 

[x]1 [x]1 x2

1.8 x 10-5 = ------------ = --------

[0.10-x]1 0.10-x

0.10 (1.8 x 10-5) = x2

 

(1.8 x 10-6) = x2 x = 1.34 x 10-3

 

Assumption: 0.10 - .00134 = 0.10 assumption holds

 

 

 

[CH3CO2H]

[H3O+1]

[CH3CO2-1]

Initial

0.10 M

0

0

Change

- .00134

0.00134

0.00134

Equilibrium

0.10

0.00134

0.00134

 

pH = -Log [H+] = -Log [0.00134] = 2.87

 

Read Section 14.8 and 14.9. Follow the seven steps on pages 560-562.

Study the worked example 14.10 pages 562-564. Then try problems 14.14 and 14.15. Try the problems at the end of the chapter #14.66-14.75 page 588.

 

Part F: Hydrolysis Calculations 10 points

What is the pH of the bleach solution which is 5.25% by weight Sodium hypochlorite, NaClO, assume density of bleach is 1.0 g/mL?

The Ka of Hypochlorous acid is 3.5 x 10-8

 

5.25 g NaClO

 

1 mole NaClO

 

1.0 g solution

 

1000 mL

 

 

-------------

X

--------------

X

----------------

X

-------------

=

0.70M

100 g solution

 

74.6 g NaClO

 

1.0 mL solution

 

1 L solution

 

 

 

NaClO (s) Na1+(aq) + ClO1- (aq)

 

Na1+(aq) + HOH No reaction (salt of a strong base)

 

ClO1- (aq) + HOH HClO + OH1-

 

Kw [HClO] [OH 1-]

Kh = ---- = ----------------------

Ka [ClO1-]

 

1.0 x 10-14 [HClO] [OH 1-]

2.9 x 10-7 = -------------- = ----------------------

3.5 x 10-8 [ClO1-]

 

 

[HClO]

[OH-1]

[ClO-1]

Initial

0

0

0.70

Change

x

x

- x

Equilibrium

x

x

0.70-x

 

[HClO] [OH 1-] x ∙x

2.9 x 10-7 = --------------------- = ---------

[ClO1-] 0.70-x

Assume x in 0.70 x is so small you can neglect it. Therefore

(0.70)( 2.9 x 10-7) = x2

2.0 x 10-7 = x2

4.5 x 10-4 = x

Assumption correct: 0.70 0.00045 is so small, that x is 0.70

pOH = - log(4.5 x 10-4 ) = 3.34 pH = 14.00-pH pH= 14.00-3.34 = 10.66

 

 

Read Chapter 14 section 14.14 page 572-576. Study the worked example 14.15 page 574 and the worked example 14.15 page 575. Look at the problems 14.90-14.95 page 589.