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   ó    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)  ó    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) à  Na1+(aq)  + ClO1- (aq)

 

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

 

ClO1- (aq) + HOH    ßà    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.