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.