CHEM 1212       Module Eleven-Chapter 17   Name:____________

 

Module Eleven: Acid/Base Equilibria  Chapter 17

Possible

Actual

 

A.   Idenification of Bronsted-Lowry acids&bases

10

 

 

B.   Writing Equilibrium Constant Expressions

10

 

 

C.   Key Terms Chapter 17                                               

10

 

 

D.   Determination of pH & pOH from concentrations Problems 

5

 

 

E.   Determination of pH of weak acids/bases Problems   

10

 

 

F.   Hydrolysis Reactions              

10

 

 

G.  Common Ion Effect Problem

10

 

 

H.  Discussion Questions Chapter 17

5

 

 

I.  Determination of pH of polyprotic acids Problem

10

 

 

K.    Acid-Base Properties of Salts

5

 

 

J.   Determination of Kw from Kc     

5

 

 

M.   Multiple Choice Applications -Chapter 17                 

10

 

 

Module Eleven Total: 

100

 

 

 

Part A: Identification of Acids and Bases                       5 points

Label the acid and the base and their corresponding conjugate pairs for the following reactions:

 

 

 

 (1)         NH4 +       +       Cl 1-      <=====>       NH3     +     HCl

 

 

 

 

(2)       CH3OH     +       C-1-       <======>      HCl       +              CH3O1-

 

 

 

 

(3)      C5H5NH+1         +   H2O         <=======>    C5H5N        +        H3O+1

 

 

 

 

 

(4)      HSeO4 1-            +     NH3        <=======>      NH4 1+         +        SeO4 2-

 

 

 

 

 

(5)      HCO3 1-      +        OH 1-         <==========>     CO3 2-       +    H2O

 

 

 

 

 

 

 


 

CHEM 1212    M-11     Sample Exam

 Part B: Equilibrium Constant Expressions of Acids and Bases        5 points

Write the appropriate equilibrium constant expressions for the following reactions they represent:

 

a.       C5H5NH+1         +   H2O         çè    C5H5N        +        H3O+1

 

 

 

Ka =

 

 

b.         NH3       +   H2O                  çè         NH4 1+       +         OH 1-        

 

 

 

Kb =

 

 

c.                   H2CO3         + H2O            ç=è             HCO3 1-      +            H3O+1

 

 

 

Ka =

 

 

 

d.         HCO3 1-   +     H2O            ç==è             CO3 2-        +         H3O+1

 

 

 

Ka2 =

 

 

 

 

 

e.     C5H5N      +   H2O         ç====è   +        OH 1-    +           C5H5N+1        

 

 

Kb =

 

 

 

f.          H2O    +         H2O            ç==è           H3O 1+      +      OH 1-

 

                        Kw  =