Matter & Motion: Winter 2001

Acid/Base Workshop Sheet Stuff

Consider a base, B. B is a weak base. B is a good base. A very good base.

1a) write the reaction that will occur when this base reacts with water.

B + H2O « BH+ + OH-

b) write the reaction that will occur when B reacts with nitric acid. What products are generated?

B + H3O+« BH+ + H2O

Note: since nitric acid is a strong acid, B will actually be reacting with hydronium, the ionization product of nitric acid in water. Another possible answer to this question, one that emphasizes the nitrate is:
B + HNO3 « BH+ + NO3-
No matter which way it is viewed however, one product is the conjugate acid, BH+. c) What is the conjugate acid of B? BH+
 

d) Imagine 0.05 mol of B are dissolved in 1.00 L of water. If the pH of the solution is 11.63, what is the Kb of B? What is the Ka of its conjugate acid?

The reaction of B with H2O is given above. The equilibrium constant, Kb, is:
   
concentration
B
BH+
OH-
initial
0.05
0
0
change
-x
x
x
equilibrium
0.05 - x
x
x

  Since you know the pH, you can calculate [H3O+] and, therefore, [OH-], which is equal to x. Substituting into the definition of Kb gives:


 
 


 
 


e) Use your result from d to determine the pH of a 0.020 M solution of the conjugate acid of B. The reaction is: BH+ + H2O «B + H3O+  
concentration
BH+
B
H3O+
initial
0.02
0
0
change
-x
x
x
equilibrium
0.02- x
x
x

x = 7.07 ´ 10-7.

pH = -log [H3O+] = 6.15


f) What would be the pH of a solution that contained 0.025 mol of B and 0.005 mol of its conjugate acid in 500 mL?
 
 




g) What would be the pH of the above solution if some joker poured 100 mL of 0.1 M HCl into it?
 
 





2. Imagine a diprotic base, B2-. Write the first and second base ionization reactions of B2-. If Kb1 = 3.5 ´ 10-4 and Kb2 = 2.0 ´ 10-8, what would be the values of Ka for the species H2B?

B2- + H2O « HB- + OH-            K = Kb1

HB- + H2O « H2B + OH-            K = Kb2

The reactions of H2B are:
H2B + H2O « HB- + H3O+            K = Ka1 = Kw/Kb2 = 5.0 ´ 10-7

HB- + H2O « H2B + OH-            K = Ka2 = Kw/Kb1 = 2.9 ´ 10-11



a) what would be the pH of a 50 mL sample of 0.100 M H2B?

Treat as a monoprotic acid with Ka = 5.0 ´ 10-7


 

x = [H3O+] = 2.2 ´ 10-4             pH = 3.65


b) what would be the pH after 25 mL of 0.100 M NaOH are added to the solution above?
 


c) After 50 mL are added?

This is a solution of the amphoteric compound, HB+.


d) After 75 mL are added?
 




e) After 100 mL are added?
 

Treat as a monoprotic base with Kb = 3.5 ´ 10-4

x = [OH-] = 3.23 ´ 10-3             pH = 11.51


f) After 125 mL are added?
 

25 mL excess 0.100 M NaOH have been added. Thus, [OH-] = 0.0143; pH = 12.15.