AP Chemistry Unit 8- Homework Problems Ka and Kb Acid & Base ...

3. For the reaction:
NH 3 (aq) + H 2 O (l) NH 4 +1 (aq) + OH -1 (aq)
In 0.0180 M NH 3 at 25 o C, the [OH -1 ] is 5.6x10 -4 M.
a) Write the equilibrium-constant expression for the reaction above.
b) Determine the pH of 0.0180 M NH 3
c) Determine the value of K b for NH 3
d) Determine the % ionization of NH 3
e) In an experiment, a 20.0 mL sample of 0.0180 M NH 3 was placed in a flask and
titrated to the equivalence point and beyond using 0.0120 M HCl
i) Determine the volume of 0.0120 M HCl that was added to reach the equivalence pt
ii) Determine the pH of the solution in the flask after a total of 15.0 mL of 0.0120 M
HCl was added
iii) Determine the pH of the solution in the flask after a total of 40.0 mL of 0.0120 M
was added.
4. For the reaction:
C 6 H 5 NH 2 (aq) + H 2 O (l) C 6 H 5 NH 3 +1 (aq) + OH -1 (aq)
a) Write the equilibrium constant expression, K b , for the reaction above.
b) A sample of C 6 H 5 NH 2 is dissolved in water to produce 25.0 mL of a 0.010 M solution.
The pH of the solution is 8.82. Calculate K b .
c) The solution prepared in part (b) is titrated with 0.10 M HCl. Calclulate the pH of the
solution when 5.0 mL of the acid has been added.
d) Calculate the pH at the equivalence point of the titration in (c)
e) The pK a values for several indicators are given below. Which is most suitable for this
titration? Justify.
Erythrosine pK a = 3
Litmus pK a = 7
Thymolphthalein pK a = 10
5. Hypochlorous acid, HOCl, is a weak acid. The K a for HOCl is:
K a = [H 3 O +1 ][OCl -1 ] = 3.2x10 -8
[HOCl]
a) Write a chemical equation showing how HOCl behaves as an acid in water.
b) Calculate the pH of a 0.175 M solution of HOCl
c) Write the net ionic equation for the reaction between HOCl and NaOH
d) In an experiment, 20.00 mL of 0.175 M HOCl is titrated with 6.55 mL of 0.435 M NaOH
i) Calculate the number of moles of NaOH added
ii) Calculate the [H 3 O +1 ] in the flask after the NaOH has been added
ii) Calculate the [OH -1 ] in the flask after the NaOH has been added