Cliffs AP Chemistry, 3rd Edition

1. A solution from Part I was determined to have a pH of 1.00. Determine the pOH, the
   [H 3 O+] and the [OH–].
   pOH = 14.00 −pH = 14.00 −1.00 = 13.00
   [H+] = 10–pH= 10–1.00= 0.100 M
   [OH–] = 10–pOH= 10–13.00= 1.00 × 10 –13M


2. A solution from Part I had an [OH–] of 2.00 × 10 –12M. Determine the pH.

pOH = −log [OH–] = −log (2.00 × 10 –12) = 11.70

pH = 14.00 – pOH = 14.00 – 11.70 = 2.30

3. The pH of a 0.100 M acetic acid solution in Part II was found to be 2.87. Calculate the Ka
   of acetic acid.
   [H 3 O+] = 10–2.87= 1.35 × 10 –3M at equilibrium
   HAc()aq++H O 23 ()lEAc()-+aq H O()aq

HAc Ac– H 3 O+

I 0.100 M 0 ~0

C −0.00135 +0.00135 +0.00135

E ~0.099 0.00135 0.00135

.

..

K.

HAc

Ac H O

0 099

0 00135 0 00135

a 18 10

(^35)
== =#
-+
^^hh-
6
77
@
AA

4. Calculate the hydrolysis constant Khfor the ammonium chloride solution from Part III.
   Assume that [NH 4 +] is the same as the initial concentration of ammonium chloride.
   From collected data, pH = 5.13, so [H 3 O+] = 10–5.13= 7.4 × 10 –6M

NH 42 ++H OENH 33 +H O+

NH 4 + NH 3 H 3 O+

I 0.100 M 0 ~0

C −7.4 × 10 –6 +7.4 × 10 –6 +7.4 × 10 –6

E ~0.100 7.4 × 10 –6 7.4 × 10 –6

4.

..

K.

NH

NH H O

0 100

74 10 74 10

h 55 10

33 ##^6610

== =#

+--

__ii

7

6 7

A

@ A

5. Calculate the theoretical pH of the original buffer solution, assuming the volume change
   due to the addition of the solid is negligible.

0 200. L#0 250. mol HAcL =0 0500. mol HAc

.

..

g NaAc

g NaAc

mol NaAc

mol NaAc

mol Ac mol Ac

1 1

400

82 034

(^11)
##=0 0488

Laboratory Experiments