- 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 - 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
- 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
- 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
- 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