Cliffs AP Chemistry, 3rd Edition

(singke) #1

  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


  1. 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



  1. 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


  1. 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










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