Z:

• HA volume sol n'
  6@=initialmoles HA-molesOHadded

(^332)
.
HA ...
0 10000L M
6@=5 00 10##---2 63 10 =237 10# -
moles A–formed = moles OH–added
(^32)
.
A..
L
mol A M
0 10000

• ==263 10# --263 10# -
  7A

5. Determine the reciprocal of the A–ion concentration for each solution.

X:^21

.

.M

900 10

(^1) 111 10

-^3 =

--
Y:
.
(^1).
139 10
2 719 10^11

• = # M-

Z:

.

(^1).
263 10
2 380 10^11

• = # M-

6. Plot the reciprocal of the A–ion concentration on the ordinate against the H 3 O+ion con-
   centration on the abscissa. Draw a straight “best-fit” line (see Figure 1).

Figure 1

7. Determine the slope of the line.
   /
   .
   ..

Slope ..

x

y

HO

A

∆

∆

∆

∆ 1

308 10

3 16 10 7 94 10

1 11 10 3 80 10

3

8

78

21

##

== = ## #

• 
  


• =

• --
  7 __ii

7

A

A

8. Determine the initial concentration of the acid, HA.

(^32)
.

..
L

mol HA M

0 10000

==500 10# - 500 10# -

9. Determine the Kafor the weak acid.

slope /A

HO

HO A

∆

1 ∆

1

3

==- 3 :

+

__+-ii

7

7

A

A

8

..

K.

HA slope

(^1) M
308 10
1
500 10
a * 2 8 649 10
: #:#
==- =# -
_i

• = concentration of HA prior to dissociation or neutralization
  120
  100
  80
  60
  40
  20
  0
  123
  1/[A-] vs. H 3 O+
  1/[A–]
  (M–1)
  [H 3 O+] x 10-7 M
  Part III: AP Chemistry Laboratory Experiments