Cliffs AP Chemistry, 3rd Edition

solutions were labeled X, Y and Z respectively. Each of the solutions were diluted to a total
volume of 100.00 mL with distilled water. The pH readings of these solutions, obtained
through the use of a calibrated pH meter, were (X) 6.50, (Y) 6.70 and (Z) 7.10.

Analysis:

1. Convert the pH of each solution to an equivalent H 3 O+concentration.
   −pH = log 10 [H 3 O+]

Step 1: Enter pH

Step 2: Change sign by pressing +/−key

Step 3: Press “2nd” then “log” or “inv” then “log”

X: 3.16 × 10 –7M H 3 O+

Y: 2.00 × 10 –7M H 3 O+

Z: 7.94 × 10 –8M H 3 O+

2. For each solution, calculate the number of moles of acid added.
   The moles of acid was constant for each solution,
   ..
   .

mL

mL

L

L

moles HA mol HA

1

50 00

1000

1

1

###0 100 =500 10-^3

3. For each solution, calculate the number of moles of OH–added.

X:

..
.

mL

mL

L

L

mol OH mol OH

1

12 00

1000

(^1) 750 10
900 10
(^24)

=

Y:

..
.

mL

mL

L

L

mol OH mol OH

1

18 50

1000

(^1) 750 10
139 10
(^23)

=

Z:^3

..^2
.

mL

mL

L

L

mol OH mol OH

1

35 00

1000

(^1) 750 10

=269 10

4. For each solution, determine the concentration of HA and of A–ion present at equilibrium.

X: HA =initial moles HA moles OH addedvolume sol n- '

• 6@

2

.

HA ...

0 10000L M

=5 00 10##--^34 - 9 00 10 =410 10# -

6@

moles A–formed = moles OH–added

-3^4

.

A..

L

mol A M

0 10000

==900 10# --900 10# -

7A

Y: HA =initial moles HA moles OH addedvolume sol n- '

• 6@

(^332)
.
HA ...
0 10000L M
=5 00 10##---1 39 10 =361 10# -
6@
moles A–formed = moles OH–added

-^32
.
A..
L

mol A M

0 10000

==139 10# --139 10# -

7A

Laboratory Experiments