Chemistry 152 Chemistry Worksheet With Answers Page 5
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1
2
3
4
5
6
7
8I
0.002125
0.01750 moles
0.015 moles
0
Neutralized -0.002125
-0.002125
+0.002125
+0.002125
Left
0
0.015375
0.017125
0.002125
(
)
16.42 A 20.0 mL sample of 0.125 M HNO
is titrated with 0.150 M NaOH. Calculate the pH
3
for at least five different points throughout the titration curve and make a sketch of the curve.
Indicate the volume at the equivalence point on your graph.
Initially, it is all just acid. Nitric acid is a strong acid.
pH = -log (0.125 M) = 0.90
The equivalence point is at:
0.125 M (20.0 mL) =0.150 M (x mL)
x=16.67 mL
Since it is a strong acid/strong base titration, the pH=7 at 16.67 mL
Between 0 mL and 16.67 mL added, it is just unneutralized strong acid, so I just subtract the
moles of base added from the original moles of acid.
Between 16.67 mL and infinity, it is just excess base, so I subtract the original moles of acid
from the moles of base added.
Hey look, Excel did all the work for me!
mmol
mL
mmol
mol
mL Base
[base]
base
acid
[acid]
acid
acid
Total
[acid]
added
original
added
original
original
original
left
volume
remaining
pH
0
0.15
0
20
0.125
2.5
2.5
20
0.125
0.90309
1
0.15
0.15
20
0.125
2.5
2.35
21 0.1119048 0.951151
5
0.15
0.75
20
0.125
2.5
1.75
25
0.07 1.154902
10
0.15
1.5
20
0.125
2.5
1
30 0.0333333 1.477121
12
0.15
1.8
20
0.125
2.5
0.7
32
0.021875 1.660052
14
0.15
2.1
20
0.125
2.5
0.4
34 0.0117647 1.929419
15
0.15
2.25
20
0.125
2.5
0.25
35 0.0071429 2.146128
16
0.15
2.4
20
0.125
2.5
0.1
36 0.0027778 2.556303
16.5
0.15
2.475
20
0.125
2.5
0.025
36.5 0.0006849 3.164353
16.55
0.15
2.4825
20
0.125
2.5 0.0175
36.55 0.0004788 3.319849
16.6
0.15
2.49
20
0.125
2.5
0.01
36.6 0.0002732 3.563481
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