question archive Experiment 1 Calculations (9p)
Experiment 1 Calculations (9p)
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Experiment 1 Calculations (9p). You wish to determine how the liquid absorption capacity of sodium polyacrylate varies as a function of the nickel ion (Ni2+) concentration. Employing the procedure outlined in the Lab Assignments 7 & 8 Guide for Experiment 1, you obtain the following:
Table LA7-1. Liquid Absorption Capacity of Sodium Polyacrylate as a Function of Nickel Ion (Ni2+) Concentration
Concentration of Ni(NO3)2 Solution (M)
Mass of Sodium Polyacrylate (g)
Incubating Volume of Ni(NO3)2 Solution (mL)
Volume of Filtrate Collected (Unabsorbed Ni(NO3)2 Solution) (mL)
Volume of Ni(NO3)2 Solution Absorbed by Sodium Polyacrylate (mL)
Volume (mL) of Solution Absorbed per Gram of Sodium Polyacrylate
0.050
0.102
40.0
36.1
0.035
0.098
40.0
35.9
0.020
0.101
40.0
34.5
0.010
0.099
40.0
31.9
0.0050
0.094
40.0
29.1
0.0010
0.098
40.0
19.7
0.000
0.103
40.0
12.5
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Answer:
Volume of Ni(NO3)2 solution absorbed by sodium polyacrylate= Incubating volume of Ni(NO3)2 - Volume of filterate collected = 40.0 - Volume of filterate collected
Volume of solution absorbed per gram of sodium polyacrylate = Volume of Ni(NO3)2 solution absorbed (mL) / mass of sodium polyacrylate (g)
| Concentration of Ni(NO3)2 solution (M) | Volume of Ni(NO3)2 solution absorbed by sodium polyacrylate (mL) | Volume (mL) of solution absorbed per gram of sodium polyacrylate |
| 0.050 | 40 - 36.1 = 3.9 | 3.9 / 0.102 = 38.235 |
| 0.035 | 40 - 35.9 = 4.1 | 4.1 / 0.098 = 41.837 |
| 0.020 | 40 - 34.5 = 5.5 | 5.5 / 0.101= 54.455 |
| 0.010 | 40 - 31.9 = 8.1 | 8.1 / 0.099 = 81.818 |
| 0.0050 | 40 - 29.1 = 10.9 | 10.9 / 0.094 = 115.957 |
| 0.0010 | 40 - 19.7 = 20.3 | 20.3 / 0.098 = 207.143 |
| 0.000 | 40 - 12.5 = 27.5 | 27.5 / 0.103 = 266.990 |
The graph is as shown
Clearly, the liquid absorption decreases exponentially with increase in the Nickel concentration.
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