Monique Gray
Experiment 8
Quantitative Analysis
Experiment 8: Separation of Cations
Introduction
The purpose of this experiment is for students to learn how an ion-exchange resin works and how to
make an ion exchange resin. Help students to understand the masking and ion-exchange technique and
understand the kind of information it could bring forth in the analytical chemistry world.
Procedure
-
Made a standard solution of 0.1M NaOH, which was standarded with KHP and 0.01 EDTA, which was
standardized with CaCO3
Set-up an ion exchange resin using Dowex Powder to make a slurry mix. Glass wool was placed in the
buret first, and then slurry was poured through. The column was washed with 0.1M HCl and then distilled
water. We then ran 10mL aliquots of unknown through resin. When sample was poured through the
resin, the H+ ions were replaced by metal cations. Thus, only the H+ came off into the flask and samples
were collected and titrated with 0.1M NaOH. This direct titration produced the concentration of H +ions in
the unknown. The 10mL of unknown was titrated with standardized EDTA to find the concentration of
Zn2+ and Mg2+. Three additional sample was calibrated to a pH of 10 with ammonia buffer and
approximately 1g of KCN was added to solution and titrated with EDTA to find the Mg 2+ concentration.
Data
KHP Masses:
1- 0.05088g – moles 0.00249
2- 0.5050g – moles 0.00247
3- 0.5133g – moles 0.00251
Trail #
1
2
3
Average Volume = 11.23mL
Amount of Sample (mL)
1mL
1mL
1mL
NaOH Added (mL)
10.9mL
11.52mL
11.28mL
Standardizing NaOH with KHP
Trial #
1
2
3
Average Molarity = 0.1016
Average Volume Added = 24.51mL
(mL) NaOH Added
24.61mL
24.30mL
24.62mL
Molarity of NaOH
0.1012M
0.1016M
0.1020M
Monique Gray
Experiment 8
Direct Titration with H+ ion and NaOH
Trial #
1
2
3
Average Volume = 12.0 mL
Volume NaOH Added (mL)
11.95mL
12.10mL
11.95mL
Calculations [H]
0.01123 𝐿 0.1016𝑀
1 𝑚𝑜𝑙 𝐻 +
1
𝑥
𝑥
𝑥
= 1.140𝑀 𝐻
1
1
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 0.001𝐿
Moles NaOH
0.1016𝑀 𝑥 0.0120𝐿 = 0.00122𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
Molarity of Hydrogen ions in solution
0.00122 𝑚𝑜𝑙
= 1.22𝑀 𝐻
0.001𝐿
Standardzing EDTA with CaCl2
Trial #
1
2
3
Average volume = 6.24mL
Added EDTA (mL)
5.74mL
6.74mL
6.24mL
Molarity
0.220M
0.1876M
0.1020M
Average Molarity = 0.2038 M
Mass of CaCO3
1- 0.0674g
2- 0.0678g
3- 0.0670g
Titration with unknown Sample
Trial #
1
2
Amount Added (mL)
15.52mL
14.69mL
pH
10.0
10.1
10.0
Monique Gray
Experiment 8
3
Average = 14.86mL
14.38mL
Trial #
1
2
3
Average volume added = 4.49mL
Grams of KCN
0.9938g
1.0248g
1.0069g
Volume EDTA added (mL)
4.54mL
4.83mL
4.101mL
[Mg]:
4.49𝑚𝐿 𝑥
0.100𝑀
1𝐿
1 𝑚𝑜𝑙 𝑀𝑔
𝑥
𝑥
= 0.000449 𝑚𝑜𝑙
1𝐿
1000𝑚𝐿 1 𝑚𝑜𝑙 𝐸𝐷𝑇𝐴
0.000449 𝑚𝑜𝑙
= 0.449𝑀
0.001𝐿
[Zn]:
14.86𝑚𝐿 𝑥 0.1000 𝑥
1𝐿
1 𝑚𝑜𝑙 𝑍𝑛
𝑥
= 0.001486 𝑚𝑜𝑙 𝑍𝑛
1000𝑚𝐿 1 𝑚𝑜𝑙 𝐸𝐷𝑇𝐴
0.001486 𝑚𝑜𝑙
= 1.486𝑀
0.001 𝐿
Conclusion
We found that the following concentration existed in the unknown solution: [H] = 1.22M, [Mg] =
0.449M, and [Zn] = 1.486. There was a lot of things we could have done better. For one, given that fact
that it was a class experiment, the results may have been hindered due to the fact that more than one
group was working on it. Thus, this could have caused error due to the inconsistency. Also, not
everybody is as precise and accurate as someone else. There could have been some math error
associated with the calculations that could have caused some error.