Experiment 8: Separation of Cations:
Jesse Diehl
Purpose:
Using two techniques (masking and ion exchange resins) separate, as a lab group, ions from unwanted
aqueous ions.
Procedure:
1-Obtain unknown sample.
2-Sample is titrated with 0.1M NaOH to find the concentration of H+ ions.
2-Pass sample through a cation exchange resin to convert or replace metal ions with H+ ions.
4-Titrate new sample with 0.1M NaOH.
5-Adjust another sample to pH=10 and titrate with 0.1M EDTA to find concentration of magnesium and
zinc.
6-Adjust another sample to pH=10 and add 1 gram of KCN with zinc masking agent to sample. Titrate
with 0.1M EDTA to endpoint.
Data:
-Standardizing NaOH:
Trial:
1
2
3
Average:
KHP(g):
0.5088
0.5050
0.5133
0.5090
mol KHP:
0.00249
0.00247
0.00251
0.00249
mL required:
24.61
24.30
24.62
24.51
-Titration with CaCl2:
Trial:
1
2
Average:
mL Required:
5.74
6.74
6.24
M EDTA:
0.2200
0.1876
0.2038
pH:
10.0
10.1
10.05
M NaOH:
0.1012
0.1016
0.1019
0.1016
-Titration of unknown with EDTA:
Trial:
1
2
3
Average:
mL required:
15.52
14.69
14.38
14.86
-Titration of H+ ion with NaOH:
Trial:
1
2
3
Average:
mL required:
11.95
12.10
11.95
12.00
-KCN:
Trial:
1
2
3
Average:
KCN (g):
0.9938
1.0248
1.0069
1.0085
mL Required:
4.54
4.83
4.61
4.66
-NaOH titration with Resin:
Trial:
1
2
3
Average:
mL Required:
10.90
11.52
11.18
11.20
Calculations:
-[Mg2+] with KCN:
4.54𝑚𝐿 ×
0.2039 𝑚𝑜𝑙
1𝐿
[𝑀𝑔2+ ] =
9.257×10−4 𝑚𝑜𝑙
0.001𝐿
×
1𝐿
1000𝑚𝐿
×
1 𝑚𝑜𝑙 𝐶𝑎2+
1 𝑚𝑜𝑙 𝐸𝐷𝑇𝐴
= 9.257 × 10−4 𝑚𝑜𝑙 𝑀𝑔2+
= 0.9257
-[Zn2+] with Sample:
15.52𝑚𝐿 𝐸𝐷𝑇𝐴 ×
0.2039 𝑚𝑜𝑙
1𝐿
×
1𝐿
1000𝑚𝐿
×
1 𝑚𝑜𝑙 𝐶𝑎2+
1 𝑚𝑜𝑙 𝐸𝐷𝑇𝐴
= 0.00316𝑚𝑜𝑙
𝑚𝑜𝑙 𝑍𝑛2+ = 9.257 × 10−4 − 0.00316 = 0.002239
[𝑍𝑛2+ ] =
0.002239 𝑚𝑜𝑙
0.001𝐿
= 2.239
Conclusion:
Through the different techniques employed to discover wanted ions in solution, it is unclear if successful
or not. The cation exchange column had issues with its manufacturing during class. The other
procedures were fairly successful.