Group A Cation Analysis Prestudy An unknown solution containing Group A and Grou
ID: 493225 • Letter: G
Question
Group A Cation Analysis
Prestudy
An unknown solution containing Group A and Group B cations is treated according to the lab procedure. At each stage below, state what each test tells you about which cations may be present, which are confirmed present, which are absent, and why.
The solution is treated with NH3 until the pH is 9 to give a colored precipitate. The precipitate was treated with NaOH-H2O2 and the mixture centrifuged and separated to give a yellow solution and a precipitate. The yellow solution was saved for the Group B cation analysis. The remaining precipitate was dissolved in HCl and the clear solution analyzed as follows.
Four drops of the clear solution from step (a) are treated with NaBiO3 to give a purple solution.
Two drops of the solution from step (a) are treated with 6M NaOH and solid SnC12 to give a black solid.
A portion of the solution from step (a) is treated with KSCN and the solution remains clear.
Explanation / Answer
Group A cations - (Bi+3, Fe+3, Mn+2)-All these ions are present in your experiment.
Group B cations - (Al+3, Cr+3, Sn+4)
1. The solution is treated with NH3 until the pH is 9 to give a colored precipitate-
This step is required to preciptate cations as hydroxide or oxide. For example if Bi3+ is present it will be preciptated as Bi(OH)3-white color. Color of preciptate generally gives which cation present
Since color of preciptate is not given we can simply say that Group A and/or group B ion may be present.
2. The precipitate was treated with NaOH-H2O2 and the mixture centrifuged and separated to give a yellow solution and a precipitate.
Treatment with NaOH-H2O2 give solution as well as preciptates , then it is possible that Group A and B ions present
Preciptate may be because of ions which do not react further with NaOH and H2O2 and
Solution, e.g. because Mn(OH)2 may react with H2O2 to form MnO2 (Mn4+)
The remaining precipitate was dissolved in HCl- Clear solution- indicates Bi+3, Fe+3, Mn+2
3. Four drops of the clear solution from step (a) are treated with NaBiO3 to give a purple solution.
Confirmation of Mn+2 Ion -By oxidation of Mn2+ to purple MnO4- using sodium bismuthate (NaBiO3) in presence of acid . Purple color confirms Mn+2 Ion
4. Two drops of the solution from step (a) are treated with 6M NaOH and solid SnC12 to give a black solid.
Confirmmatory test for Bi3+ Bismuth with sodium hydroxide will form Bi(OH)3(s), then it is reduced to Bi metal (solid) in presence of basic Sn2+ solution, the presence of black color preciptate confirms Bi. In our experiment also Black solid is formed which confirms Bi+3 as per equation below.
Sn2+cl2- + NaOH-----------Sn(OH)3-
Bi(OH)3(s) + Sn(OH)3 + OH----------------- Bi(s) black color----Bi+3 confirmed
5. A portion of the solution from step (a) is treated with KSCN and the solution remains clear.
This test is for Fe+3, if on treating solution from step (a) with KSCN give red brown solution - Fe3+ confirmed.
Our inference- Clear soution means Fe3+ absent
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