1) 4 drops of 6M HCl solution are added to the unknown solution and no precipita
ID: 830673 • Letter: 1
Question
1) 4 drops of 6M HCl solution are added to the unknown solution and no precipitate
2) The pH of the solution from (1) is raised to 9.5 by the addition of 6M NH3 solution. A cloudy, red-brown reaction mixture results
3) The solution from (2) is centrifuged, and the colorless supernatant liquid is decanted into a separate container for later analysis. The precipitate is rinsed, and 6M NaOH solution is added. The mixture is heated and then centrifuged, resulting in a red-brown precipitate and a clear, colorless supernatant liquid.
4) The precipitate from (3) is dissolved in 6M HCl solution, the pH is adjusted to 5, and 0.1M K4Fe(CN)6 solution is added to the precipitate, resulting in the formation of a dark blue precipitate in a blue solution
5) When 6M HCl, aluminon, and 6M NH3 solutions are added to the second supernatant liquid from (3), a clear, red solution results.
6) When 0.5M (NH4)2C2O4 solution is added to the first supernatant liquid from(2), a clear, colorless solution results
7) The solution from (6) is heated to dryness with HNO3, and the residue dissolved in 6M HCl solution. The solution remains clear and colorless following the adjustment of the pH of the solution to 12 with the addition of 6M and 1M NaOH solution
8) When 6M CH3COOH followed by 0.1M Na3Co(NO2)6 solutions are added to the solution from (7), a yellow precipitate forms
Explanation / Answer
1)HCl solution are added to the unknown solution and no precipitate
absence of group 1 - absence of Pb Ag Hg (they should give ppt so their absence confirm)
2)The pH of the solution from (1) is raised to 9.5 by the addition of 6M NH3 solution. A cloudy, red-brown reaction mixture results
Presence of group III - Specifically Ferric salts
Fe(NO3)3 + NH4OH -----------> Fe(OH)3 + NH4NO3
It is due to formation of Fe(OH)3 which give red brown ppt
3)The solution from (2) is centrifuged, and the colorless supernatant liquid is decanted into a separate container for later analysis. The precipitate is rinsed, and 6M NaOH solution is added.
This is done to clean the ppt
4)The precipitate from (3) is dissolved in 6M HCl solution, the pH is adjusted to 5, and 0.1M K4Fe(CN)6 solution is added to the precipitate, resulting in the formation of a dark blue precipitate in a blue solution
Fe(OH)3 + K4Fe(CN)6 ------> Fe?(CN)?? + KOH
It is due to formation of Iron(III) hexacyanoferrate(II). Parisian blue.
5) When 6M HCl, aluminon, and 6M NH3 solutions are added to the second supernatant liquid from (3), a clear, red solution results.
This is doing to check the presence of other ions in group III
A red, gelatinous precipitate (sometimes called a red lake) indicates Al3+.
It is due to adsorption of Al3+ on aluminon
6) When 0.5M (NH4)2C2O4 solution is added to the first supernatant liquid from(2), a clear, colorless solution results
To confirm absence of group iv that is Barium, Calcium, Strauntium
7)The solution from (6) is heated to dryness with HNO3, and the residue dissolved in 6M HCl solution. The solution remains clear and colorless following the adjustment of the pH of the solution to 12 with the addition of 6M and 1M NaOH solution
8) 6M CH3COOH followed by 0.1M Na3Co(NO2)6 solutions are added to the solution from (7), a yellow precipitate forms.
This is doing to confirm the presence of group V - K
K+ + Na3Co(NO2)6 -------------> K3Co(NO2)6 + Na+
formation of K3Co(NO2)6 make yellow ppt
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.