Answer:
Present - Cr3+
Absent - Fe3+ and Al3+, Ni2+
In doubt- none
Step-by-step explanation:
qualitative analysis is used to detect the presence of specific ions in aqueous solution.
Different analytical schemes have have been devised that are based on various ways of defining major groups. there are 5 groups which are:
Group I (Ag+, Pb2+, Hg2+) cations
Group II (Cu2+, Bi3+, Cd2+, Hg2+, As3+, Sb3+, Sn4+) cations
Group III (Al3+, Cr3+, Fe3+, Ni2+)
Group IV (Mg2+, Ca2+, Sr2+, Ba2+) cations
Group V (Na+, K+, NH4+) cations
The four group III cations are Cr+3, Al+3, Fe+3, and Ni+2.
The first step in analysis involves separating the ions into two subgroups by treating the solution with NaOH and NaOCl. The hypochlorite ion oxidizes Cr(III) to a higher,more stable oxidation state (namely Cr(VI)) which is soluble:
2 Cr+3(aq) + 3 OCl-1(aq) + 10 OH-1(aq) → 2 CrO4-2(aq) + 3 Cl-1(aq) + 5 H2O(l)
In addition, Al+3 forms a soluble hydroxo-complex ion in the presence of excess hydroxide: Al+3(aq) + 4 OH-1(aq) → Al(OH)4-1(aq)
In contrast, Ni+2 and Fe+3 do not readily form hydroxo-complexes and are not oxidized by hypochlorite. They forminsoluble hydroxides under these conditions: Ni+2(aq) + 2 OH-1(aq) → Ni(OH)2(green solid)
Fe+3(aq) + 3 OH-1(aq) →Fe(OH)3(red solid)
To separate aluminum from chromium, the solution containing CrO4-2 and Al(OH)4- is acidified to destroy the hydroxo-complex forming: Al(OH)4-1(aq) + 4 H+ → Al+3(aq) + 4 H2O(l) (aq)
Treatment with aqueous ammonia gives a gelatinous white precipitate of aluminum hydroxide. The concentrationof hydroxide in ammonia is too low to form the hydroxo-complex:
Al+3(aq) + 3 NH3(aq) + 3 H2O(l) → 3 NH4+(aq) + Al(OH)3(white solid)
The chromate ion remains in solution. It can be tested and confirmed by precipitation as yellow BaCrO4: Ba+2(aq) + CrO4-2(aq) → BaCrO4(yellow solid)
The BaCrO4 precipitate dissolves in acid. The solution is then treated with H2O2 to produce a deep blue color due tothe presence of a peroxo-compound, probably CrO5:2 BaCrO4(s) + 4 H+ + 4 H2O2(aq) → 2 Ba+2(aq) + 6 H2O(l) + 2 CrO5(blue, aq) (aq)
The mixed precipitate of Ni(OH)2 and Fe(OH)3 can be dissolved by adding a strong acid. The Ni+2 and Fe+3 ions can be separated by adding ammonia. Ni+2 is converted to the deep-blue complex Ni(NH3)6+2 which stays in solution.The Fe+3 ion does not readily form a complex and re-precipitates as;
Fe(OH)3: Ni+2(aq) + 6 NH3(aq) → Ni(NH3)6+2(blue, aq) Fe+3(aq) + 3 NH3(aq) + 3 H2O(l) → 3 NH4+(aq) + Fe(OH)3(red solid)Confirm the presence of Ni+2 by adding dimethylglyoxime (DMG), C4H8N2O2, to give a deep rose precipitate:
Ni+2(aq) + 2 C4H8N2O2(aq) → 2 H+ + Ni(C4H7N2O2)2(rose solid) (aq)Confirm the presence of Fe+3 by dissolving Fe(OH)3 in a strong acid and adding KSCN to form a blood-red FeSCN+2 complex ion: Fe+3(aq) + SCN-(aq) → FeSCN+2(red, aq).
In the experiment, The yellow solution formed during the analysis is unaffected by treatment with ammonia. However, in acid medium, a blue solution is formed which confirms the presence of Chromium III in the solution.