Use equations to explain the following observations
1. When aq. NaOH is added to aq. iron(II) sulphate, green precipitate is obtained instantaneously, which rapidly turns brown on exposure to air.
2. On shaking solid copper (I) iodide with aq. ammonia, it dissolves to give a colourless solution, which gradually turns deep blue on standing. Addition of conc. HCl to the resultion solution changes its colour to green
AL CHEM (urgent!!!)
2010-09-07 7:27 am
回答 (2)
2010-09-07 6:46 pm
✔ 最佳答案
1) The green precipitate is iron (II) hydroxide:Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s)
Upon exposure to air, iron (II) hydroxide is readily oxided to brown iron (III) hydroxide:
4Fe(OH)2(s) + O2(g) + 2H2O(l) → 4Fe(OH)3(s)
2) Copper (I) iodide readily dissolves in aq. ammonia, forming colourless complex:
CuI(s) + 2NH3(aq) →[Cu(NH3)2]+(aq) + I-(aq)
Since copper (I) is unstable, it is readily oxiddized to copper (II):
Cu+(aq) → Cu2+(aq) + e
and it forms a deep blue complex with ammonia ligands:
Cu2+(aq) + 6NH3(aq) →[Cu(NH3)6]2+(aq)
Upon adding conc. HCl, aq. copper (II) chloride is formed and with adequate amount of chloride ions, green complex is formed:
CuCl2(aq) + 2Cl-(aq) → CuCl4-(aq)
2010-09-07 11:11:41 補充：
Thx for the opinion from 己式庚辛
參考: Myself
2010-09-07 6:59 pm
CuCl4(-) is yellow in color.
the green color is due to partially "chlorinated" copper(II) complex, or due to mixture of blue copper(II) ion and yellow CuCl4(-) complex.
the green color is due to partially "chlorinated" copper(II) complex, or due to mixture of blue copper(II) ion and yellow CuCl4(-) complex.
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