Now consider an unknown that might contain Al3+, Mn2t, and Pb2 c, d One of these

Question

Now consider an unknown that might contain Al3+, Mn2t, and Pb2 c, d One of these three cations is easily separated from the other two, forming precipitates with two different reagents that have no effect on the other 2 cations. Based on this information and Part 4, decide on the first step for this separation scheme and fill in the blanks below. How would you separate the remaining two cations? (Hint: Look carefully at the list of complex ions. One cation forms a complex in the presence of excess of a reagent that will precipitate out the other.) Fill in the remaining blanks on the scheme. f. What final step must be performed (involving the remaining cation at g)? Why? What must be done to the cation at g before this final step is performed? Why? g.

Explanation / Answer

Based on the solubility we can separate the metal salts. If Ksp is low, the solubility will be low and it will form precipitate (ppt). look at the first problem....

Al3+, Mn2+ and Pb2+---- 1) from solubility of common salts, we can observe that, PbCl2 is ppt and Al2Cl6 and MnCl2 will be in solution state. So by trating with Cl-, we can separate Pb2+ from AL3+ and Mn2+.

so a is Cl- and b is PbCl2, c is Al2Cl6 and d is MnCl2

2) Now we can separate Al3+ and Mn2+ by addition of excess NaOH solution. When excess OH- used Al3+gives complex structure which is in solution state. The complex is Na[Al(OH)4]. When MnCl2 treated with NaOH we will get insoluble Mn(OH)2. (look at the Ksp value of Mn(OH)2, it is very low that means insoluble but for Na[Al(OH)4], it is very high)

So we can decide e isOH-, f is [Al(OH)4]- ande g is   Mn(OH)2.

3) Na[Al(OH)4] is hydrated form of NaAl2O.2H2O so drying the solution will giveNaAl2O crystals. Inorder to get Na[Al(OH)4], we need to heat the sample when Al3+ is treating with NaOH. And make sure that Pb2+ wont present. Otherwise it wll also participate in reaction.

Q2) I follows as question 1.

a is Cl- which gives b AgCl (ppt), and in soltion BaCl2, FeCl3 and NiCl2.

To the remaining solution if you add SO42-, BaSO4 will precipitate out i.e.

c is SO42- d is BaSO4, e is Fe2(SO4)3 and f is NiSO4

If you add NH4OH to Ni2+ and Fe3+, Fe will precipaitate out as Fe(OH)3 but Ni2+ forms soluble Ni(NH3)6 complex.

so g is NH4OH and h is Fe(OH)3 and i is Ni(NH3)6.

The presence of Ni2+ in the aqueous solution is confirmed by adding dimethylglyoxime (C4H8N2O2) resulting in the formation of a rose red precipitate:

Ni2+ (aq) + 2 C4H8N2O2 (aq) Ni(C4H7N2O2)2 (s) + 2 H+ (aq)

so j is  Ni(C4H7N2O2)2

reaction:

Ni2+ (aq) + 6 NH3 (aq) Ni(NH3)6 2+ (aq)

Fe3+ (aq) + 3 NH3 (aq) + 3 H2O (l) Fe(OH)3 (s) + 3 NH4 + (aq)

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