State one reagent that can be used to distinguish btwn each of the pairs of ions and in each case state what is observed if each ion is treated with the named reagent.a)Pb2+(aq) and Al3+ b)SO4 2- and SO3 2-
a. Add HCl. Pb^2+ + 2Cl^- ==> PbCl2(white ppt). No rxn with Al^3+
b. Add conc'd HCl. sulfate no rxn. sulfite will get SO2 fumes given off
OR
add BaCl2. sulfate gives white ppt of BaSO4. sulfite gives white ppt of BaSO3. Then add dilute HCl. BaSO3 dissolves. BaSO4 does not.
a) To distinguish between Pb2+ and Al3+ ions, we can use sodium hydroxide (NaOH) as a reagent.
- Pb2+(aq) + NaOH(aq) → Pb(OH)2(s) (white precipitate)
- Al3+(aq) + NaOH(aq) → Al(OH)3(s) (white precipitate)
When Pb2+ reacts with NaOH, a white precipitate of lead(II) hydroxide (Pb(OH)2) is formed. On the other hand, when Al3+ reacts with NaOH, a white precipitate of aluminum hydroxide (Al(OH)3) is formed.
b) To distinguish between SO4 2- and SO3 2- ions, we can use barium chloride (BaCl2) as a reagent.
- SO4 2-(aq) + BaCl2(aq) → BaSO4(s) (white precipitate)
- SO3 2-(aq) + BaCl2(aq) → no reaction
When SO4 2- reacts with BaCl2, a white precipitate of barium sulfate (BaSO4) is formed. However, when SO3 2- reacts with BaCl2, no reaction occurs, and no precipitate is formed.
a) To distinguish between Pb2+ (lead) and Al3+ (aluminum) ions, you can use the reagent called sodium hydroxide (NaOH).
When a solution containing Pb2+ ions is treated with NaOH, a white precipitate (solid) of lead hydroxide (Pb(OH)2) is formed. This is because lead hydroxide is insoluble in water.
On the other hand, when a solution containing Al3+ ions is treated with NaOH, no precipitate is formed. Aluminum hydroxide (Al(OH)3) is amphoteric, meaning it can act as both an acid and a base. It forms a soluble complex with excess NaOH, resulting in a colorless solution.
Therefore, by observing the formation of a white precipitate or no precipitate when treated with NaOH, you can distinguish between Pb2+ and Al3+ ions.
b) To differentiate between SO4 2- (sulfate) and SO3 2- (sulfite) ions, you can use the reagent called acidified potassium dichromate (K2Cr2O7/H2SO4).
When a solution containing SO4 2- ions is treated with acidified potassium dichromate, no change is observed. This is because sulfate ions are already in their highest oxidation state (+6) and cannot be further oxidized by the dichromate ion.
However, when a solution containing SO3 2- ions is treated with acidified potassium dichromate, the orange color of the dichromate ion changes to green. This is because the sulfite ion can be oxidized by the dichromate ion to form sulfate ions, thus reducing the dichromate ion from its orange to green state.
Therefore, by observing the lack of color change or the change in color from orange to green when treated with acidified potassium dichromate, you can distinguish between SO4 2- and SO3 2- ions.