Question

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Answer 1

During separation of ions we react the aqueous solution of these ions with another reagent solution. Now after reaction with the new solution, if a precipitate is formed, then one of the ions from our taken solution has separated. We can take out the precipitate compound of these ion by filtering. But if no precipitate is formed upon reaction, then our reagent solution is not capable for separation.

Ex- For separation of Ca⁺² and Pb⁺²

1- With Na₂SO₄

The reaction between aqueous solution of Na₂SO₄ and Ca⁺² will be

Na₂SO_4(aq) + Ca⁺²_(aq) -------------> CaSO_4(aq)

And reaction between aqueous solution of Na₂SO₄ and Pb⁺² will be

Na₂SO_4(aq) + Pb⁺²_(aq) -------------> PbSO_4(s)

Here the PbSO4 formed is insoluble in water. Thus it remains in a solid state as a precipitate in the solution.

So if we take Na₂SO₄ as a reagent solution, then it can be used to separate Ca⁺² and Pb⁺² . Where the Pb⁺² will form a precipitate compound and can be filtered out and the Ca⁺² will remain in the solution.

2-

1- With KI

The reaction between aqueous solution of KI and Ca⁺² will be

_2KI(aq) + Ca⁺²_(aq) -------------> CaI_2(aq)

And reaction between aqueous solution of KI and Pb⁺² will be

_2KI(aq) + Pb⁺²_(aq) -------------> PbI_2(aq)

Here after reaction, both the compounds of Ca⁺² and Pb⁺² formed are soluble in water.

Hence KI cant be used for the separation process of Ca⁺² and Pb⁺²

3- With K₂SO₄

The reaction between aqueous solution of K₂SO₄ and Ca⁺² will be

K₂SO_4(aq) + Ca⁺²_(aq) -------------> CaSO_4(aq)

And reaction between aqueous solution of Na₂SO₄ and Pb⁺² will be

K₂SO_4(aq) + Pb⁺²_(aq) -------------> PbSO_4(s)

Here the PbSO4 formed is insoluble in water. Thus it remains in a solid state as a precipitate in the solution.

So if we take K₂SO₄ as a reagent solution, then it can be used to separate Ca⁺² and Pb⁺² . Where the Pb⁺² will form a precipitate compound and can be filtered out and the Ca⁺² will remain in the solution.