Question

An excess of sodium hydroxide solution was added to two test tubes with different solutions. In the
first tube with a solution of a substance X that led to the formation of a white precipitate. In the
second tube with a solution of a substance Y that led to the formation and further dissolution of a
precipitate of the following substances, decide which substances are X and Y.


pls help me out ​

Answer 1

Answer:

A white precipitate is formed when sodium hydroxide is added to solutions containing aluminum ions (Al+3), calcium ions (Ca+2) or magnesium ions (Mg+2).

The net ionic reactions, ignoring the spectator ions are:

Al+3 (aq) + 3OH-1 (aq) → Al(OH)3 (s)

Ca+2 (aq) + 2OH-1 (aq) → Ca(OH)2 (s)

Mg+2 (aq) + 2OH-1 (aq) → Mg(OH)2 (s)

Each metallic hydroxide has (s) after its formula, which indicates it's a solid. Since all of these precipitates are white, we have to do a second test to narrow down which metallic ion is in our original solution. All we have to do is add more sodium hydroxide, and if the precipitate dissolves, it's aluminum hydroxide. Calcium hydroxide and magnesium hydroxide won't dissolve upon the addition of more sodium hydroxide.

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