Question

Aluminium and potassium hydroxide reacts with water to form the potassium salt X. Then with addition of sulphuric acid in presence of water it is converted into a precipitate known as alum. Identify X.
[A] $KAl{{\left( OH \right)}_{4}}$
[B] $Al{{\left( OH \right)}_{3}}$
[C] ${{K}_{2}}Al{{\left( OH \right)}_{4}}$
[D] None of these

Answer 1

To solve this you have to start by finding the salt X that is produced upon reaction of aluminium and potassium hydroxide with water. To check it you have to know its reaction with aqueous sulphuric acid. If you get a precipitate of $KAl{{\left( S{{O}_{4}} \right)}_{2}}\cdot 12{{H}_{2}}O$ then you can be sure that the X is the salt formed in the reaction.

Complete step by step solution:
To solve this we have to see the reaction of aluminium and potassium hydroxide with water.
We know that potassium hydroxide is KOH. In presence of aluminium and water, potassium hydroxide reacts to form potassium tetra hydroxy aluminate (III) and di-hydrogen. This is a salt of potassium and aluminium.
The reaction here is a complexation reaction as we obtain aluminium – potassium complex.
Here, hydrogen undergoes reduction and aluminium undergoes oxidation. We can write the balanced chemical reaction as-
$2Al+2KOH+6{{H}_{2}}O\to 2KAl\left( O{{H}_{4}} \right)+3{{H}_{2}}$
Here we can see that the salt X thus obtained $KAl{{\left( OH \right)}_{4}}$. Now let’s see its reaction with aqueous sulphuric acid.
Initial addition of sulphuric acid will give us a precipitate of aluminium hydroxide. Further addition of sulphuric acid will lead to dissolution of aluminium hydroxide and formation of aluminium sulphate. Upon cooling, it will give us the precipitate known as alum. We can write the reactions as-

& 2KAl{{(OH)}_{4}}+{{H}_{2}}S{{O}_{4}}\to Al{{(OH)}_{3}}+{{K}_{2}}S{{O}_{4}}+2{{H}_{2}}O \\\ & 2Al{{(OH)}_{3}}+{{H}_{2}}S{{O}_{4}}\to A{{l}_{2}}{{(S{{O}_{4}})}_{3}}+6{{H}_{2}}O \\\ & {{K}_{2}}S{{O}_{4}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}+24{{H}_{2}}O\to KAl{{\left( S{{O}_{4}} \right)}_{2}}\cdot 12{{H}_{2}}O \\\ \end{aligned}$$ We can understand from the above discussion that $KAl{{\left( OH \right)}_{4}}$ is the salt obtained. **Therefore, the correct answer is option [A] $KAl{{\left( OH \right)}_{4}}$.** **Note:** Alum is also known as potassium aluminium sulphate dodecahydrate. The reaction that we have shown above is the method that we use for production of alum from aluminium from scrap. Alum is used widely as a dye for fabrics, also for making paper and pickles. It is also used for purifying water.