Question

A colourless crystalline salt X is soluble in dilute HCl. On adding NaOH solution, it gives a white precipitate which is insoluble in excess of NaOH. What is the value of X?
A. Al₂(SO₄)₃
B. ZnSO₄
C. MgSO₄
D. SnCl₂

Answer 1

Hint: First clue is that it is colourless in nature and crystalline too. Also ,it reacts with NaOH when access is limited whereas it goes unreactive in its excessive amount. It suggests that its acidic and basic radicals are highly polar .

Complete Step by Step Answer:
Firstly, form an equation which shows its reactivity with HCl.
MgSO₄ + HCl ⇋ MgCl₂ + H₂SO₄

In this reaction, magnesium sulphate reacts with hydrochloride and gives magnesium chloride which will further react with sodium hydroxide.

Now, it is required that this MgCl₂ reacts with NaOH. Hence, the final product will come out of the whole procedure, which itself is a proof of presence of Mg in the salt X.
So, MgCl₂ + 2NaOH → Mg(OH)₂ + 2NaCl

In this reaction, magnesium chloride reacts with sodium hydroxide which yields magnesium hydroxide as the final product of the required equation. So, it has been confirmed that Mg is present in the given salt X. Therefore, the salt X will be MgSO₄.
From the above options, option (C) is the correct option.

Note: It should be remembered that MgSO₄ is a colourless substance in the given options. So, it will be easy to figure out the required compound needed. Besides, it satisfies the property of crystalline nature.