Question

Aqueous solution of a salt $(A)$ forms a dense white precipitate with $BaCl_2$ solution. The precipitate dissolves in dilute $HCl$ to produce a gas $(B)$ which decolourises acidified $KMnO_4$ solution $A$ and $B$ respectively are :

• A. $BaSO_3, SO_2$
• B. $BaSO_4, H_2S$
• C. $BaSO_3, H_2S$
• D. $BaSO_4, SO_2$

Answer 1

The Correct Option is A

The corresponding reactions are

$BaCl _{2}+ Na _{2} SO _{3} \rightarrow BaSO _{3}+2 NaCl$
$BaSO _{3}+2 HCl \rightarrow BaCl _{2}+ SO _{2}+ H _{2} O$

Salt $A$ is $Na _{2} SO _{3}$ and the dense white precipitate is $BaSO _{3}$ (Barium sulfite).

Gas $B$ is $SO _{2}$. It decolourises acidified $KMnO _{4}$ solution because of its reducing nature.

The corresponding reaction is:

$2 KMnO _{4}+5 SO _{2}+2 H _{2} O \rightarrow K _{2} SO _{4}+2 MnSO _{4}+2 H _{2} SO _{4}$

Answer 2

The reaction described in the question involves the formation of a white precipitate with BaCl₂. The formation of a white precipitate with BaCl₂ suggests the presence of sulfate ions (\( \text{SO}_4^{2-} \)), forming barium sulfate (\( \text{BaSO}_4 \)): \[ \text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4 \, (\text{white precipitate}) \] Next, the precipitate dissolves in dilute HCl, which releases a gas (B). The fact that this gas decolourises acidified KMnO₄ solution indicates that the gas is sulfur dioxide (SO₂), which is a reducing agent and can reduce potassium permanganate (KMnO₄) to Mn²⁺, decolourising the solution: \[ \text{BaSO}_4 + 2\text{HCl} \rightarrow \text{BaCl}_2 + \text{H}_2\text{SO}_3 \] \[ \text{H}_2\text{SO}_3 \rightarrow \text{SO}_2 + \text{H}_2\text{O} \] Thus, the correct answer is (A), where the salt (A) is BaSO₄ and the gas (B) is SO₂.