Question

Given below are two statements:
Statement-I: The gas liberated on warming a salt with dil H₂SO₄, turns a piece of paper dipped in lead acetate into black, it is a confirmatory test for sulphide ion.
Statement-II: In statement-I the colour of paper turns black because of formation of lead sulphite. In the light of the above statements, choose the most appropriate answer from the options given below:

• A. Both Statement-I and Statement-II are false
• B. Statement-I is false but Statement-II is true
• C. Statement-I is true but Statement-II is false
• D. Both Statement-I and Statement-II are true.

Answer 1

The Correct Option is C

The problem asks us to evaluate two statements regarding a chemical test for the sulphide ion and choose the most appropriate answer describing their validity.

Concept Used:

This problem is based on the principles of qualitative inorganic analysis, specifically the confirmatory test for the sulphide anion (\(S^{2-}\)). The key chemical reactions are:

1. The reaction of a sulphide salt with a dilute acid to liberate hydrogen sulphide (\(H_2S\)) gas.
2. The reaction of hydrogen sulphide gas with lead(II) acetate solution to form a precipitate.

Knowledge of the chemical formulas and physical properties (especially color) of the products, such as lead sulphide (\(PbS\)) and lead sulphite (\(PbSO_3\)), is required.

Step-by-Step Solution:

Step 1: Analyze Statement-I.

Statement-I describes a procedure and its conclusion. Let's break it down:

• Reaction 1: Warming a salt with dil H₂SO₄. If the salt contains the sulphide ion (\(S^{2-}\)), it will react with the acid to produce hydrogen sulphide gas, which has a characteristic smell of rotten eggs. \[ S^{2-}(aq) + 2H^+(aq) \longrightarrow H_2S(g) \]
• Reaction 2: The liberated gas turns a piece of paper dipped in lead acetate into black. The gas, \(H_2S\), reacts with lead(II) acetate (\(Pb(CH_3COO)_2\)) solution. \[ H_2S(g) + Pb(CH_3COO)_2(aq) \longrightarrow PbS(s) \downarrow + 2CH_3COOH(aq) \]
• Observation and Conclusion: The product of this reaction is lead(II) sulphide (\(PbS\)), which is a black solid precipitate. The formation of this distinct black precipitate is a standard and reliable confirmatory test for the presence of the sulphide ion.

Therefore, Statement-I is a factually correct description of the confirmatory test for sulphide ions.

Conclusion for Statement-I: Statement-I is true.

Step 2: Analyze Statement-II.

Statement-II provides an explanation for the observation in Statement-I. It claims that the paper turns black because of the formation of lead sulphite (\(PbSO_3\)).

• As established in Step 1, the black precipitate formed is lead sulphide, with the chemical formula \(PbS\).
• Lead sulphite, with the chemical formula \(PbSO_3\), is a white solid. It is formed from the reaction of a sulphite ion (\(SO_3^{2-}\)) with a soluble lead salt.
• The chemical species mentioned in Statement-II, lead sulphite, is incorrect.

Therefore, Statement-II provides an incorrect reason for the observation.

Conclusion for Statement-II: Statement-II is false.

Final Computation & Result:

Based on the analysis:

• Statement-I is correct.
• Statement-II is incorrect.

Therefore, the most appropriate answer is that Statement-I is true but Statement-II is false.

Answer 2

Step 1. Identify the Reaction Involved: When a salt containing sulphide ion is treated with dilute H₂SO₄, hydrogen sulphide gas (H₂S) is liberated.

Step 2. Reaction with Lead Acetate: H₂S gas reacts with lead acetate solution present on the paper, resulting in the formation of black lead sulphide (PbS):

(CH₃COO)₂Pb + H₂S → PbS + 2CH₃COOH

Step 3. Explanation of Statements: Statement-I is correct, as the blackening of lead acetate paper confirms the presence of sulphide ions.
Statement-II is incorrect because it suggests that the paper turns black due to formation of “lead sulphite,” which is incorrect. The actual black compound is lead sulphide (PbS).