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

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).