Question

In lead acetate test of sulphur detection, Lassaigne extract is acidified with ___________ and then lead acetate solution is added.
(A) Oxalic acid
(B) Hydrochloric acid
(C) Sulphuric acid
(D) Acetic acid

Answer 1

Hint: The lassaigne extract is used for the qualitative analysis of the organic compounds that contain Nitrogen, Oxygen, and Phosphorus. In lassaigne’s test, the extract is prepared by fusing sodium metal in a red hot tube and is used for the detection.

Complete answer:
For the detection of the sulphur atom, we have to prepare the lassaigne’s extract first. For that, we have to take sodium metal in the test tube and it is heated until the sodium metal melts. After this stage is reached, a small amount of substance is added to the test tube and then it is heated till it becomes red hot. After the test tube becomes red hot it is put in a china dish with water in it and crushed. The content in the dish is boiled till it becomes half of its initial value and filtered after that. The filtrate is transferred to a test tube. This filtrate is called lassaigne’s extract.
The lassaigne’s extract formed is of sodium sulphide which is formed by the reaction of sulphur and sodium. The equation for this is:
 $ 2Na + S \to N{a_2}S $
This extract is divided into two parts to perform two tests. One is lead acetate test and the other is sodium nitroprusside test. In the lead acetate test lassaigne extract is acidified with acetic acid because the lead acetate that is formed is soluble in the solution and it does not interfere with the test. On the other hand, if we use $ Dil.\,HCl $ or $ {H_2}S{O_4} $ , it will form precipitate and it will interfere with the test.
Hence, we use acetic acid for the detection of sulphur in lassaigne’s extract. Therefore, option (D) is correct.

Note:
When you add acetic acid and lead acetate in lassaigne’s extract it forms a black precipitate of lead sulphide which confirms the presence of sulphur atom in the compound. The equation of this reaction is given as:
 $ {S^{2 - }} + P{b^{2 + }} \to PbS $.