![]() ![]() Therefore, the color of a precipitate resulting from the formation of Ag2S is answer choice (D) black. This black precipitate is silver sulfide. If sulfide ions are present in the solution, as they would be in a solution of sodium sulfide, they will react with the silver ions to produce a black precipitate. In the confirmatory test described by the question, silver nitrate is added to a solution of the unknown salt. Then, a confirmatory test is generally performed to further verify the results from the primary test. ![]() This black precipitate is lead(II) sulfide, a product in the reaction between hydrogen sulfide and lead(II) acetate. The presence of hydrogen sulfide and, thus, sulfide ions in the salt is confirmed when a black precipitate is formed on the paper. The presence of hydrogen sulfide gas is detected using a paper moistened with lead(II) acetate. The reaction between the sulfide anions in the salt and the hydrogen cations from the acid produces the foul-smelling gas hydrogen sulfide. The primary test for sulfide ions in an unknown salt is conducted by first adding a dilute acid, such as hydrochloric acid. What is the color of the precipitate resulting from the formation of Ag2S? (A) White, (B) yellow, (C) brown, or (D) black.Īdding silver nitrate to a solution can be used as a confirmatory test to determine if sulfide ions are present. Silver iodide (AgI) Silver iodide is a yellow precipitate. ![]() Silver bromide is soluble in concentrated ammonia solution and give the colourless coordination complex, Ag(NH 3) 2 +. But, as products KNO 3 can be formed in the aqueous phase while forming silver bromide precipitate. 1)The initial disappearance of silver nitrate in water indicates a chemical. It can be prepared easily by mixing silver nitrate with KBr. Adding AgNO3 to a Na2S solution forms Ag2S. 3)The color change of the solution indicates a chemical reaction occurred. ![]()
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