Analytical ChemistryHard
Question
Which of the following statement(s) is/are correct ?
Options
A.Yellow precipitated of silver arsenite is soluble in both nitric acid and ammonia.
B.Potassium cyanide when added in very small quantity to copper sulphate solution, produces first yellow precipitate which quickly converts in to white precipitate.
C.Black precipitate of BiI3 turns orange on heating with water.
D.White precipitate of Bi(OH)3 turns yellowish brown, when boiled.
Solution
(A) Correct.
(B) Cu2+ + 2CN- → Cu(CN)2↓ (yellow)
2Cu(CN)2↓ → 2Cu(CN)2↓ (white) + (CN)2 ↑
(C) BiI3↓ + H2O
BiOI ↓ (orange) + 2HI
(D) Bi(OH)3↓
BiO.OH↓ (yellowish white) + H2O
(B) Cu2+ + 2CN- → Cu(CN)2↓ (yellow)
2Cu(CN)2↓ → 2Cu(CN)2↓ (white) + (CN)2 ↑
(C) BiI3↓ + H2O
(D) Bi(OH)3↓
Create a free account to view solution
View Solution FreeMore Analytical Chemistry Questions
A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after...A metal nitrate solution reacts with dilute hydrochloric acid to give a white precipitate which is soluble in concentrat...Concentrated aqueous ammonia dissolve(s) which of the following completely ?...To increase significantly the concentration of free Zn2+ ion in a solution of the complex ion [Zn(NH3)4]2+ Zn2+ (aq) + 4...Which one of the following can be used in place of NH4Cl for the identification of the third group radicals?...