A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially so
luble in hot water. The supernatant and the precipitate are placed into two separate test tubes. The solid remaining after treatment with hot water turns black on addition of 6 M NH3. The supernatant is tested with K2CrO4 and a yellow precipitate forms. No other precipitates are observed while performing the procedure. Which of the ions are present and which are absent? State your reasoning with equations
Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: .
Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
Secondly, addition of liquid ammonia would form a precipitate with silver: ; Silver hydroxide at higher temperatures decomposes into black silver oxide: .
Thirdly, we also know we have in the mixture, since addition of potassium chromate produces a yellow precipitate: . The latter precipitate is yellow.
It happens because particles of gas are in constant random motion. Thus they can collide with the walls of the container causing pressure on the walls.
; Silver hydroxide at higher temperatures decomposes into black silver oxide:
.
in the mixture, since addition of potassium chromate produces a yellow precipitate:
. The latter precipitate is yellow.