Question

A solution of ammonium Chromium sulphate was mixed with an equal amount of ammonium hydroxide, releases which colour chromium precipitate?
A.White
B.Yellow
C.Blue/green
D.Black

Answer 1

Hint:This question can be solved from knowledge of analytical chemistry and knowledge of the colour of cations. Wet test for cation analysis is done on the basis of the solubility of different cations and their precipitation by different solvents. Depending on their solubility and reactivity towards certain solvents, these cations have been categorized into certain categories. Chromium sulphate reacts with ammonium hydroxide in a double displacement reaction, where chromium is present in $ + 3$ oxidation state.

Complete step by step answer:
The molecular formula for Ammonium Chromium Sulphate $\left[ {N{H_4}Cr{{\left( {S{O_4}} \right)}_2}.12{H_2}O} \right]$ and here Chromium is in the $ + 3$ oxidation state. It is a double salt that dissociates in water to release $C{r^{ + 3}}$ ions. In presence of ammonium hydroxide, Chromium is precipitated to form Chromium hydroxide by the following reaction:
$C{r^{ + 3}} + N{H_4}OH \to Cr{\left( {OH} \right)_3} \downarrow + N{H_4}^ + $
This chromium hydroxide is insoluble in sodium hydroxide and greenish is colour.

So, the correct answer is option C, Blue/Green.

Note:
Ammonium chromium sulphate is formed when crystals of ammonium sulphate and chromium sulphate are dissolved together in hot water and stirred well. The solution on cooling gives crystals of ammonium chromium sulphate.Chromium hydroxide does not dissolve in excess of ammonium hydroxide and boiling the precipitate only leads to re-precipitating.Chromium ions are precipitated by sodium hydroxide as well but adding excess of sodium hydroxide leads to the formation of soluble $Cr{(OH)_4}^ - $
In a basic solution, $C{r^{ + 3}}$is oxidized to $C{r^{ + 6}}$by hydrogen peroxide by the following reaction:
$2Cr{(OH)_4}^ - + 3{H_2}{O_2} + 2O{H^ - } \to 2Cr{O_4}^ - + 8{H_2}O$
A solution of barium chloride is then added which precipitates the chromate to from barium chromate:
$BaC{l_2} + Cr{O_4}^ - \to BaCr{O_4}^ - \downarrow + C{l^ - }$