A) \[{{K}_{2}}C{{O}_{3}}\]
B) KI
C) KBr
D) \[{{K}_{3}}P{{O}_{4}}\]
Correct Answer: B
Solution :
[b] KI reacts with \[CuS{{O}_{4}}\] solution to produce cuprous iodide (white precipitate) and \[{{I}_{2}}\](which gives brown colour) Iodine reacts with hypo \[(N{{a}_{2}}{{S}_{2}}{{O}_{3}}5{{H}_{2}}O)\] solution. Decolourisaiton of solution shows the appearance of white precipitate. \[2CuS{{O}_{4}}+4KI\to 2{{K}_{2}}S{{O}_{4}}+\]\[\underset{\begin{smallmatrix} Cuprous\,\,iodide\,\, \\ \,\,(white\,\,ppt.) \end{smallmatrix}}{\mathop{2CuI}}\,+\underset{\begin{smallmatrix} (Brown\,\,colour\,\, \\ \,\,\,\,\,in\,\,solution) \end{smallmatrix}}{\mathop{{{I}_{2}}}}\,\] \[2N{{a}_{2}}{{S}_{2}}{{O}_{3}}+{{I}_{2}}\to \underset{\begin{smallmatrix} Sod.\,\,tetrathionate\,\, \\ \,\,\,\,(colourless) \end{smallmatrix}}{\mathop{N{{a}_{2}}{{S}_{4}}{{O}_{6}}}}\,+2NaI\]
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