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JEE Main & Advanced Chemistry Analytical Chemistry Question Bank Critical Thinking Analytical

  • question_answer
    A white crystalline substance dissolves in water. On passing \[{{H}_{2}}S\] in this solution, a black precipitate is obtained. The black precipitate dissolves completely in hot \[HN{{O}_{3}}\]. On adding a few drops of conc. \[{{H}_{2}}S{{O}_{4}}\] a white precipitate is obtained. This precipitate is that of [CPMT 1990]

    A) \[BaS{{O}_{4}}\]

    B) \[SrS{{O}_{4}}\]

    C) \[PbS{{O}_{4}}\]

    D) \[CdS{{O}_{4}}\]

    Correct Answer: C

    Solution :

    \[P{{b}^{+2}}\ +\ {{H}_{2}}S\ \xrightarrow[\text{acidic}]{}\ \underset{\text{(Black ppt}\text{.)}}{\mathop{PbS\downarrow \ }}\,+\ {{H}_{2}}\uparrow \] \[3PbS+8HN{{O}_{3}}\to 3Pb{{(N{{O}_{3}})}_{2}}+2NO+3S\]\[+4{{H}_{2}}O\] \[Pb{{(N{{O}_{3}})}_{2}}\ +\ {{H}_{2}}S{{O}_{4}}\ \to \underset{\text{(white ppt}\text{.)}}{\mathop{\ PbS{{O}_{4}}}}\,\downarrow \ +\ 2HN{{O}_{3}}\uparrow \]


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