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NEET Chemistry Analytical Chemistry Question Bank Self Evaluation Test - Analytical Chemistry

  • question_answer
    A solution containing \[A{{s}^{3+}},C{{d}^{2+}},N{{i}^{2+}}\] and \[Z{{n}^{2+}}\] is made alkaline with dilute \[N{{H}_{4}}OH\] and treated with \[{{H}_{2}}S\]. The precipitate obtained will consist of

    A) \[A{{s}_{2}}{{S}_{3}}\] and CdS  

    B) CdS.NiS and ZnS

    C) NiS and ZnS    

    D) Sulphide of all ions

    Correct Answer: D

    Solution :

    [d] \[A{{s}^{3+}}\] and \[C{{d}^{2+}}\] are the radicals of group II, whereas \[N{{i}^{2+}}\And Z{{n}^{2+}}\] are the radicals of group IV. The solubility product of group IV radicals is higher as compared to group II. \[N{{H}_{4}}OH\] increases the ionisation of \[{{H}_{2}}S\] by removing \[{{H}^{+}}\]of \[{{H}_{2}}S\] as unionisable water. \[{{H}_{2}}S\rightleftharpoons 2{{H}^{+}}+{{S}^{2-}};\] \[{{H}^{+}}+O{{H}^{-}}\xrightarrow{{}}{{H}_{2}}O\] Thus excess of sulphide ions are present which leads to the precipitation of all the four ions. Note: \[HCl\] decreases ionisjation of \[{{H}_{2}}S\] whereas \[N{{H}_{4}}OH\] increases the ionisation of \[{{H}_{2}}S\].


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