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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2010

  • question_answer
    \[1\,\,d{{m}^{3}}\] solution containing \[{{10}^{-5}}\] moles each of \[C{{l}^{-}}\] ions and \[CrO_{4}^{2-}\] ions is treated with \[{{10}^{-4}}\] moles of silver nitrate. Which one of the following observations is made? \[[{{K}_{sp}}A{{g}_{2}}Cr{{O}_{4}}=4\times {{10}^{-12}}]\]

    A) Precipitation does not occur

    B) Silver chromate gets precipitated first

    C) Silver chloride gets precipitated first

    D) Both silver chromate and silver chloride start precipitating simultaneously

    Correct Answer: C

    Solution :

    For precipitation, ionic product > solubility product \[({{K}_{sp}})\] For                         \[A{{g}_{2}}Cr{{O}_{4}}\] ionic product \[={{[A{{g}^{+}}]}^{2}}[CrO_{4}^{-}]\]                 \[={{({{10}^{-4}})}^{2}}({{10}^{-5}})={{10}^{-13}}\] \[{{K}_{sp}}\] of \[A{{g}_{2}}Cr{{O}_{4}}=4\times {{10}^{-12}}\] Here,\[{{K}_{sp}}>IP\] Thus, no precipitate is obtained. For\[AgCl\], ionic product \[=[A{{g}^{+}}]\,[C{{l}^{-}}]\]                                 \[=[{{10}^{-4}}]\,[{{10}^{-5}}]\]                                 \[={{10}^{-9}}\] \[{{K}_{sp}}\,(AgCl)=1\times {{10}^{-10}}\] Here,                     \[IP>{{K}_{sp}}\] So, precipitate will form. Thus, silver chloride gets precipitated first.


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