Tejas posted an Question
June 28, 2020 • 18:20 pm 30 points
  • IIT JAM
  • Chemistry (CY)

Why are group iv salts not precipitated when group ii test is carried out (on addition of h2s) in identification of basic radical ?

Why are group 4 salts not precipitated when group 2 test is carried out (on addition of H2S) in salt analysis?

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    Shivaji khadake

    Hello Student Solubility product of sulphides of II group radicals are less than solubility product of sulphides of IV group. So to pricipitate II group radicals as sulphides little S2-are sufficient. In presence of Dil HCl little S2- are released due to the common ion effects. In IV group the ionisation of H2S is favoured by the use of NH4OH ok for asking this question if you have any doubt you can ask me

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    Why you have marked Abuse

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    Say Shivaji

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    oh sorry may happen by mistake. i did not do

  • Suman Kumar Best Answer

    See H2S is a weak acid. It will dissociate very weakly but Nh4Oh will help it in more dissociation. This reaction will be in forward direction

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    NH4OH increase ph of solution (Ph of solution greater than pka of H2S) and hence H2S exist as S2- in solution....

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    is this correct, suman sir?

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    No Simun NH4OH will consume the H+ ion of H2S hence reaction will go in forward direction

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    H2S will dissociate into H+ and S²- but it will stop. When NH4OH is added it will consume it's H+ making H2S more to dissociate

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    I get.. thanks...🙏

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    Dinesh khalmaniya 1 best-answer

    Solubility product of sulphides of II group radicals are less than solubility product of sulphides of IV group. So to pricipitate II group radicals as sulphides little S2-are sufficient. In presence of Dil HCl little S2- are released due to the common ion effects. In IV group the ionisation of H2S is favoured by the use of NH4OH

  • Suman Kumar best-answer

    Solubility product of sulphides of II group radicals are less than solubility product of sulphides of IV group. So to pricipitate II group radicals as sulphides little S2-are sufficient. In presence of Dil HCl little S2- are released due to the common ion effects. In IV group the ionisation of H2S is favoured by the use of NH4OH

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    how's ionization of H2S favoured by NH4OH?

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    H2S = H+. +. S²-

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