In the presence of $HCl$,$H_2S$ results in the precipitation of Group-$2$ elements but not Group-$4$ elements during qualitative analysis. It is due to

  • A
    higher concentration of $S^{2-}$
  • B
    higher concentration of $H^{+}$
  • C
    lower concentration of $S^{2-}$
  • D
    lower concentration of $H^{+}$

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Calculate $[S^{2-}]$ and $[HS^{-}]$ of the solution which contains $0.1 \ M \ H_2S$ and $0.3 \ M \ HCl$.
[$K_{a1} = 1.0 \times 10^{-7}$ and $K_{a2} = 1.3 \times 10^{-13}$ for $H_2S$]

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