$H_{2} S$ is a dibasic weak acid with dissociation constant $K a_{1} = 10^{- 7}$ and $K a_{2} = 10^{- 14}$ . The concentration of different chemical species produced by $H_{2} S$ depends upon the $p H$ of the solution in which it is present. Which of the following statement are correct

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If pH of solution is 12 then $H S^{-}$ concentration will be more than $H_{2} S$ and $S^{2 -}$

If pH of solution is 6 then $H_{2} S$ concentration is more than $H S^{-}$ and $S^{2 -}$

If pH of solution is 13 then $S^{2 -}$ concentration will be more than $H S^{-}$ and . $H_{2} S$

If pH of solution is 2 then $H_{2} S$ concentration is more than $H S^{-}$ and $S^{2 -}$

answer is A, B, C.

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Detailed Solution

a) $10^{- 7} = 10^{- 2} \frac{[H S^{-}]}{[H_{2} S]} \Rightarrow \frac{[H S^{-}]}{[H_{2} S]} = 10^{- 5}$
$10^{- 21} = 10^{- 4} \frac{[S^{2 -}]}{[H_{2} S]} \Rightarrow \frac{[S^{2 -}]}{[H_{2} S]} = 10^{- 17}$
b) $10^{- 7} = 10^{- 6} \frac{[H S^{-}]}{[H_{2} S]} \Rightarrow \frac{[H S^{-}]}{[H_{2} S]} = 10^{- 7}$
$10^{- 21} = 10^{- 12} \frac{[S^{2 -}]}{[H_{2} S]} \Rightarrow \frac{[S^{2 -}]}{[H_{2} S]} = 10^{- 9}$
C) $10^{- 7} = 10^{- 12} \frac{[H S^{-}]}{[H_{2} S]} \Rightarrow \frac{[H S^{-}]}{[H_{2} S]} = 10^{5}$