A substance $A_{2} H_{4}$ dissolved in water to form aqueous solution. If its base dissociation constant are $K_{b_{1}} = 4 \times 10^{- 5}$ and $K_{b_{2}} = 10^{- 8}$ respectively then which of the options are correct regarding approximate concentration of species when 0.1M aqueous solution $A_{2} H_{4}$ is taken :

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$[{OH}^{-}] = 2 \times 10^{- 3} M$

$[A_{2} H_{6}^{2 +}] = 10^{- 8} M$

$[A_{2} {H_{5}}^{+}] = 2 \times 10^{- 3} M$

$[H^{+}] = 5 \times 10^{- 12} M$

answer is A, B, C, D.

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Detailed Solution

$\begin{array}{l} {[A_{2} H_{4}]}_{0} = [A_{2} H_{4}] + [A_{2} H_{5}^{+}] + [A_{2} H_{6}^{2 +}] \\ K_{b_{1}} = \frac{[A_{2} H_{5}^{+}] [{OH}^{-}]}{[A_{2} H_{4}]} \\ [A_{2} H_{5}^{+}] = [{OH}^{-}] = \sqrt{4 \times 10^{- 5} \times 0.1} = 2 \times 10^{- 3} M \\ \Rightarrow [H^{+}] = 5 \times 10^{- 12} M \\ K_{b_{1}} \times K_{b_{2}} = \frac{[A_{2} H_{6}^{2 +}] {[{OH}^{-}]}^{2}}{[A_{2} H_{4}]} \\ \Rightarrow [A_{2} H_{6}^{2 +}] = \frac{4 \times 10^{- 5} \times 10^{- 8} \times 0.1}{{(2 \times 10^{- 3})}^{2}} = 10^{- 8} M \end{array}$