0.1M H2S solution in which enough HCl has been added to produce [H+] equal to 2×10−4 M. In this solution [S−2] concentration is found to be y×10−15M.Find out the value of y_______(Given for H2S : K1=10−7,K2=10−14 )

H2S⇄ H++HS−1[K1] HS−⇄ H++S−2[K2]----------------------------------------------------- H2S⇄ 2H++S−2[K] K=K1.K2=10−21K=[H+]2[S−2][H2S] K=(2×10−4)2[S−2]0.1 [S−2]=2.5×10−15M

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0.1M $H_{2} S$ solution in which enough HCl has been added to produce $[H^{+}]$ equal to $2 \times 10^{- 4}$ M. In this solution $[S^{- 2}]$ concentration is found to be $y \times 10^{- 15} M .$
Find out the value of y_______
(Given for $H_{2} S$ : $K_{1} = 10^{- 7}, K_{2} = 10^{- 14}$ )

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answer is 2.5.

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Detailed Solution

$H_{2} S \underset{}{⇄} H^{+} + H S^{- 1} [K_{1}]$
$H S^{-} \underset{}{⇄} H^{+} + S^{- 2} [K_{2}]$
-----------------------------------------------------
$H_{2} S \underset{}{⇄} 2 H^{+} + S^{- 2} [K]$
$K = K_{1} . K_{2} = 10^{- 21}$
$K = \frac{{[H^{+}]}^{2} [S^{- 2}]}{[H_{2} S]}$
$K = \frac{{(2 \times 10^{- 4})}^{2} [S^{- 2}]}{0.1}$
$[S^{- 2}] = 2.5 \times 10^{- 15} M$