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A solution contains 0.1MH(2)S and 0.3MHC...

A solution contains `0.1MH_(2)S` and `0.3MHCl` .Calculate the conc.of `S^(2-)` and `HS^(-)` ions is solutions.Given `K_(a_(1))` and `K_(a_(2))` are `10^(-7)`and `1.3xx10^(-13)` respectively.

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To solve the problem, we need to calculate the concentrations of the sulfide ion \( S^{2-} \) and the bisulfide ion \( HS^{-} \) in a solution containing \( 0.1 \, M \, H_2S \) and \( 0.3 \, M \, HCl \). Given the dissociation constants \( K_{a1} \) and \( K_{a2} \) for \( H_2S \) are \( 10^{-7} \) and \( 1.3 \times 10^{-13} \) respectively, we can follow these steps: ### Step 1: Determine the concentration of \( H^+ \) ions Since \( HCl \) is a strong acid, it will completely dissociate in solution: \[ [H^+] = [HCl] = 0.3 \, M \] ...
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