What are the concentration of H_(3) O^(+), HS^(-)and s^(2-) is 0.01MH_(2) Ssolution, given H_(2) S + H_(2) hArr H_(3) O^(+) + HS^(-), K_(a_(1)) = 1.1 xx 10^(-7) HS^(-) + H_(2) O hArr H_(3) O^(+) + S^(-) , K_(a_(2)) = 1.0 xx 10^(- 34 )

What are the concentration of H_(3) O^(+), HS^(-)and s^(2-) is 0.01MH_

1.	What are the concentration of H_(3) O^(+), HS^(-)and s^(2-) is 0.01MH_(2) Ssolution, given H_(2) S + H_(2) hArr H_(3) O^(+) + HS^(-), K_(a_(1)) = 1.1 xx 10^(-7) HS^(-) + H_(2) O hArr H_(3) O^(+) + S^(-) , K_(a_(2)) = 1.0 xx 10^(- 34 )
Answer» Solution :By ostwald DILUTION law, `[H_(3) O^(+)]` and `[HS^(-)]` due to first STEP IONIZATION are `[H_(3) O^(+)]= [HS^(-)] = sqrt( Ka_(1) C )` `= sqrt(1.1 XX 10^(-7) xx 0.1) =1 xx 10^(-4) M ` For second step ionization `HS^(-) + H_(2) O rarr H_(3) O^(+) + S^(2-)` Since `K_(a_(2)) lt lt lt K_(a)` hence ionization is very-very small and `[HS^(-)]` remains constant `K_(a_(2)) = ([ H_(3) O^(+)][S^(-2)])/( [HS^(-)])` `:. 1. 0 xx 10^(-14) = ( 1.0 xx 10^(-14) [ S^(2-)])/(1.0 xx 10^(-4))` `:. [S^(2-) ] = 1.0 xx 10^(-14) M `