The first ionization constant of `H_(2)S` is `9.1 xx 10^( -8)`. Calculate the concentration of `HS^(-)` ion in its 0.1 M solution. How will this concentration be affected, if the solution is 0.1 M in HCl also? If the second dissociation constant of `H_(2)S` is `1.2 xx 10 ^(-13)` , then calculate the concentration of `S^(2-)` under both conditions. Select these four answers from the choices given below.

Options

A.
B.
C.
D. None of these

Answer

A

Explanation

(i) `H_(2)S +H_(2)O^)(+) hArr H_(3)O^(+) +HS^(-)`
`K_(a_(1)) =([H_(3)O^(+)][HS^(-)])/( [H_(2)S]) `
`[H_(3) O^(+) ] = [HS^(-)] = sqrt(K_(a_(1)) *C) `
`=sqrt( 9.1 xx 10^(-8) xx 0.1) = 9.54 xx 10^(-5) M`
(ii) In the presence of 0.1 M HCl, `[H_(3)O^(+)]=0.1M`
`K_(a_(1)) = ([H_(2)O^(+)] [HS^(-)])/([H_(2)S])`
`9.1 xx 10^(-8) =([0.1] [HS^(-)])/([0.1])`
`[HS^(-)]=9.1 xx 10^(-8)M`.
Hence, the concentration of `[HS^(-)]` is decreased in the presence of 0.1 M HCl due to the common-ion effect.
(iii) For the second dissociation constant,
`HS^(-) +H_(2)O hArr H_(3)O^(+) +S^(2-) ` [In the absence of HCI]
`[HS^(-)] = [H_(3)O^(+)] `
`K_(a_(2))=([H_(3)O^(+)][S^(2-)])/([HS^(-)])`
`[S^(-2)] = Ka_(2) = 1.2 xx 10^(-13)`
(iv) In the presence of 0.1 M HCI,
`K_(a_(2)) = ([H_(2)O^(+)] [S^(2-)])/( [HS^(-)]) = 1.2 xx 10^(-13) = ([ 0.1 ][ S^(2-)])/( [9.1 xx 10^(-8)]) `
`[S^(2-)] = 1.092 xx 10^(-19) M`.