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

Explanation

`H_2S ` ionises in two stages . The first stage of ionisation is as follows:
` {:( , H_2S , hArr, H^(+) , +, HS^(-)),(" Initial conc" , 0.1M ,, - ,,- ),(" Conc. at eq" , (0.1-x)M,,x,,x):}`
` " " K_(a1) = 9.1 xx 10 ^(-8) ` (given )
According to the law of equilibrium ,
` K_(a1) = ([H^(+)][HS^(-)])/([H_2S]) `
or ` " " 9.1 xx 10 ^(-8) = (x.x)/(0.1-x) = (x^(2))/(0.1) `
` " " ` ( as x is very small and 0.1 - x = 0.1)
or ` " " x =(9.1xx 10 ^(-8) xx 0.1 )^(1//2) = 9.54 xx 10 ^(-5)`
Hence ` [HS^(-) ] ` in 0.1 M solution = x = ` 9.54 xx 10 ^(-5)`
In the presence of 0.1 M HCl :
In the presence of 0.1 M HCl . Suppose `H_2S ` dissociates to an extent of yM. Therefore,
` " " [H_2S]=0.1 -y ~~ 0.1 M, [H^(+)] = y+ 0.1 = 0.1 M`
and ` " " [HS^(-) ] =yM`
(Since in the presence of HCl, the dissociations of `H_2S ` will further get suppressed )
` therefore " " K_(a1) = (0.1 xx y )/(0.1) = 9.1 xx 10 ^(-8)`
or ` " " y= 9.1 xx 10 ^(-8) M`
Calculation of ` [S^(2-)]` :
` " " H_2S overset( K_(a1))hArr H^(+) HS^(-) , K_(a_1) = 9.1 xx 10 ^(-8)`
` HS^(-) overset( K_(a_2)) hArr H^(+) +S^(2-) , K_(a_2) =1.2 xx 10 ^(-3)`
The overall dissociations can be expressed as follows.
` H_2S hArr 2H^(+) S^(2-)`
` K_a = ([H^(+) ]^(2) [S^(2-)])/([H_2S]) `
` K_a= K_(a_1) xx K_(a_2) = 9.1 xx 10 ^(-8) xx 1.2 xx 10 ^(-13)`
` = 1.1 xx 10 ^(-20)`
In the absence of 0.1 M HCl :
` {:( ,H_2S , hArr, 2H^(+) , +, S^(2-) ),( " Initial conc. " , 0.1 M, ,- ,, -),(" Conc. at eq " , 0.1-x,, 2x.,, x):}`
` or " x. =((1.1 xx 10 ^(-20) xx 0.1)/(4))^(1//3) = 6.5 xx 10 ^(-8) M.`
Hence , `[S^(2-) ]` in the absence of `0.1 M HCl = 6.5 xx 10 ^(-8) M`.
In the presence of 0.1 M HCl :
Suppose `H_2S ` dissociates to an extent of . M.
` therefore " " [H_2S ) = 0.1 -y. ~~ 0.1 [H^(+) ] = 2y . + 0.1 ~~ 0.1 `
and ` " " [S^(2-) ]=y. `
` therefore " " K_a= ((0.1)^(2) xx y.)/(0.1 ) ~~1.1 xx 10 ^(-20) `
` or " " y.= (1.1 xx 10 ^(-20) )/(0.1) = 1.1 xx 10 ^(-19)`
Hence ,` [S^(2) ]` in the presence of 0.1 M HCl
` " " = 1.1 xx 10 ^(-19) M.`