The first ionization constant of `H_(2)S` is `9.1xx10^(-8)`. Calculate the concentration of `HS^(Θ)` 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 if `H_(2)S` is `1.2xx10^(-13)`, calculate the concentration of `S^(2-)` under both conditions.

Explanation

`{:(,H_2S hArr , HS^(-)+,H^+),("Initial conc.", 0.1 M, 0,0),("After dissociation", 0.10-x, x,x):}`
where x is the amount of `H_2S` dissociated
`K_(a_1)=([HS^-][H^+])/([H_2S])=9.1xx10^(-8)`
or `(x xx x)/(0.10-x)=9.1xx10^(-8)`
Assuming x < < 0.10
`x^2/0.10=9.1xx10^(-8)`
`x^2=9.1xx10^(-8)xx0.1=9.1xx10^(-9)`
`therefore x=(9.1xx10^(-2))^(1//2) =9.54xx10^(-5)`
In the presence of 0.1 M HCl, suppose the amount of `H_2S` dissociated is y so that at equilibrium.
`[H_2S]=0.1-y approx 0.1, [H^+]=0.1 y approx 0.1` M , `[HS^-]=yM`
`K_a=(0.1xxy)/0.1=9.1xx10^(-8)`
or `y=9.1xx10^(-8)` .
To calculate concentration of `S^(2-)` ion, we are to consider the second dissociation constat :
`H_2S overset(K_(a_1))hArr H^(+) + HS^-`
`HS^(-) overset(K_(a_2))hArr H^(+)+S^(2-)`
Overall reaction
`H_2S hArr 2H^(+) +S^(2-)`
`K_a=K_(a_1)xxK_(a_2)=9.1xx10^(-8)xx1.2xx10^(-13)`
`=1.092xx10^(-20)`
`K_a=([H^+]^2[S^(2-)])/([H_2S])=1.092xx10^(-20)`
Now, in 0.1 M solution of `H_2S` :
`[H^+]=2x, [S^(2-)]=x, [H_2S]=0.1 -x approx 0.1`
`((2x)^2xxx)/0.1=1.092xx10^(-20)`
`4x^3=1.092xx10^(-21)`
`x^3=0.273xx10^(-21)`
or `x=(0.273xx10^(-21))^(1//3)`
`=6.49xx10^(-8)`
In presence of 0.1 HCl , suppose `[S^(2-)]` =y , then `[H^+]` = 0.1+y `approx` 0.1 M, `[H_2S] = 0.1 - y approx ` 0.1 M
`K_a=((0.1)^2xxy)/0.1=1.09xx10^(-20)`
or `y=1.09xx10^(-19)` M . .