Question

The first ionization constant of H_2­­S is 9.1 × 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 H₂S is 1.2 × 10^–13, calculate the concentration of S^2– under both conditions.

Answer 1

(i) To calculate the concentration of HS^– ion:

Case I (in the absence of HCl):

Let the concentration of HS^– be x M.

	H2S	↔	H+	+	HS-
Ci	0.1		0		0
Cf	0.1 - x		x		x

Then, `"K"_("a"_1)` = `(["H"^(+)]["HS"^(-)])/["H"_2"S"]`

`9.1 xx 10^(-8) = ((x)(x))/(0.1 - x)`

`(9.1 xx 10^(-8)) (0.1 - x) = x^2`

Taking 0.1 - x M ; 0.1M we have `(9.1xx10^(-8))(0.1) =  x^2`

`x = sqrt(9.1 xx 10^(-9))`

`= 9.54 xx 10^(-5)`M

`=> ["HS"^(-)] = 9.54 xx 10^(-5) "M"`

Case II (in the presence of HCl):

In the presence of 0.1 M of HCl, let `["HS"^(-)]` be yM

Then	H2S	↔	H+	+	HS-
Ci	0.1		0		0
Cf	0.1 - y		y		y

 

Also	HCl	↔	H+	+	Cl-
			0.1		0.1

Now `"K"_("a"_1)  = (["HS"^(-)]["H"^(+)])/["H"_2S]`

`"K"_("a"_1) = (["y"] (0.1 + "y"))/(0.1 - "y")`

`9.1 xx 10^(-8) = (y xx 0.1)/0.1 `     (∵ 0.1 - y ; 0.1 M)  (and 0.1 + y; 0.1 M)

`9.1 xx 10^(-8) = "y"`

`=>["HS"^(-)] = 9.1 xx 10^(-8)`

(ii) To calculate the concentration of [`"S"^(2-)`]

Case I (in the absence of 0.1 M HCl):

`"HS"^(-) ↔ "H"^(+) + "S"^(2-)`

`["HS"^(-)] = 9.54 xx 10^(-5) "M"` (From first ionization, case I)

Let `["S"^(2-)]` be X

Also `["H"^(+)] = 9.54  xx 10^(-5) "M"` (From first ionization case i)

`"K"_("a"_2) = (["H"^+]["S"^(2-)])/(["HS"^(-)])`

`"K"_("a"_2) = ((9.54 xx 10^(-5))("X"))/(9.54 xx 10^(-5))`

`1.2 xx 10^(-1) = "X" = ["S"^(2-)]`

Case II (in the presence of 0.1 M HCl):

Again, let the concentration of HS^– be X' M.

`["HS"^(-)] = 9.1 xx 10^(-8) "M"` (From first ionization, case II)

`["H"^+] = 0.1 "M"` (From HCl, case II)

`["S"^(2-)] = "X'"`

Then `"K"_("a"_2) = (["H"^+]["S"^(2-)])/["HS"^(-)]`

`1.2 xx 10^(-13) = ((0.1)(X'))/(9.1 xx 10^(8))`

`10.92 xx 10^(-21)` = 0.1 X'

`(10.92 xx 10^(-21))/0.1 = X'`

`"X'" = (1.092 xx 10^(-20))/0.1`

`= 1.092 xx 10^(-19) "M"`

`=> "K"_("a"_1) = 1.74 xx 10^(-5)`