Question

Calculate the concentration of S^2- in an aqueous solution of 0.274 M hydrosulfuric acid, H₂S (aq).

[S^2-] = _____ M.

Answer 1

Sol :-

ICE table for the first dissociation of H₂S is :

..................................H₂S <-------------------------> HS^-.............................+.........................H⁺  

Initial (I)....................0.274 M...............................0.0 M.....................................................0.0 M

Change (C).................-y.......................................+y..........................................................+y

Equilibrium (E).........(0.274-y) M.............................y M.......................................................y M

Expression of K_a1 is :

K_a1 = [HS^-].[H⁺]/[H₂S]

8.9 x 10^-8 = y²/(0.274-y)

y <<<0.274, So neglect y as compare to 0.274

y² = 2.44 x 10^-8

y = 1.56 x 10^-4

So, [H⁺] = [HS^-] = y = 1.56 x 10^-4 M

Again, second dissociation is :

..................................HS^- <-------------------------> S^2-.............................+.........................H⁺  

Initial (I)..................1.56 x 10^-4 M.....................0.0 M.....................................1.56 x 10^-4 M

Change (C).................-y.......................................+y.....................................................+y

Equilibrium (E).........(1.56 x 10^-4 -y) M.............................y M.............................(1.56 x 10^-4  +y) M

Expression of K_a2 is :

K_a2 = [S^2-].[H⁺]/[HS^-]

1.0 x 10^-19 = y(1.56 x 10^-4  +y) / (1.56 x 10^-4 -y)

y <<<1.56 x 10^-4 , So neglect y as compare to 1.56 x 10^-4  

y = 1.0 x 10^-19

Hence, concentration of S^2- = y = 1.0 x 10^-19 M