The equation for the first dissociation is:
H₂S(aq) + H₂O(l) ⇄ HS⁻(aq) + H₃O⁺(aq)
The acid dissociation constant, Ka1 is 9.0 x 10⁻⁸
Construct ICE table and obtain their equilibrium concentrations:
H₂S(aq) + H₂O(l) ⇄ HS⁻(aq) + H₃O⁺(aq)
I (M): 0.1 0 0
C (M): -x +x +x
E (M): 0.1 -x x x
So:
9.0 x 10⁻⁸ =
![\frac{X^{2} }{0.1-x}](https://tex.z-dn.net/?f=%20%5Cfrac%7BX%5E%7B2%7D%20%7D%7B0.1-x%7D%20)
x = 9.4 x 10⁻⁵
From the equilibrium table:
[H₃O⁺] = x = 9.4 X 10⁻⁵ M
[HS⁻] = x = 9.4 X 10⁻⁵ M
The equation for the second dissociation of the acid is:
HS⁻(aq) + H₂O(l) ⇄ S⁻²(aq) + H₃O⁺(aq)
The acid dissociation constant Ka2 is 1.0 x 10⁻¹⁷
Construct ICE table and obtain their equilibrium concentrations:
HS⁻(aq) + H₂O(l) ⇄ S²⁻(aq) + H₃O⁺(aq)
I (M): 9.4 x 10⁻⁵ 0 9.4 x 10⁻⁵
C (M): -x +x +x
E (M): 9.4 x 10⁻⁵ -x x 9.4 x 10⁻⁵ + x
So:
1.0 x 10⁻¹⁷ =
![\frac{(x)(9.4 x 10^{-5} + x) }{(9.4 x 10^{-5} - x) }](https://tex.z-dn.net/?f=%20%5Cfrac%7B%28x%29%289.4%20x%2010%5E%7B-5%7D%20%2B%20x%29%20%7D%7B%289.4%20x%2010%5E%7B-5%7D%20-%20x%29%20%7D%20)
x = 1.0 x 10⁻¹⁷ M
Therefore the equilibrium concentrations are as follows:
[S²⁻] = x = 1.0 x 10⁻¹⁷ M
[H₃O⁺] = 9.4 x 10⁻⁵ + 1.0 x 10⁻¹⁷ M = 9.4 x 10⁻⁵ M