Question

Calculate [h3o+] and [s2−] in a 0.10 m solution of the diprotic acid hydrosulfuric acid. (for hydrosulfuric acid ka1 = 9.0 × 10−8 and ka2 = 1.0 × 10−17.) enter your answers in scientific notation.

Answer 1

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 = $\frac{X^{2} }{0.1-x}$
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 = $\frac{(x)(9.4 x 10^{-5} + x) }{(9.4 x 10^{-5} - x) }$
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