What is another name for the sugars organisms use for energy?
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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 =
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 =
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
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 =
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 =
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