Answer:
A. [OH⁻] = 2.188x10⁻³M
B. pKb = 6.02
Explanation:
When hydrazine is in equilbrium with water, its reaction is:
N₂H₄(aq) + H₂O(l) ⇄ HN₂H₄⁺(aq) + OH⁻(aq)
Where Kb, is defined as the ratio between concentrations in equilibrium of the species, thus:
Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]
A. From pH, you can find [OH⁻], thus:
pH = -log [H⁺]
11.34 = -log [H⁺]
4.57x10⁻¹² = [H⁺]
As 1x10⁻¹⁴ = [OH⁻] [H⁺]
1x10⁻¹⁴ / 4.57x10⁻¹² = [OH⁻]
B. Concentrations in equilibrium of the species are:
[N₂H₄] = 5.0M - X
[HN₂H₄⁺] = X
[OH⁻] = X
Where X is reaction coordinate
As [OH⁻] = 2.188x10⁻³M
X = 2.188x10⁻³M
Replacing:
[N₂H₄] = 5.0M - 2.188x10⁻³M = 4.9978M
[HN₂H₄⁺] = 2.188x10⁻³M
[OH⁻] = 2.188x10⁻³M
Replacing in Kb expression:
Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]
Kb = [2.188x10⁻³M] [2.188x10⁻³M] / [4.9978M]
Kb = 9.577x10⁻⁷
pKb is defined as -log Kb
pKb = -log 9.577x10⁻⁷
