Answer:
Fraction of association = 0.01
pH = 11
Explanation:
A weak base, B, is in equilibrium with water as follows:
B(aq) + H2O(l) ⇄ BH⁺(aq) + OH⁻(aq)
Where Kb is defined as:
Kb = 1.00x10⁻⁵ = [BH⁺] [OH⁻] / [B]
Some B will react producing BH⁺ and OH⁻:
[BH⁺] = X
[OH⁻] = X
[B] = 0.100M - X
As Kb <<< [B] we can say:
[B] ≈ 0.100M
Replacing:
1.00x10⁻⁵ = [X] [X] / [0.100]
1.00x10⁻⁶ = X²
X = 1x10⁻³M = [BH⁺] = [OH⁻]
The fraction of association is [BH⁺] / [B] = 1x10⁻³M / 0.100M = 0.01
As [OH⁻] = 1x10⁻³M, pOH = -log[OH⁻] = 3
pH = 14- pOH
<h3>pH = 11</h3>
