Find the pH and fraction of association of a 0.100 M solution of the weak base B with Kb = 1.00 x 10^5.

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

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explanation:

the answer is   c.) 1575 j.

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