Question

In the laboratory, a general chemistry student measured the pH of a 0.486 M aqueous solution of triethanolamine, C6H15O3N to be 10.740. Use the information she obtained to determine the Kb for this base.

Answer 1

Kb = 6.22x10⁻⁷

Step-by-step explanation:

Triethanolamine, C₆H₁₅O₃N, is in equilibrium with water:

C₆H₁₅O₃N(aq) + H₂O(l) ⇄ C₆H₁₅O₃NH⁺(aq) + OH⁻(aq)

Kb is defined from concentrations in equilibrium, thus:

Kb = [C₆H₁₅O₃NH⁺] [OH⁻] / [C₆H₁₅O₃N]

The equilibrium concentration of these compounds could be written as:

[C₆H₁₅O₃N] = 0.486M - X

[C₆H₁₅O₃NH⁺] = X

[OH⁻] = X

pH is -log [H⁺], thus, [H⁺] = 10^-pH = 1.820x10⁻¹¹M

Also, Kw = [OH⁻] ₓ [H⁺];

1x10⁻¹⁴ = [OH⁻] ₓ [H⁺]

1x10⁻¹⁴ = [OH⁻] ₓ [1.820x10⁻¹¹M]

5.495x10⁻⁴M = [OH⁻], that means X = 5.495x10⁻⁴M

Replacing in Kb formula:

Kb = [5.495x10⁻⁴M] [5.495x10⁻⁴M] / [0.486M-5.495x10⁻⁴M]

Kb = 6.22x10⁻⁷