Question

Ethanolamine, CH_NO, is a clear, viscous, hygroscopic liquid with an ammonia-like odor. It is used to remove Co, and H2S from natural gas but it is also found in polishes, hair waving solutions, and in emulsifiers (often in pharmaceuticals). Overexposure to this compound can cause irritation to the eyes, skin and respiratory system as well as lethargy. If a 0.25M solution of this compound is found to be 1.125 % ionized, determine the Kb and the Ka for this substance.

Answer 1

Initial concentration of ethanolamine = 0.25 M

percent ionization = 1.125

fractional ionization = (percent ionization) / 100

fractional ionization = (1.125) / 100

fractional ionization = 0.01125

[OH^-] = (fractional ionization) * (initial concentration of ethanolamine)

[OH^-] = (0.01125) * (0.25 M)

[OH^-] = 0.00281 M

K_b = [OH^-]² / (initial concentration of ethanolamine - [OH^-])

K_b = (0.00281 M)² / (0.25 M - 0.00281 M)

K_b = 3.20 x 10^-5

K_a for conjugate acid = (K_w) / (K_b)

K_a for conjugate acid = (1.0 x 10^-14) / (3.20 x 10^-5)

K_a for conjugate acid = 3.12 x 10^-10