Ethanamine, CH3CH2NH2, is a weak base. pKa(CH3CH2NH3+) = 10.6
(a) (i) Write an equation to show the reaction of ethanamine CH3CH2NH2 with water.
CH3CH2NH2 + H2O ⇌ CH3CH2NH3+ + OH-
(ii) Explain, with reference to the equation and the species in solution, what is meant by the terms weak and base.
In the equation, ethanamine (CH3CH2NH2) reacts with water to form ethanammonium ion (CH3CH2NH3+) and hydroxide ion (OH-). A weak base is a substance that partially dissociates in water, meaning it does not completely ionize. In this case, ethanamine only partially reacts with water to form the ethanammonium ion and hydroxide ion.
(b) (i) Determine the pKb for ethanamine.
pKb = 14 - pKa
pKb = 14 - 10.6
pKb = 3.4
(ii) Calculate the pH of a 0.109 mol L-1 solution of ethanamine.
To calculate the pH, we need to find the pOH first.
pOH = -log10[OH-]
pOH = -log10(0.109)
pOH = 0.962
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 0.962
pH = 13.038