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Ch.16 - Acid-Base Equilibria
Chapter 16, Problem 71

Calculate the molar concentration of OH- in a 0.075 M solution of ethylamine (C2H5NH2); Kb = 6.4 * 10^-4. Calculate the pH of this solution.

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1
Identify the reaction: Ethylamine (C2H5NH2) is a weak base and will react with water to form C2H5NH3+ and OH-. Write the equilibrium expression for this reaction.
Write the expression for the base dissociation constant (Kb): Kb = [C2H5NH3+][OH-] / [C2H5NH2].
Set up an ICE (Initial, Change, Equilibrium) table to determine the concentrations of the species at equilibrium. Assume the initial concentration of OH- and C2H5NH3+ is 0, and the change in concentration for C2H5NH2 is -x, for OH- and C2H5NH3+ is +x.
Substitute the equilibrium concentrations from the ICE table into the Kb expression: Kb = (x)(x) / (0.075 - x).
Solve for x, which represents the concentration of OH- at equilibrium. Use the approximation method if x is small compared to the initial concentration, then calculate the pH using the relation pH = 14 - pOH, where pOH = -log[OH-].