Answer to Question #108489 in General Chemistry for Beverlie

Question

1. Calculate the pH of a 1.08×10-2 M solution of the decongestant ephedrine hydrochloride if the pKb of ephedrine (its conjugate base) is 3.86.

Answer 1

The acid-base equilibrium is:

EphH⁺ = Eph + H⁺

The constant is Ka = Kw/Kb, where pKw = 14,

pKa = 14 - pKb = 10.14.

The equilibrium concentration of [H⁺] is equal to [Eph]. The equilibrium concentration of [EphH⁺] = 1.08*10^-2 - [H⁺].

The equilibrium constant is very small (10^-10.14) and the concentration of EphH⁺ is by 8 orders of magnitude higher (1.08*10^-2), that is why the degree of dissociation is negligible and the equilibrium concentration of [EphH⁺] is close its initial concentration 1.08*10^-2.

Thus, Ka = 10^-10.14 = [H⁺]²/[EpH⁺] = [H⁺]²/(1.08*10^-2),

in terms of logarithms:

pH = (pKa + 2 - lg1.08)/2 = 6.05