What is the pH of a 0.010 M triethanolammonium chloride, (HOC₂H₂) ₃NHCl, solution?
K_b, ((HOC₂H₂) ₃N) = 5.9 × 10^-7

Question

What is the pH of a 0.010 M triethanolammonium chloride, (HOC₂H₂)₃NHCl, solution? K_b, ((HOC₂H₂)₃N) = 5.9 × 10^-7

Quizplus · Accepted Answer

The pH of the solution can be calculated using the equilibrium expression for the dissociation of the triethanolammonium cation: (HOC₂H₂)₃NH2+ (aq) ⇌ (HOC₂H₂)₃NH+ (aq) + H+ (aq) The equilibrium constant for this reaction, Kb, is related to the given equilibrium constant, Ka, for the conjugate acid, ammonium ion, by Kw/Ka, where Kw is the ion product constant for water. Kw = 1.0 × 10^-14 Kb = Kw/Ka = (1.0 × 10^-14)/(5.9 × 10^-7) = 1.69 × 10^-8 Using the quadratic formula to solve for [H+], we obtain: [H+] = 2.28 × 10^-5 M pH = -log[H+] = 4.64 Therefore, the best choice is B.

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