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

First, let's write the chemical equation for the dissociation of triethanolammonium chloride:

(HOCH2CH2)3NCl ⇔ (HOCH2CH2)3NH+ + Cl-

The acid dissociation constant, Ka, for this reaction is given by: 
Ka = ([H+][HOCH2CH2)3NH+])/[HOCH2CH2)3NCl]) 
In a 0.010 M solution of triethanolammonium chloride, the initial concentration of HOCH2CH2)3NCl is 0.010 M, and the initial concentrations of HOCH2CH2)3NH+ and Cl- are both zero. At equilibrium, let x be the concentration of H+ and HOCH2CH2)3NH+. Then, the concentration of Cl- will also be x.

Using the Ka value given, we can set up an expression for x: 
5.9 x 10^-11 = (x^2)/(0.010-x)

Solving this equation gives x = 7.73 x 10^-6 M.

To find the pH, we take the negative logarithm of the H+ concentration: 
pH = -log[H+] = -log(7.73 x 10^-6) = 4.89

Therefore, the best answer is B, with a pH of 4.89.

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