Question

The hydronium ion concentration of an aqueous solution of 0.359 M triethanolamine (a weak base with the formula C₆H₁₅O₃N) is ...

Answer 1

kb for triethanolamine = 5.8*10^-7

C6H15O3N dissociates as:

C6H15O3N -----> C6H15O3NH+ + OH-

0.359 0 0

0.359-x x x

Kb = [C6H15O3NH+][OH-]/[C6H15O3N]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.8*10^-7)*0.359) = 4.563*10^-4

since c is much greater than x, our assumption is correct

so, x = 4.563*10^-4 M

use:

[H+] = (1.0*10^-14)/[OH-]

[H+] = (1.0*10^-14)/4.563*10^-4

[H+] = 2.19*10^-11 M

Answer: 2.19*10^-11 M