Answer:
[tex]Kb=6.16x10^{-7}[/tex]
Explanation:
Hello,
In this case, given the pH of the base, we can compute the pOH as shown below:
[tex]pOH=14-pH=14-10.781=3.219[/tex]
Next, we compute the concentration of the hydroxyl ions when the triethanolamine is dissociated:
[tex]pOH=-log([OH^-])[/tex]
[tex][OH^-]=10^{-pOH}=10^{-3.219}=6.04x10^{-4}M[/tex]
Then, by writing the equilibrium expression for the dissociation of triethanolamine we have:
[tex]Kb=\frac{[OH^-][C6H14O2N^+]}{[C6H15O3N ]}[/tex]
That is suitable for the direct computation of Kb, knowing that based on the ICE procedure, [tex]x[/tex] equals the concentration of hydroxyl ions that was previously, computed, therefore, we have:
[tex]Kb=\frac{6.04x10^{-4}M*6.04x10^{-4}M}{0.592M-6.04x10^{-4}M}\\ \\Kb=6.16x10^{-7}[/tex]
Regards.
