Answer:
pH = 10.11
Explanation:
Hello there!
In this case, since it is possible to realize that this base is able to acquire one hydrogen atom from the water:
[tex]C_{18}H_{21}NO_4+H_2O\rightleftharpoons C_{18}H_{21}NO_4H^+OH^-[/tex]
We can therefore set up the corresponding equilibrium expression:
[tex]Kb=\frac{[C_{18}H_{21}NO_4H^+][OH^-]}{[C_{18}H_{21}NO_4]}[/tex]
Which can be written in terms of the reaction extent, [tex]x[/tex]:
[tex]Kb=\frac{x^2}{0.00500M-x}=3.39x10^{-6}[/tex]
Thus, by solving for [tex]x[/tex] we obtain:
[tex]x_1=-0.000132M\\\\x_2=0.0001285M[/tex]
However, since negative solutions are now allowed, we infer the correct [tex]x[/tex] is 0.0001285 M; thus, the pOH can be computed:
[tex]pOH=-log(x)=-log(0.0001285)=3.89[/tex]
And finally the pH:
[tex]pH=14-pOH=14-3.89\\\\pH=10.11[/tex]
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