Question

Oxycodone (C18H21NO4), a narcotic analgesic, is a weak base with pKb = 5.47. Calculate the pH of a .00500 M oxycodone solution.

Answer 1

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:

$C_{18}H_{21}NO_4+H_2O\rightleftharpoons C_{18}H_{21}NO_4H^+OH^-$

We can therefore set up the corresponding equilibrium expression:

$Kb=\frac{[C_{18}H_{21}NO_4H^+][OH^-]}{[C_{18}H_{21}NO_4]}$

Which can be written in terms of the reaction extent, $x$:

$Kb=\frac{x^2}{0.00500M-x}=3.39x10^{-6}$

Thus, by solving for $x$ we obtain:

$x_1=-0.000132M\\\\x_2=0.0001285M$

However, since negative solutions are now allowed, we infer the correct $x$ is 0.0001285 M; thus, the pOH can be computed:

$pOH=-log(x)=-log(0.0001285)=3.89$

And finally the pH:

$pH=14-pOH=14-3.89\\\\pH=10.11$

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