Question

Oxycodone (C18H21NO4), a narcotic analgesic, is a weak base with pKb = 5.47. Calculate the pH of a .00250 M oxycodone solution.

Answer 1

Answer: The pH of a 0.00250 M oxycodone solution is 9.96

Explanation:

$C_{18}H_{21}NO_4\rightarrow C_{18}H_{20}NO_3^++OH^-$

 cM                     0  M         0 M

$c-c\alpha$        $c\alpha$         $c\times \alpha$

So dissociation constant will be:

$K_b=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.00250 M and $pK_b$ = 5.47

$pK_b=-log(K_b)$

$K_b=3.38\times 10^{-6}$

Putting in the values we get:

$3.38\times 10^{-6}=\frac{(0.00250\times \alpha)^2}{(0.00250-0.00250\times \alpha)}$

$(\alpha)=0.036$

$[OH^-]=c\times \alpha$

$[OH^-]=0.00250\times 0.0369=9\times 10^{-5}$

Also $pOH=-log[OH^-]$

$pOH=-log[9\times 10^{-5}]=4.04$

$pH+pOH=14$

$pH=14-404=9.96$

Thus pH of a 0.00250 M oxycodone solution is 9.96