Answer: The pH of a 0.00250 M oxycodone solution is 9.96
Explanation:
[tex]C_{18}H_{21}NO_4\rightarrow C_{18}H_{20}NO_3^++OH^-[/tex]
cM 0 M 0 M
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\times \alpha[/tex]
So dissociation constant will be:
[tex]K_b=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Give c= 0.00250 M and [tex]pK_b[/tex] = 5.47
[tex]pK_b=-log(K_b)[/tex]
[tex]K_b=3.38\times 10^{-6}[/tex]
Putting in the values we get:
[tex]3.38\times 10^{-6}=\frac{(0.00250\times \alpha)^2}{(0.00250-0.00250\times \alpha)}[/tex]
[tex](\alpha)=0.036[/tex]
[tex][OH^-]=c\times \alpha[/tex]
[tex][OH^-]=0.00250\times 0.0369=9\times 10^{-5}[/tex]
Also [tex]pOH=-log[OH^-][/tex]
[tex]pOH=-log[9\times 10^{-5}]=4.04[/tex]
[tex]pH+pOH=14[/tex]
[tex]pH=14-404=9.96[/tex]
Thus pH of a 0.00250 M oxycodone solution is 9.96
