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

To calculate the pH of a .00250 M oxycodone solution, we need to determine the concentration of hydrogen ions [H+]. Using the pKb value of 5.47, we can calculate the pKa value as 14 - 5.47 = 8.53. Plugging in the pKa value, we find that the pH of the oxycodone solution is 8.73.

Step-by-step explanation:

Oxycodone (C18H21NO4) is a weak base with pKb = 5.47. To calculate the pH of a .00250 M oxycodone solution, we need to determine the concentration of hydrogen ions [H+].

To do this, we can use the equation pKw = pKa + pKb, where pKw = 14 (the ionization constant of water), and pKa is the negative logarithm of the acid dissociation constant, which can be calculated from the pKb value. In this case, the pKb value is 5.47, so we can calculate pKa as 14 - 5.47 = 8.53.

Knowing the pKa value, we can calculate [H+] using the equation [H+] = 10^(-pH). Plugging in the pKa value of 8.53, we get [H+] = 10^(-8.53) = 1.86 x 10^(-9). Taking the negative logarithm of [H+], we get pH = -log(1.86 x 10^(-9)) = 8.73.

Answer 2

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

Step-by-step 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* \alpha`

So dissociation constant will be:

`K_b=((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* 10^(-6)`

Putting in the values we get:

`3.38* 10^(-6)=((0.00250* \alpha)^2)/((0.00250-0.00250* \alpha))`

`(\alpha)=0.036`

`[OH^-]=c* \alpha`

`[OH^-]=0.00250* 0.0369=9* 10^(-5)`

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

`pOH=-log[9* 10^(-5)]=4.04`

`pH+pOH=14`

`pH=14-404=9.96`

Thus pH of a 0.00250 M oxycodone solution is 9.96