Codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) is a weak organic base. A \(5.0 \times 10^{-3} \mathrm{M}\) solution of codeine has a \(\mathrm{pH}\) of \(9.95 .\) Calculate the value of \(K_{b}\) for this substance. What is the \(\mathrm{p} K_{b}\) for this base?

Since we already have the pH value, we will first convert it to pOH by using the relationship: \[ \mathrm{p}H + \mathrm{p}OH = 14 \] Next, we can calculate the concentration of hydroxide ions, \([\mathrm{OH}^-]\), using the pOH value as follows: \[ [\mathrm{OH}^-] = 10^{-\mathrm{p}OH} \]

The base dissociation equation for codeine can be written as: \[ \mathrm{C}_{18}\mathrm{H}_{21}\mathrm{NO}_{3} + \mathrm{H}_2\mathrm{O} \rightleftarrows \mathrm{C}_{18}\mathrm{H}_{20}\mathrm{NO}_{3}^{-} + \mathrm{OH}^- \]

The expression for the base dissociation constant, \(K_b\), can be described as: \[ K_b = \frac{[\mathrm{C}_{18}\mathrm{H}_{20}\mathrm{NO}_{3}^-][\mathrm{OH}^-]}{[\mathrm{C}_{18}\mathrm{H}_{21}\mathrm{NO}_{3}]} \]

In order to calculate the value of \(K_b\), we need the equilibrium concentrations of all species. Let's assume that the initial concentration of the hydroxide ions is negligible, and for codeine, \([\mathrm{C}_{18}\mathrm{H}_{21}\mathrm{NO}_{3}] = 5.0 \times 10^{-3}\) M, as given in the exercise. As codeine dissociates, the equilibrium concentration of the negatively charged codeine ion, \([\mathrm{C}_{18}\mathrm{H}_{20}\mathrm{NO}_{3}^-]\), becomes equal to \([\mathrm{OH}^-]\). Now, we can plug these values into the \(K_b\) expression: \[ K_b = \frac{[\mathrm{C}_{18}\mathrm{H}_{20}\mathrm{NO}_{3}^-][\mathrm{OH}^-]}{[\mathrm{C}_{18}\mathrm{H}_{21}\mathrm{NO}_{3}]} \] Use the value of \(\mathrm{p}OH\) calculated in Step 1 and find \([\mathrm{OH}^-]\).

Once we've calculated the value of \(K_b\), we can determine the \(p K_{b}\) value as: \[ \mathrm{p}K_b = -\log_{10}(K_b) \] By plugging in the calculated value for \(K_b\) into this equation, we can find the \(p K_{b}\) value for the base codeine.