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 pH of 9.95. Calculate the value of \(K_{b}\) for this substance. What is the \(\mathrm{p} K_{b}\) for this base?

First, we need to find the pOH of the solution, which can be done by using the relationship between pH and pOH: \(pH + pOH = 14\) Given pH is 9.95, so \(pOH = 14 - pH\) \(pOH = 14 - 9.95\) \(pOH = 4.05\)

Next, we will convert pOH to the hydroxide ion concentration (\([\mathrm{OH}^{-}]\)), by using the formula: \(pOH = -\log{[\mathrm{OH}^{-}]}\) Therefore, we have: \([\mathrm{OH}^{-}] = 10^{-pOH}\) \([\mathrm{OH}^{-}] = 10^{-4.05}\)

Since codeine is a weak base, it will react with the water molecules to produce hydroxide ions and its conjugate acid: \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\mathrm{H}^{+} + \mathrm{OH}^{-}\) At equilibrium, let the concentration of \(\mathrm{OH}^{-}\) be \(x\), the concentration of \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\mathrm{H}^{+}\) will also be x, and the initial concentration of \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\) will decrease by x. So the concentrations will become: \([\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}] = 5.0 \times 10^{-3} - x\) \([\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\mathrm{H}^{+}] = x\) \([\mathrm{OH}^{-}] = x\) Since we have already calculated \([\mathrm{OH}^{-}]\) from Step 2, we have \(x = [\mathrm{OH}^{-}] = 10^{-4.05}\).

Now that we have all the equilibrium concentrations, we can find the value of \(K_{b}\) using the formula: \(K_{b} = \frac{[\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\mathrm{H}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}]}\) Plugging in the values: \(K_{b} = \frac{ (10^{-4.05})(10^{-4.05})}{ 5.0 \times 10^{-3}}\) Calculate the value of \(K_{b}\) with these concentrations.

Finally, to find the \(pK_{b}\), we use the formula: \(\mathrm{p} K_{b} = -\log{K_{b}}\) Substitute the value of \(K_{b}\) from Step 4 and, you will get the value of \(\mathrm{p} K_{b}\).