Question

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?

Short answer

The value of \(K_b\) for codeine is approximately \(8.8 \times 10^{-11}\), and its \(pK_b\) value is approximately 10.05.

Step-by-step solution

Convert pH to pOH

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 \(pOH = 14 - pH\) \(pOH = 14 - 9.95\) \(pOH = 4.05\)

Calculate the Hydroxide ion concentration

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

Calculate the concentration of conjugate acid

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 \(x = [\mathrm{OH}^{-}] = 10^{-4.05}\).

Find the value of \(K_{b}\)

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.

Calculate the \(\mathrm{p} K_{b}\)

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 \(\mathrm{p} K_{b}\).

Key concepts

Weak Base

A weak base, like codeine, partially dissociates in water. This means only a small amount of the base breaks into ions. In the case of codeine, it interacts with water to form hydroxide ions (\( \text{OH}^- \)) and a conjugate acid When codeine (\( \text{C}_{18} \text{H}_{21} \text{NO}_{3} \)) is added to water, the following reaction occurs:\[\text{C}_{18} \text{H}_{21} \text{NO}_{3} + \text{H}_2\text{O} \rightleftharpoons \text{C}_{18} \text{H}_{21} \text{NO}_{3}\text{H}^+ + \text{OH}^-\]This equilibrium involves both the base itself and its products. Since weak bases do not fully dissociate, the concentration of hydroxide ions will be much smaller than the initial concentration of the base. It is this partial dissociation that determines the equilibrium dynamics we need to analyze.

pH and pOH Relationship

Understanding the pH and pOH relationship is crucial in solving problems involving weak bases. The pH scale measures how acidic or basic a solution is. It ranges from 0 to 14, where lower values are acidic, higher values are basic, and 7 is neutral.
However, for bases like codeine, we often discuss pOH instead.The pH and pOH are related by the equation:\[pH + pOH = 14\]This formula helps us convert a known pH into the pOH, and vice versa. For instance, if a solution has a pH of 9.95, the pOH can be calculated as 14 - 9.95, equating to 4.05. Each step thus interconnects by means of these crucial relationships, guiding us towards the analysis of base reactions.

Equilibrium Concentration

Determining the equilibrium concentration of ions is fundamental in calculating the ionization constant (\( K_b \)) for a weak base. At equilibrium, the concentrations of the products and reactants stabilize.From an initial concentration, changes occur as the reaction progresses towards equilibrium.
For codeine, equilibrium concentrations are described by the equation:

• Initial concentration of codeine: \( 5.0 \times 10^{-3} \text{M} \)
• Concentration of \( \text{OH}^- \): \( x \)
• Concentration of conjugate acid \( \text{C}_{18} \text{H}_{21} \text{NO}_{3}\text{H}^+ \): \( x \)
• Concentration of codeine at equilibrium: \( 5.0 \times 10^{-3} - x \)

Using the concentration of \( \text{OH}^- \), which is \( 10^{-4.05} \), we define \( x \) as the hydroxide ion concentration. Once \( x \) is determined, it aids in understanding how much has dissociated, providing insight into the reaction’s equilibrium state.

pK_b Calculation

The calculation of \( pK_b \) is a measure of the base's strength. Lower values of \( pK_b \) indicate stronger bases. The first step involves determining the \( K_b \), which is the equilibrium constant of the base dissociation equation.
The equation for \( K_b \) is:\[K_b = \frac{[\text{C}_{18} \text{H}_{21} \text{NO}_{3}\text{H}^+][\text{OH}^-]}{[\text{C}_{18} \text{H}_{21} \text{NO}_{3}]}\]By substituting the equilibrium concentrations into this equation, you can calculate the exact value of \( K_b \). After finding \( K_b \), \( pK_b \) is calculated using the formula:\[pK_b = -\log{K_b}\]This logarithmic transformation offers an easier way to discuss base strength. Calculating \( pK_b \) helps properly categorize the base's relative dissociation propensity, aiding in conceptualizing the behavior within solution dynamics.