Question

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

Short answer

The value of \(K_b\) for codeine is \(\frac{10^{4.05}(10^{-4.05} - 5.0\times10^{-3})}{(5.0\times10^{-3})}\). The \(pK_b\) for codeine is \(-\log_{10}\left(\frac{10^{4.05}(10^{-4.05} - 5.0\times10^{-3})}{(5.0\times10^{-3})}\right)\).

Step-by-step solution

Determine the concentration of hydroxide ions

Given the pH of the codeine solution as 9.95, we can find the concentration of hydroxide ions, using the following equation: \[ [\mathrm{OH}^{-}] = 10^{(14 - \mathrm{pH})} \] Now, let's calculate the concentration of hydroxide ions: \[ [\mathrm{OH}^{-}] = 10^{(14 - 9.95)} = 10^{4.05} \]

Calculate the change in concentration

Let \(x\) be the change in concentration of \(\mathrm{OH}^-\) due to the ionization of codeine which is initially \(5.0 \times 10^{-3}M\). Thus, at equilibrium, we have: \[ \begin{cases} [\mathrm{OH}^-] = 10^{4.05}+x\\ [\mathrm{Codeine}] = 5.0 \times 10^{-3} - x \end{cases} \] Now, we can use these expressions in the \(K_b\) expression for codeine ionization.

Calculate the Kb expression

The ionization of codeine in water can be represented by the following equation: \[ \mathrm{Codeine} + \mathrm{H_2O} \leftrightharpoons \mathrm{CodeineOH^{+}} + \mathrm{OH^{-}} \] Now, using the equilibrium concentrations, we have: \[ K_b = \frac{[\mathrm{CodeineOH^{+}}][\mathrm{OH}^{-}]}{[\mathrm{Codeine}]} \] Since the change in the concentration of codeine is very small comparing to its initial concentration, we can assume \(x<<5.0\times10^{-3}M\). Thus, we have: \[ K_b = \frac{x(10^{4.05}+x)}{5.0 \times 10^{-3}} \] Since \(x<<10^{4.05}\), we can simplify this equation by ignoring the \(x^2\) term as it's negligible.

Find the value of Kb

Now, we can solve for \(x\): \[ K_b = \frac{x(10^{4.05})}{5.0 \times 10^{-3}} \] \[ x = K_b \times \frac{5.0 \times 10^{-3}}{10^{4.05}} \] Using the initial relationship from Step 2, we have: \[ 10^{4.05}+x = 10^{4.05}+K_b \times \frac{5.0 \times10^{-3}}{10^{4.05}} \] Now, solve for \(K_b\): \[ K_b = \frac{10^{4.05}(10^{-4.05} - 5.0\times10^{-3})}{(5.0\times10^{-3})} \]

Determine the pKb value

Once we have the value of \(K_b\), we can find the \(pK_b\) using the following equation: \[ \mathrm{pK}_{b} = -\log_{10}(K_{b}) \] Now, calculate the \(pK_b\) for codeine: \[ \mathrm{pK}_{b} = -\log_{10}\left(\frac{10^{4.05}(10^{-4.05} - 5.0\times10^{-3})}{(5.0\times10^{-3})}\right) \]

Key concepts

Acid-Base Equilibria

Understanding acid-base equilibria is crucial when studying chemistry, especially when analyzing the behavior of weak bases like codeine. The equilibrium in an acid-base reaction refers to the point at which the rate of the forward reaction (base reacting with water to form its conjugate acid and hydroxide ions) is equal to the rate of the reverse reaction (the conjugate acid donating a proton to hydroxide ions to reform the base and water). In the case of weak bases, this equilibrium lies far to the left, meaning that only a small fraction of the base is ionized in solution. It's this dynamic balance and the concentration of the resulting ions that play a key role in understanding the base's behavior in solution.

Hydroxide Ion Concentration

The hydroxide ion concentration, represented as \[\mathrm{OH}^{-}\], is an essential part of the puzzle for calculating the basicity of a compound in water. It directly affects the pH of the solution, which is a measure of how acidic or basic the solution is. In the context of the provided exercise, determining the hydroxide ion concentration involves using the pH, and understanding its inverse relationship with the hydroxide concentration helps to further make sense of the ionization process of the weak base in water.

pH and pOH Relationship

The pH and pOH of a solution are interconnected and are key to understanding how acidic or basic a solution is. The pH measures the concentration of hydrogen ions \(\mathrm{H}^{+}\) and the pOH measures the concentration of hydroxide ions \(\mathrm{OH}^{-}\). They are related by the equation \(\mathrm{pH} + \mathrm{pOH} = 14\) in aqueous solutions at 25°C. This relationship is derived from the ion product constant for water and is a pivotal concept in determining other properties of acids and bases, like their ionization constants.

Weak Base Ionization

Ionization of a weak base in water is a partial process. This means that not all the base molecules will react to form hydroxide ions. The base ionization is represented by an equilibrium expression that helps to understand the extent to which a base will donate electrons to form hydroxide ions. Calculating the degree of ionization or solutions such as those containing codeine can reveal a lot about the properties of the base and the solution's pH.

Equilibrium Constant

The equilibrium constant, represented by the symbol \(K\), indicates the extent of a reaction at equilibrium and is temperature-dependent. For the reaction of a weak base with water, we use the \(K_{b}\) value, which is the base ionization constant. This value is a ratio of the concentration of the products to the reactants, excluding water as it is the solvent. The \(K_{b}\) helps to predict the degree of base ionization and is crucial in calculating other parameters like \(pK_{b}\), allowing us to quantify the strength of the base.

Polyatomic Ions

Polyatomic ions, such as the codeine ion formed during the weak base ionization of codeine, consist of two or more atoms bonded together that carry a net positive or negative charge. Understanding these ions is important for balancing equations and calculating ionization constants in acid-base chemistry. They play a significant role in the properties of the compounds they form and are essential in biological, environmental, and industrial chemical processes.