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 can be calculated using the expression \(K_{b} = \frac{(10^{-(14 - 9.95)})^2}{5.0 \times 10^{-3}}\). After finding the \(K_{b}\), the \(\mathrm{p}K_{b}\) can be determined using the formula \(\mathrm{p}K_{b} = -\log_{10}{K_{b}}\). Evaluating these expressions will give the desired results for \(K_{b}\) and \(\mathrm{p}K_{b}\).

Step-by-step solution

Find the concentration of hydroxide ions

The first thing we need to do is find the concentration of hydroxide ions (\([\mathrm{OH}^{-}]\)) in the solution. We're given the pH of the solution, but we're looking for the concentration of hydroxide ions, so we need to find the \(\mathrm{pOH}\) and then convert it to the hydroxide ion concentration. $$ \mathrm{pOH} = 14 - \mathrm{pH} $$ So, $$ \mathrm{pOH} = 14 - 9.95 $$

Calculate the concentration of hydroxide ions

Now, we can calculate the concentration of hydroxide ions using the following formula: $$ [\mathrm{OH}^{-}] = 10^{-\mathrm{pOH}} $$ Substituting the value of \(\mathrm{pOH}\) calculated in step 1: $$ [\mathrm{OH}^{-}] = 10^{-(14 - 9.95)} $$

Define the base dissociation equilibrium constant expression

We know that the base dissociation for codeine, \(\mathrm{C_{18}H_{21}NO_{3}}\), is given by the following equation: $$ \mathrm{C_{18}H_{21}NO_{3} + H_2O} \rightleftharpoons \mathrm{C_{18}H_{21}NO_{3}H^{+} + OH^{-}} $$ So, the equilibrium constant expression for this reaction is: $$ K_{b} = \frac{[\mathrm{C_{18}H_{21}NO_{3}H^{+}][\mathrm{OH}^{-}]}{[\mathrm{C_{18}H_{21}NO_{3}}]} $$

Calculate the value of \(K_{b}\) using the concentrations

Since codeine is a weak base, we can assume that the concentration of codeine ([codeine]) before the reaction is equal to the initial concentration. $$ [\mathrm{C_{18}H_{21}NO_{3}}] = 5.0 \times 10^{-3} \mathrm{M} $$ After the reaction, we're left with \(x\) moles of \(\mathrm{C_{18}H_{21}NO_{3}H^{+}}\) and \(\mathrm{OH}^{-}\), where \(x = 10^{-(14 - 9.95)}\). Then, the expression for \(K_{b}\) becomes: $$ K_{b} = \frac{x^2}{5.0 \times 10^{-3} - x} $$ By assuming that \(x\) is small compared to \(5.0 \times 10^{-3}\), we can simplify the equation as follows: $$ K_{b} = \frac{x^2}{5.0 \times 10^{-3}} $$ Now substituting the value of \(x\) which is the concentration of hydroxide ions: $$ K_{b} = \frac{(10^{-(14 - 9.95)})^2}{5.0 \times 10^{-3}} $$

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

Finally, we can calculate the \(\mathrm{p}K_{b}\) using the relationship between \(\mathrm{p}K_{b}\) and \(K_{b}\): $$ \mathrm{p}K_{b} = -\log_{10}{K_{b}} $$ Substitute the value of \(K_{b}\) calculated in step 4: $$ \mathrm{p}K_{b} = -\log_{10}{\frac{(10^{-(14 - 9.95)})^2}{5.0 \times 10^{-3}}} $$ At this point, you can evaluate the numerical values in steps 1, 2, 4, and 5 to get the desired results for \(K_{b}\) and \(\mathrm{p}K_{b}\).

Key concepts

Weak Base

A weak base is a type of chemical base that does not fully ionize in solution, meaning that not all the base molecules lose a proton to form a hydroxide ion \(\mathrm{(OH^-)}\). This partial ionization leads to an equilibrium state between the base, its conjugate acid, and the hydroxide ions.

Weak bases are important because they help to maintain the pH of a solution without fully neutralizing it. They are used in various applications, such as buffering solutions to maintain stable pH levels. Codeine is an example of a weak organic base, where its base form \(\mathrm{C_{18}H_{21}NO_{3}}\) exists in equilibrium with its ions.

The base dissociation constant \(K_b\) is a measure of the strength of a base in solution. It is used to express the equilibrium concentration ratio of the products over reactants. The lower the \(K_b\) value, the weaker the base. Determining the value of \(K_b\) for substances like codeine helps us understand how it will behave in a solution.

pH and pOH

Understanding both pH and pOH is essential in dealing with acids and bases in chemistry. These values are logarithmic measures reflecting the concentration of hydrogen ions \(\mathrm{(H^+)}\) and hydroxide ions \(\mathrm{(OH^-)}\) in a solution.

is a scale used to specify how acidic or basic a water-based solution is. The pH scale ranges from 0 to 14, with 7 being neutral. Values less than 7 represent an acidic solution, whereas values greater than 7 indicate a basic solution.

When calculating the properties of a weak base like codeine, we often need to determine the pOH, which is linked to the pH through the equation \(\text{pOH} = 14 - \text{pH}\). This relation allows us to find the concentration of hydroxide ions, necessary for the calculation of the base dissociation constant \(K_b\).

• To find the concentration of \(\mathrm{OH^-}\), calculate it using \(\text{pOH}\) as \(\[ \mathrm{[OH^-] = 10^{-pOH}} \]\).

Having a grasp of both pH and pOH is crucial for properly describing the behavior of weak bases in solution, which ultimately ties into their equilibrium characteristics.

Chemical Equilibrium

Chemical equilibrium occurs in a reversible reaction when the rates of the forward and backward reactions are equal, causing the concentrations of both reactants and products to remain constant. In the case of weak bases, this dynamic state is vital as only a portion of the base ionizes in solution.

Codeine, being a weak base, exemplifies a state of equilibrium where the equilibrium expression of its dissociation can be written as follows:

• The equilibrium expression is \( K_b = \frac{[\mathrm{C_{18}H_{21}NO_{3}H^+}][\mathrm{OH^-}]}{[\mathrm{C_{18}H_{21}NO_{3}}]} \).

In such expressions, the constant \(K_b\) reflects how much the base dissociates.

The key to understanding chemical equilibrium in the context of weak bases like codeine lies in recognizing that while a small amount will dissociate to produce \(\mathrm{C_{18}H_{21}NO_{3}H^+}\) and \(\mathrm{OH^-}\), much of the substance remains undissociated, maintaining a balance between the reactants and products. This concept is critical to predicting the extent of a reaction and how various conditions can shift the equilibrium to favor either ionization or formation of the base.