Problem 73 - Chemistry: The Central Science

Question

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: $$ \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ A \(0.035 \mathrm{M}\) solution of ephedrine has a \(\mathrm{pH}\) of 11.33. (a) What are the equilibrium concentrations of \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}, \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}\), and \(\mathrm{OH}\) ? (b) Calculate \(K_{b}\) for ephedrine.

Step-by-Step Solution

Answer

The equilibrium concentrations for C10H15ON, C10H15ONH⁺, and OH⁻ are approximately 0.03286 M, 2.14 × 10⁻³ M, and 2.14 × 10⁻³ M, respectively. The value of Kb for ephedrine is approximately \(1.39 \times 10^{-5}\).

Step 1: Write the initial concentration

We are given that the initial concentration of ephedrine C10H15ON is 0.035 M. # Step 2: Calculate pOH #

Step 2: Calculate pOH

Since the pH of the solution is given as 11.33, we can calculate pOH using the relationship: pOH = 14 - pH pOH = 14 - 11.33 = 2.67 # Step 3: Calculate hydroxide ion concentration #

Step 3: Find the hydroxide ion concentration

Using the pOH value from step 2, we can calculate the hydroxide ion concentration: \[ OH^{-} = 10^{-pOH} \] \[ OH^{-} = 10^{-2.67} \approx 2.14 \times 10^{-3} \, M \] # Step 4: Set up the ICE table #

Step 4: Set up the ICE table

An ICE table (I = initial, C = change, E = equilibrium) will help us to visualize the changes in concentrations as the reaction proceeds. | Species | Initial (M) | Change (M) | Equilibrium (M) | |---------------|-------------|-----------------|-----------------| | C10H15ON | 0.035 | -[OH⁻] | 0.035-[OH⁻] | | C10H15ONH+ | 0 | +[OH⁻] | [OH⁻] | | OH⁻ | 0 | +[OH⁻] | 2.14×10⁻³ M | # Step 5: Calculate the equilibrium concentrations #

Step 5: Calculate the equilibrium concentrations

From the ICE table, we can determine the equilibrium concentrations: C10H15ONH⁺: [OH⁻] = 2.14 × 10⁻³ M C10H15ON: 0.035 - [OH⁻] = 0.035 - 2.14 × 10⁻³ ≈ 0.03286 M Now we have the equilibrium concentrations for C10H15ON, C10H15ONH⁺, and OH⁻. # Step 6: Calculate Kb #

Step 6: Calculate Kb

We can now use these equilibrium concentrations to calculate Kb, the base ionization constant for ephedrine: \[ K_b = \dfrac{[\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}]} \] \[ K_b = \dfrac{(2.14 \times 10^{-3})(2.14 \times 10^{-3})}{0.03286} \approx 1.39 \times 10^{-5} \] The value of Kb for ephedrine is approximately \(1.39 \times 10^{-5}\).

Key Concepts

ICE Table

When dealing with chemical equilibria, an ICE table is incredibly helpful. It stands for Initial, Change, and Equilibrium, representing the different stages of the concentrations of substances involved in a chemical reaction.
The ICE table allows one to track how the concentrations of reactants and products change from the initial state to the equilibrium state in a reaction.
Let's break down each component:

• Initial: This part shows the starting concentrations of all chemical species before any reaction occurs. For ephedrine, the starting concentration of the base, C₁₀H₁₅ON, is listed as 0.035 M, while the concentrations of products (C₁₀H₁₅ONH⁺ and OH^-) are initially 0 M since the reaction has not started.
• Change: This section represents the change in the concentrations as the system moves towards equilibrium. Typically, these changes are expressed in terms of a variable, often using 'x' to denote the amount of reactant that reacts. For simplification, here it's represented as −[OH^-] for the decrease in reactants and +[OH^-] for the formation of products.
• Equilibrium: This part indicates the concentrations of all species once equilibrium is reached. The concentration at equilibrium for each species is calculated using the initial concentration and the change. For example, if C₁₀H₁₅ON initially was 0.035 M and changed by −[OH^-], its equilibrium concentration becomes (0.035 - [OH^-]).

Using an ICE table simplifies solving equilibrium problems by organizing your data and calculations systematically.

Equilibrium Concentrations

Equilibrium concentrations are the amounts of reactants and products that remain after a chemical reaction has reached equilibrium.
In our example, knowing the equilibrium concentrations allows us to make other meaningful calculations, like determining ionization constants.
Here's how we find them:

• From the ICE table, it's straightforward to determine these concentrations by combining initial values and their respective changes recorded in the reaction's progress.
• For C₁₀H₁₅ON, the equilibrium concentration is found by subtracting the amount that reacts (2.14 × 10⁻³ M) from the initial concentration (0.035 M). This results in an equilibrium concentration of approximately 0.03286 M.
• For the products C₁₀H₁₅ONH⁺ and OH^-, since they start from zero, their equilibrium concentrations equal the change in concentration, 2.14 × 10⁻³ M.

Equilibrium concentrations are important because they reveal how far a reaction proceeds, and with this information, we can calculate the strength of the base via its ionization constant.

Base Ionization Constant

The base ionization constant, denoted as \(K_b\), measures the strength of a base in solution.
It tells us how well a base accepts protons or, in this case of ephedrine, how well it forms hydroxide ions in water.
Understanding \(K_b\) is crucial for predicting and calculating the behavior of weak bases like ephedrine.
Here's a look at how it's calculated:

• Based on the equilibrium concentrations obtained from the ICE table, \(K_b\) is derived using the formula:\[K_b = \dfrac{[\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}]}\]
• Plugging the equilibrium values into the formula: \(C_{10}H_{15}ONH^+\) is 2.14 × 10⁻³ M, \(OH^-\) is 2.14 × 10⁻³ M, and \(C_{10}H_{15}ON\) is 0.03286 M.
• Calculation gives us:\[K_b = \dfrac{(2.14 \times 10^{-3})(2.14 \times 10^{-3})}{0.03286} \approx 1.39 \times 10^{-5}\]

The resultant \(K_b\) value shows that ephedrine is a weak base, as its tendency to produce hydroxide ions in solution is relatively low. The information from \(K_b\) is beneficial in applications such as predicting the pH of solutions containing weak bases.