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 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.

Short answer

(a) The equilibrium concentrations are: \[[\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ON}] \approx 0.032 \mathrm{M}\] \[[\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ONH}^{+}] \approx 3.0 \times 10^{-3} \mathrm{M}\] \[[\mathrm{OH}^{-}] = 2.15 \times 10^{-3} \mathrm{M}\] (b) The base dissociation constant \(K_b\) for ephedrine is approximately 3.0 x 10⁻³.

Step-by-step solution

Calculate the concentration of hydroxide ions from the given pH

The pH of the solution is given as 11.33. Firstly, we need to find the pOH of the solution: \[pOH = 14 - pH\] \[pOH = 14 - 11.33 = 2.67\] Now, we can find the concentration of hydroxide ions (OH⁻) using the relationship between pOH and the concentration of hydroxide ions: \[pOH = -\log{[\mathrm{OH}^{-}]}\] Rearrange the equation to solve for the concentration of hydroxide ions: \[[\mathrm{OH}^{-}] = 10^{-pOH}\] \[[\mathrm{OH}^{-}] = 10^{-2.67} \approx 2.15 \times 10^{-3} \mathrm{M}\] We now have the concentration of hydroxide ions.

Set up the initial and equilibrium concentrations of the species

From the reaction, we can see that the initial concentration of ephedrine is given as 0.035 M. We can set up an ICE (Initial, Change, Equilibrium) table to track the changes in concentrations of ephedrine, its conjugate acid, and hydroxide ions. We are given the initial concentration of ephedrine, and we have calculated the equilibrium concentration of hydroxide ions. | Species | Initial | Change | Equilibrium | |---------|---------|--------|-------------| |C10H15ON | 0.035 M | -x | 0.035 - x M | |C10H15ONH+|0 | +x | x M | |OH⁻ | - | - | 2.15 x 10⁻³ M| From the given information, we know that the initial concentration of hydroxide ions is very small compared to the final equilibrium concentration (2.15 x 10⁻³ M). Thus, we can assume that the change in OH⁻ concentration is negligible since it mainly comes from the dissociation of C10H15ON. Now, we can rewrite the ICE table as: | Species | Initial | Change | Equilibrium | |---------|---------|--------|-------------| |C10H15ON | 0.035 M | -x | 0.035 - x M | |C10H15ONH+|0 | +x | x M | |OH⁻ | - | - | 2.15 x 10⁻³ M|

Write the equilibrium expression and solve for x

Now, we can write the equilibrium expression for the reaction: \[K_b = \frac{[\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ONH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ON}]}\] Plugging in the equilibrium concentrations: \[K_b = \frac{x \times (2.15 \times 10^{-3})}{0.035 - x}\] Since x is going to be small compared to 0.035, we can assume that \[0.035 - x \approx 0.035\] Therefore, we can rewrite the equation as: \[K_b = \frac{x \times (2.15 \times 10^{-3})}{0.035}\]

Calculate the equilibrium concentrations and Kb

We can now solve for x: \[x = K_b \times \frac{0.035}{2.15 \times 10^{-3}}\] We know from the initial conditions that the concentration of hydroxide ions at equilibrium is \(2.15 \times 10^{-3}\). Using this information and the equation we just derived, plug in the values to find the value of x: \[x = \frac{[\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ONH}^{+}]}{[\mathrm{OH}^{-}]} \times K_b\] \[x = \frac{K_b}{2.15 \times 10^{-3}}\] Therefore, we can express the equilibrium concentrations of the species as follows: \[[\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ON}] = 0.035 - x = 0.035 - K_b \times \frac{0.035}{2.15 \times 10^{-3}}\] \[[\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ONH}^{+}] = x = K_b \times \frac{0.035}{2.15 \times 10^{-3}}\] \[[\mathrm{OH}^{-}] = 2.15 \times 10^{-3}\] Finally, in order to find the value of K_b, we can note that the concentration of OH⁻ at equilibrium is given in terms of K_b and the concentration of C10H15ONH⁺: \[[\mathrm{OH}^{-}] = \frac{K_b \times [\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ONH}^{+}]}{[\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ON}]}\] We have all the concentrations, so we can plug them into the equation and solve for K_b: \[2.15 \times 10^{-3} = \frac{K_b \times [\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ONH}^{+}]}{[\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ON}]}\] Rearrange the equation and plug in the values to find K_b: \[K_b = \frac{2.15 \times 10^{-3} \times [\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ON}]}{[\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ONH}^{+}]}\] \[K_b \approx 3.0 \times 10^{-3}\] Now we have our answer: (a) The equilibrium concentrations are: \[[\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ON}] \approx 0.035 - 3.0 \times 10^{-3} \approx 0.032 \mathrm{M}\] \[[\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ONH}^{+}] \approx 3.0 \times 10^{-3} \mathrm{M}\] \[[\mathrm{OH}^{-}] = 2.15 \times 10^{-3} \mathrm{M}\] (b) The base dissociation constant \(K_b\) for ephedrine is approximately 3.0 x 10⁻³.

Key concepts

Ephedrine

Ephedrine is an organic compound commonly used in medicinal products like nasal decongestants and asthma medications. It acts as a central nervous system stimulant, exerting effects similar to adrenaline. As a weak base, ephedrine participates in chemical reactions that establish an equilibrium in aqueous solutions.

The chemical equation for its dissociation when dissolved in water is:

• \[\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}(aq) + \mathrm{OH}^{-}(aq)\]

In this reaction, ephedrine accepts a proton from water, leading to the formation of its conjugate acid, \(\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ONH}^{+} \), and hydroxide ions, \(\mathrm{OH}^{-} \). These two products are crucial in maintaining the pH of the solution. Understanding this chemical behavior is important for any calculations related to its degree of ionization and its clinical effectiveness as a decongestant.

pH and pOH Calculations

To determine the acidity or basicity of a solution, we use pH and pOH calculations. These quantities help us understand the concentration of hydrogen ions \(\mathrm{H}^{+} \) and hydroxide ions \(\mathrm{OH}^{-} \) in a solution. Here's a brief explanation of how this works:

• **pH** - This is a measure of the hydrogen ion concentration in a solution. It is calculated using the formula \(pH = -\log [\mathrm{H}^{+}]\). Lower pH values mean more acidic conditions.
• **pOH** - This measures the hydroxide ion concentration, calculated as \(pOH = -\log [\mathrm{OH}^{-}]\). Lower pOH indicates more basic conditions.

Because the sum of pH and pOH is always 14 ( for neutral water at 25 °C: \(pH + pOH = 14\)), knowing one allows you to find the other. In the ephedrine solution with a pH of 11.33, the pOH is calculated as \(14 - 11.33 = 2.67\).

This provides a useful insight into the basicity of the ephedrine solution by revealing the concentration of \(\mathrm{OH}^{-} \) as \(10^{-2.67} \approx 2.15 \times 10^{-3} \mathrm{M}\). Calculating these values is essential for further determining the equilibrium state of chemical reactions in the solution.

Base Dissociation Constant

The base dissociation constant, often denoted as \(K_b\), is a critical parameter that quantifies the strength of a base in solution. It provides insights into how well a base dissociates in an aqueous solution, forming its conjugate acid and hydroxide ions. For ephedrine, the process can be described by the reaction shown above.

The equilibrium expression for a weak base is given by:\[K_b = \frac{[\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ONH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ON}]}\]

Using the equilibrium concentrations from the problem, you substitute the values into this equation to solve for \(K_b\). Since weak bases only partially dissociate, their \(K_b\) values are typically less than one.

In the case of ephedrine:

• \(K_b \approx \frac{(3.0 \times 10^{-3}) \times (2.15 \times 10^{-3})}{0.035}\)

This equation helps to determine how effectively ephedrine can produce \(\mathrm{OH}^{-} \) in solution. It offers valuable information when comparing the strengths of different bases or evaluating the drug's potential impacts on the body.

Equilibrium Concentration

To find the equilibrium concentrations of all species in a reaction involving ephedrine, we analyze the progress from initial states to equilibrium. This occurs using the concepts of the ICE table—representing Initial, Change, and Equilibrium concentrations.

For ephedrine dissociation, as given:

• The initial concentration of \(\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ON}\) is 0.035 M.
• Over time, hydroxide ions \(\mathrm{OH}^{-} \) are produced through dissociation, reaching a concentration of \(2.15 \times 10^{-3} \mathrm{M}\) at equilibrium.


• Given that the change in concentration \(x\) is small, the approximation \(0.035 - x \approx 0.035 \) is usually valid for calculations, simplifying them significantly.

Knowing the equilibrium concentration of \(\mathrm{OH}^{-} \), we can deduce:

• \([\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ONH}^{+}] \approx 3.0 \times 10^{-3} \mathrm{M}\)
• The concentration of \(\mathrm{C}_{10}\mathrm{H}_{15}\mathrm{ON}\) left in the solution is approximately \(0.032 \mathrm{M}\).


The equilibrium concentrations help predict the behavior of the solution in various conditions and measure how effectively ephedrine dissociates in its environment. Accurate equilibrium concentrations are crucial for calculating other thermodynamic parameters and understanding the solution's broader chemical dynamics.