Question

A \(0.50 \mathrm{M}\) solution of \(\mathrm{CH}_{3} \mathrm{COOH}\) has \(\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]=3.0 \times 10^{-3} \mathrm{M}\). What are the values of: \(\begin{array}{lr}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)_{0} & = \\\ {\left[\mathrm{CH}_{3} \mathrm{COOH}\right]} & = \\ {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]} & = \\ K_{\mathrm{a}} & =\end{array}\)

Short answer

\(\left[\mathrm{CH}_{3} \mathrm{COOH}\right]_{0} = 0.50 \mathrm{M}\), \(\left[\mathrm{CH}_{3} \mathrm{COOH}\right] = 0.497 \mathrm{M}\), \(\left[\mathrm{H}_{3}\mathrm{O}^{+}\right] = 3.0 \times 10^{-3} \mathrm{M}\), and \(K_{a} = 1.81 \times 10^{-5}\)

Step-by-step solution

Determine the Initial Acetic Acid Concentration

The exercise provides the initial concentration of \(\mathrm{CH}_{3}\mathrm{COOH}\), which is 0.50 M. Therefore, \(\left[\mathrm{CH}_{3} \mathrm{COOH}\right]_{0} = 0.50 \mathrm{M}\).

Calculating the Final Concentration of Acetic Acid

Since acetic acid donates a proton to become \(\mathrm{CH}_{3}\mathrm{COO}^{-}\), the amount of acetic acid will decrease by the same amount as the increase in \(\mathrm{CH}_{3}\mathrm{COO}^{-}\) concentration. As per the details provided in the exercise, \(\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right] = 3.0 \times 10^{-3} \mathrm{M}\). So after the reaction, the concentration of \(\mathrm{CH}_{3}\mathrm{COOH}\) will be the initial concentration minus the increase in \(\mathrm{CH}_{3}\mathrm{COO}^{-}\), which results in: \(\left[\mathrm{CH}_{3} \mathrm{COOH}\right] = 0.50 \mathrm{M} - 3.0 \times 10^{-3} \mathrm{M} = 0.497 \mathrm{M}\).

Calculating the Hydronium Ion Concentration

The hydronium ion (\(\mathrm{H_{3}O}^{+}\)) concentration will be equal to \(\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]\) at equilibrium since one acetic acid molecule donates a proton to become one \(\mathrm{CH}_{3}\mathrm{COO}^{-}\) and one \(\mathrm{H_{3}O}^{+}\). Therefore, \(\left[\mathrm{H_{3}O}^{+}\right] = 3.0 \times 10^{-3} \mathrm{M}\).

Calculating the Ka Value

The Ka can be calculated using the equilibrium concentrations from the previous steps. The equation Ka = [\(\mathrm{CH}_{3}\mathrm{COO}^{-}\)][\(\mathrm{H_{3}O}^{+}\)] / [\(\mathrm{CH}_{3}\mathrm{COOH}\)] is used. Substituting the given values, Ka = \(3.0 \times 10^{-3} \mathrm{M}\) * \(3.0 \times 10^{-3} \mathrm{M}\) / 0.497 M, gives a Ka value of \(1.81 \times 10^{-5}\).

Key concepts

Acetic Acid Equilibrium

The concept of acetic acid equilibrium is fundamental for students working with weak acids in chemistry. At equilibrium, acetic acid (\( CH_3COOH \)), a weak organic acid, partially dissociates into acetate ions (\( CH_3COO^- \)) and hydronium ions (\( H_3O^+ \)).

The dissociation can be represented by the equation: \[ CH_3COOH \rightleftharpoons CH_3COO^- + H_3O^+ \]
This reaction reaches a state where the rate of the forward reaction (acetic acid dissociating) equals the rate of the reverse reaction (acetate and hydronium ions recombining). The extent of this reaction is quantified by the acid dissociation constant, or Ka. Understanding the equilibrium concept is critical, as it lays the foundation for predicting the behavior of acids and bases in solution and is essential for various applications, including buffer solutions and titrations.

Chemistry Education

In chemistry education, it's pivotal to approach complex topics, such as acid-base equilibria, in an accessible manner, ensuring that foundational concepts are well understood before moving on to intricate applications. A solid educational strategy might involve breaking down the principle of acid equilibrium into observable steps and explaining the mathematical reasoning behind each step with clarity.

For instance, using visual aids and interactive models can significantly improve the learning process. Students should be encouraged to practice determination of concentrations and pH, engage in lab experiments, and apply their knowledge to real-world scenarios, fostering a comprehensive understanding of the subject matter.

Chemical Concentration Calculation

The chemical concentration calculation plays a central role in understanding chemical reactions in solution. In the context of acid-base reactions, it involves determining the molar concentration of all species at equilibrium.

When calculating concentrations, keeping track of the moles of reactants and products is essential. In our exercise example, we started with a 0.50 M solution of acetic acid and determined the concentration of acetate ions to be 3.0 x 10^-3 M. This figure is key to finding the remaining acetic acid and hydronium ion concentrations. Such calculations are not only critical for academic purposes but also in practical applications like pharmaceuticals, environmental monitoring, and industrial processes. Being adept at these calculations ensures proper reagent mixing and the ability to predict reaction outcomes.

Ka Value Determination

Determining the Ka value, which stands for the acid dissociation constant, helps us understand the strength of an acid in solution. It is a ratio that compares the product of the concentrations of the dissociated species (in this case, acetate and hydronium ions) to the concentration of the undissociated species (acetic acid).

The equation for Ka is given by: \[ Ka = \frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]} \] Using the concentrations derived from the balanced chemical equation and the stoichiometry of the dissociation, we can calculate the Ka. For example, if the concentration of \( CH_3COO^- \) and \( H_3O^+ \) is each 3.0 x 10^-3 M and the concentration of \( CH_3COOH \) is 0.497 M, the Ka is calculated to be 1.81 x 10^-5. This calculation is essential for chemists to predict the extent of a reaction and the resulting pH of the solution, which has vital applications across different fields, from industrial chemistry to biochemistry.