Question

A 0.0560 -g quantity of acetic acid is dissolved in enough water to make \(50.0 \mathrm{~mL}\) of solution. Calculate the concentrations of \(\mathrm{H}^{+}, \mathrm{CH}_{3} \mathrm{COO}^{-},\) and \(\mathrm{CH}_{3} \mathrm{COOH}\) at equilibrium. \(\left(K_{\mathrm{a}}\right.\) for acetic acid \(=\) \(\left.1.8 \times 10^{-5} .\right)\)

Short answer

The equilibrium concentrations of \( H^{+} \), \( CH_{3}COO^{-} \), and \( CH_{3}COOH \) are \( 7.57 * 10^{-4} M \), \( 7.57 * 10^{-4} M \), and \( ~0.0178 M \), respectively.

Step-by-step solution

Identifying the Ionization Reaction and Its Equilibrium Expression

The ionization reaction of \( CH_{3}COOH \) (acetic acid) in water is given by: \( CH_{3}COOH \leftrightarrow H^{+} + CH_{3}COO^{-} \). The equilibrium expression for this reaction in terms of concentrations is given by the law of mass action as: \( K_{a} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} \), where \( [X] \) indicates the molar concentration of \( X \).

Calculating Initial Concentration of Acetic Acid

The initial molar concentration of acetic acid before any ionization occurs can be calculated using its mass and the volume of the solution. As molarity = moles/volume, the initial concentration of acetic acid, [\( CH_{3}COOH_{i} \)], is = 0.0560 g / 60.052 g/mol (molar mass of acetic acid) / 0.050 L (volume of solution in liters) = 0.0186 M.

Applying the ICE Method and Solving for Equilibrium Concentrations

The ICE method refers to Initial, Change, Equilibrium, which helps in keeping track of the concentrations at various stages of the reaction. Applying ICE method, \([CH_{3}COOH]_{eq} = [CH_{3}COOH_{i}] - x = 0.0186 - x \), where \( x \) is the amount of \( CH_{3}COOH \) ionized into \( H^{+} \) and \( CH_{3}COO^{-} \). So, \([H^{+}]_{eq} = [CH_{3}COO^{-}]_{eq} = x \). Substituting these into the equation for \( K_{a} \), we get \( 1.8 * 10^{-5} = \frac{x * x}{0.0186 - x} \). Solving the above quadratic equation (assuming \( x^2 << 0.0186x \), which can be justified due to the smallness of K), one obtains \( x = 7.57 * 10^{-4} \). Therefore, the equilibrium concentrations of \( H^{+} \), \( CH_{3}COO^{-} \), and \( CH_{3}COOH \) are \( 7.57 * 10^{-4} M \), \( 7.57 * 10^{-4} M \), and \( 0.0186 - 7.57 * 10^{-4} M \) respectively.

Key concepts

Ionization Reaction

Ionization reactions are fundamental in chemistry, especially when dealing with weak acids such as acetic acid (\( \text{CH}_3\text{COOH} \)). In the realm of chemistry, ionization is the process in which a molecule dissociates into ions when dissolved in water.
For acetic acid, the ionization reaction is represented as:

• \[\text{CH}_3\text{COOH} \leftrightarrow \text{H}^+ + \text{CH}_3\text{COO}^-\]

Here, the acetic acid molecule partially dissociates, producing hydrogen ions (\( \text{H}^+ \)) and acetate ions (\( \text{CH}_3\text{COO}^- \)). This equilibrium determines how the ions and the undissociated molecules coexist in the solution.

Recognizing and writing down the ionization reaction is the first step in understanding the acid's behavior in water. This reaction involves the reversible conversion of the acid to its constituents, which sets the stage for calculating equilibrium concentrations.

Equilibrium Expression

The concept of equilibrium is crucial when discussing reactions like the ionization of acetic acid. Once the reaction reaches equilibrium, the rate of ionization matches the rate of recombination.
This scenario allows us to establish an equilibrium expression using the law of mass action. For acetic acid, the equilibrium expression derived from the ionization reaction is:

• \[ K_{a} = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \]

Here, \( K_{a} \) is the acid dissociation constant and provides a measure of how well the acid dissociates in water. The brackets indicate the molar concentrations of the species at equilibrium. The numerator accounts for the products of ionization, whereas the denominator includes the concentration of the un-ionized acid.

Understanding and setting up the equilibrium expression is vital because it lets us calculate the concentrations of all species, given the equilibrium constant \( K_{a} \), which in this case is \( 1.8 \times 10^{-5} \).

ICE Method

The ICE method is a technique used to track the concentration changes during a chemical reaction, especially useful for reactions like the ionization of acetic acid, which reaches a dynamic equilibrium.
The acronym ICE stands for Initial, Change, Equilibrium, describing the states of concentrations throughout the process.
To apply the ICE method:

• Initial concentrations: Start with the concentration of the acetic acid solution before ionization. Based on given data, we initially have \( [\text{CH}_3\text{COOH}] = 0.0186 M \), with zero for both \( [\text{H}^+] \) and \( [\text{CH}_3\text{COO}^-] \).
• Change in concentrations: As the reaction progresses, a small part \( x \) of the acid will ionize. So, \( [\text{CH}_3\text{COOH}] \) decreases by \( x \) while \( [\text{H}^+] \) and \( [\text{CH}_3\text{COO}^-] \) increase by \( x \).
• Equilibrium concentrations: Finally, use these expressions to calculate the equilibrium concentrations: \([\text{CH}_3\text{COOH}] = 0.0186 - x\), \([\text{H}^+] = x\), and \([\text{CH}_3\text{COO}^-] = x\).

Plugging these into the equilibrium expression allows us to solve for \( x \), confirming the system's equilibrium state. This structured approach simplifies our understanding and ensures accuracy in solving for chemical equilibria.