Question

Use the information provided in the table to determine the equilibrium constant \(K\) for these weak acid solutions. $$\begin{array}{|c|c|c|c|}\hline \text{ Weak acid solution } & \text{ \(\left[\mathrm{H}^{+}\right]\) } & \text{ \(K\) } \\ \hline \text{ 0.10 M chloroacetic acid } & \text{ 0.0118 M } & \text{ } \\ \hline \text{ 0.10 M formic acid } & \text{ 0.00424 M } & \text{ } \\ \hline \text{ 0.10 M acetic acid } & \text{ 0.00134 M } & \text{} \\ \hline \text{ 0.050 M acetic acid0 } & \text{ 0.000949 M }& \text{ }\\\ \hline\end{array}$$

Short answer

Chloroacetic acid: \(1.58 \times 10^{-3}\), Formic acid: \(1.88 \times 10^{-4}\), Acetic acid (0.10 M): \(1.82 \times 10^{-5}\), Acetic acid (0.050 M): \(1.83 \times 10^{-5}\).

Step-by-step solution

Understand the Formula for the Equilibrium Constant

The equilibrium constant for a weak acid, denoted as \( K_a \), is defined by the expression \( K_a = \frac{[H^+][A^-]}{[HA]} \), where \([H^+]\) is the concentration of hydrogen ions, \([A^-]\) is the concentration of the conjugate base, and \([HA]\) is the initial concentration of the weak acid minus \([H^+]\).

Set up the Equation for Chloroacetic Acid

For chloroacetic acid with a concentration of 0.10 M and \([H^+] = 0.0118 M\), assume \([A^-] = [H^+] = 0.0118 M\). Thus, \([HA] = 0.10 - 0.0118\). Plug these values into \( K_a \): \[ K_a = \frac{(0.0118)(0.0118)}{0.10 - 0.0118} \]

Calculate the Equilibrium Constant for Chloroacetic Acid

Calculate \( [HA] = 0.10 - 0.0118 = 0.0882 \) M, and substitute back into the equation: \[ K_a = \frac{(0.0118)(0.0118)}{0.0882} = 1.58 \times 10^{-3} \]

Set up the Equation for Formic Acid

For formic acid with \([H^+] = 0.00424 M\), assume \([A^-] = 0.00424 M\) and \([HA] = 0.10 - 0.00424\). Substitute into \( K_a \): \[ K_a = \frac{(0.00424)(0.00424)}{0.10 - 0.00424} \]

Calculate the Equilibrium Constant for Formic Acid

Calculate \( [HA] = 0.10 - 0.00424 = 0.09576 \) M, and substitute: \[ K_a = \frac{(0.00424)(0.00424)}{0.09576} = 1.88 \times 10^{-4} \]

Set up the Equation for Acetic Acid (0.10 M)

For acetic acid (0.10 M) with \([H^+] = 0.00134 M\), set \([A^-] = 0.00134 M\) and \([HA] = 0.10 - 0.00134\). Substitute into \( K_a \): \[ K_a = \frac{(0.00134)(0.00134)}{0.10 - 0.00134} \]

Calculate the Equilibrium Constant for Acetic Acid (0.10 M)

Calculate \( [HA] = 0.10 - 0.00134 = 0.09866 \) M, and substitute: \[ K_a = \frac{(0.00134)(0.00134)}{0.09866} = 1.82 \times 10^{-5} \]

Set up the Equation for Acetic Acid (0.050 M)

For acetic acid (0.050 M) with \([H^+] = 0.000949 M\), set \([A^-] = 0.000949 M\) and \([HA] = 0.050 - 0.000949\). Substitute into \( K_a \): \[ K_a = \frac{(0.000949)(0.000949)}{0.050 - 0.000949} \]

Calculate the Equilibrium Constant for Acetic Acid (0.050 M)

Calculate \( [HA] = 0.050 - 0.000949 = 0.049051 \) M, and substitute: \[ K_a = \frac{(0.000949)(0.000949)}{0.049051} = 1.83 \times 10^{-5} \]

Key concepts

Weak Acids

Weak acids are fascinating because they do not completely ionize in water. This means that a significant percentage of these acids remain "intact" even when dissolved. Unlike strong acids, where virtually all the molecules dissociate into ions, weak acids exist in a balance between the non-ionized acid and the ions it produces.
So, why does this matter? Understanding weak acids is crucial for learning about pH, buffer solutions, and many real-world applications like food chemistry and drugs. For instance, acetic acid in vinegar is a weak acid, making vinegar less harsh and more useful in cooking and cleaning.
Key characteristics of weak acids include:

• Partial ionization in solution.
• Presence of an equilibrium between the undissociated molecules and ions.
• Typically lower values of the acid dissociation constant, or \(K_a\).

Recognizing these properties helps in comprehending how weak acids behave in various chemical reactions and solutions.

Chemical Equilibria

Chemical equilibria are at the core of understanding how reactions progress in closed systems. When a reaction reaches equilibrium, the rates of the forward and backward reactions are equal, so the concentrations of reactants and products remain constant over time.
The fascinating thing about equilibria in the context of weak acids is the dynamic balance they maintain between ionization and recombination. This means that even after the system reaches equilibrium, individual molecules of acid and its ions continuously convert back and forth.
Consider this analogy: it's like a bustling train station where people constantly enter and exit trains, but the number of people in the station stays the same. This dynamic nature is what makes equilibria so intriguing.
To navigate chemical equilibria effectively:

• Understand the concept of dynamic balance.
• Use the equilibrium constant (\(K\)) to evaluate the extent of the reaction.
• Remember that changes in conditions (like concentration, temperature, or pressure) can disturb the equilibrium.

Knowing these aspects helps predict and control chemical reactions, essential in fields ranging from environmental science to materials engineering.

Acid Dissociation Constant

The acid dissociation constant, denoted as \(K_a\), provides valuable insight into the strength of an acid. For weak acids, \(K_a\) measures how well the acid ionizes in solution, telling us the equilibrium concentration of products over reactants.
A typical scenario involves calculating \(K_a\) with the format: \[ K_a = \frac{[H^+][A^-]}{[HA]} \] Here:

• \([H^+]\) represents the concentration of hydrogen ions.
• \([A^-]\) indicates the concentration of the conjugate base.
• \([HA]\) is the concentration of the unionized acid.

The value of \(K_a\) helps chemists determine an acid's ionization degree under specific conditions. Smaller \(K_a\) values indicate weaker acids, meaning the equilibrium heavily favors the undissociated form.
Learning about \(K_a\) equips you to tackle problems in analytical chemistry and understand how acids behave in physiological, industrial, and environmental contexts.