Question

What are the major species present in 0.250 \(M\) solutions of each of the following acids? Calculate the pH of each of these solutions. a. \(\mathrm{HNO}_{2}\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

Short answer

In a 0.250 M solution of HNO2, the major species present are HNO2, its conjugate base NO2-, and water (H2O) molecules. The pH of this solution is approximately 1.35. In a 0.250 M solution of CH3CO2H (HC2H3O2), the major species present are CH3CO2H (HC2H3O2), its conjugate base CH3CO2- (C2H3O2-), and water (H2O) molecules. The pH of this solution is approximately 2.63.

Step-by-step solution

Identifying Major Species of HNO2

HNO2, known as nitrous acid, is a weak acid. In a 0.250 M solution of HNO2, the major species present will be HNO2 itself, along with its conjugate base NO2-, and water (H2O) molecules.

Identifying Major Species of CH3CO2H (HC2H3O2)

CH3CO2H (HC2H3O2), known as acetic acid, is also a weak acid. In a 0.250 M solution, the major species present will be CH3CO2H (HC2H3O2) itself, along with its conjugate base CH3CO2- (C2H3O2-), and water (H2O) molecules. Now, let's find the pH for each solution.

Calculate pH for HNO2 solution

First, we need the Ka value for HNO2. The Ka value for HNO2 is \(4.50 \times 10^{-4}\). Now, set up the equilibrium expression: \(Ka = \frac{[H^{+}][NO_{2}^{-}]}{[HNO_{2}]}\) Since the problem states the initial concentration of HNO2 is 0.250 M, we can use the following concentrations changes at equilibrium: - HNO2: 0.250 - x - H+: x - NO2-: x Plugging in the values, we get: \(4.50 \times 10^{-4} = \frac{x^2}{0.250-x}\) Solve for x to find the hydrogen ion concentration, [H+]. Then, calculate the pH as follows: \(pH = -\log[H^{+}]\)

Calculate pH for CH3CO2H (HC2H3O2) solution

For acetic acid, the Ka value is \(1.76 \times 10^{-5}\). Set up the equilibrium expression: \(Ka = \frac{[H^{+}][CH_{3}CO_{2}^{-}]}{[CH_{3}CO_{2}H]}\) Since the initial concentration of CH3CO2H (HC2H3O2) is 0.250 M, we can use the following concentrations changes at equilibrium: - CH3CO2H (HC2H3O2): 0.250 - x - H+: x - CH3CO2- (C2H3O2-): x Plugging in the values, we get: \(1.76 \times 10^{-5} = \frac{x^2}{0.250-x}\) Solve for x to find the hydrogen ions concentration, [H+]. Then, calculate the pH as follows: \(pH = -\log[H^{+}]\) Once you've calculated the [H+] for each solution and found the pH values, the problem is solved.

Key concepts

Weak Acids

Weak acids are a fundamental concept in acid-base equilibria, crucial for understanding how substances dissociate and interact in water. Unlike strong acids, which completely dissociate into ions in solution, weak acids only partially dissociate.
This means that in a solution of a weak acid, both the undissociated form of the acid and its dissociated ions are present. For instance, in a solution of nitrous acid (HNO2) or acetic acid (CH3CO2H, also known as HC2H3O2), most of the acid molecules remain in the undissociated form.
Only a small portion ionizes, producing hydrogen ions (H+) and the conjugate base (NO2- for HNO2 or CH3CO2- for CH3CO2H). This partial dissociation is what characterizes weak acids.
The extent of dissociation is determined by the acid dissociation constant, known as Ka, which quantifies the strength of the acid. The smaller the Ka, the weaker the acid, indicating less dissociation. In summary, understanding weak acids involves recognizing that these acids do not fully ionize in solution, which has profound effects on their chemical behavior and the pH of the solutions they form.

pH Calculation

Calculating the pH of a weak acid solution is a central skill in chemistry, providing insights into the acidity of a solution.
The pH is defined as the negative logarithm of the hydrogen ion concentration: \( pH = -\log[H^+] \). Because weak acids do not fully dissociate, calculating their pH involves using an equilibrium expression.The steps to calculate pH start with the acid's equilibrium expression, such as:

• For HNO2: \( Ka = \frac{[H^+][NO_2^-]}{[HNO_2]} \)
• For CH3CO2H: \( Ka = \frac{[H^+][CH_3CO_2^-]}{[CH_3CO_2H]} \)

Next, you assume an initial concentration (often given) and then denote the change in concentration from dissociation as \( x \).
You write the equilibrium concentration in terms of \( x \) and solve for \( x \) using the known Ka value. Solving this gives the concentration of H+, which can be used to find the pH.
For weak acids like HNO2 and CH3CO2H, this method accurately determines the pH by accounting for the partial dissociation that occurs.

Equilibrium Expressions

Equilibrium expressions are mathematical equations used to describe the balance between products and reactants in a reversible reaction.
They are indispensable in calculating the extent of reactions and determining the concentrations of each species involved.In the context of weak acids, the equilibrium expression is built around the acid dissociation constant, \( Ka \).
This constant combines the concentrations of the dissociated ions and the undissociated acid at equilibrium, representing how far a reaction proceeds.For a weak acid like HNO2, the equilibrium expression is:\[ Ka = \frac{[H^+][NO_2^-]}{[HNO_2]} \]Similarly, for acetic acid (CH3CO2H):\[ Ka = \frac{[H^+][CH_3CO_2^-]}{[CH_3CO_2H]} \]In practice, setting up the equilibrium expression involves using initial concentrations and equilibrium changes (represented by \( x \)), allowing you to solve for \( x \) and use that to find equilibrium concentrations.
This helps chemists calculate the specific quantities of ions and molecules present, which is crucial for understanding the reaction's behavior and properties at equilibrium.
Mastering equilibrium expressions is thus essential for anyone studying acid-base chemistry.