Question

An aqueous solution contains a mixture of 0.0500\(M \mathrm{HCOOH}\) \(\left(K_{\mathrm{a}}=1.77 \times 10^{-4}\right)\) and 0.150\(M \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\left(K_{\mathrm{a}}=\right.\) \(1.34 \times 10^{-5} ) .\) Calculate the \(\mathrm{pH}\) of this solution. Because both acids are of comparable strength, the \(\mathrm{H}^{+}\) contribution from both acids must be considered.

Short answer

The pH of the aqueous solution containing both 0.0500 M formic acid and 0.150 M propanoic acid is approximately 2.48. This was determined by calculating the H+ ion contributions from both acids, considering their comparable strength, and using their respective Ka values and initial concentrations.

Step-by-step solution

Identify the equilibrium expressions for both acids

Write the equilibrium expressions for both acids, using their respective dissociation constants (Ka): For formic acid (HCOOH): \( K_{a1} = \dfrac{[\mathrm{H}^{+}][\mathrm{HCOO}^{-}]}{[\mathrm{HCOOH}]} \) For propanoic acid (CH3CH2COOH): \( K_{a2} = \dfrac{[\mathrm{H}^{+}][\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{COO}^{-}]}{[\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{COOH}]} \)

Set up ICE tables for both acids

For each acid, set up an Initial-Change-Equilibrium (ICE) table to find the equilibrium concentrations of H+ ions. We will denote the H+ ion contribution from formic acid as x, and the contribution from propanoic acid as y. Formic Acid: Initial: [H+]=x, [HCOO-]=0, [HCOOH]=0.0500-x, Change: -x, +x, -x, Equilibrium: 0, x, 0.0500-x Propanoic Acid: Initial: [H+]=y, [CH3CH2COO-]=0, [CH3CH2COOH]=0.150-y Change: -y, +y, -y, Equilibrium: 0, y, 0.150-y Now we will use these equilibrium concentrations in the Ka expressions to set up two equations to later solve for x and y.

Set up the equilibrium constant equations for both acids

Insert the equilibrium concentrations from the ICE tables into the Ka expressions for both acids: For formic acid: \( K_{a1} = 1.77 \times 10^{-4} = \dfrac{x \cdot x}{0.0500 - x} \) For propanoic acid: \( K_{a2} = 1.34 \times 10^{-5} = \dfrac{y \cdot y}{0.150 - y} \)

Solve for x and y, the H+ ion contributions from both acids

Solve each equilibrium constant equation for the H+ ion contributions (x and y) from both acids. For simpler problems, the quadratic formula could be used. In this case, as both acids have a comparable strength, we will solve them using numerical methods or a calculator, obtaining: x ≈ 2.85 × 10^(-3) M y ≈ 4.84 × 10^(-4) M

Calculate the total H+ ion concentration and pH

Now that we have the H+ ions' contributions from both acids, we can find the total H+ concentration in the solution by adding x and y: [H+] = x + y ≈ 2.85 × 10^(-3) M + 4.84 × 10^(-4) M ≈ 3.33 × 10^(-3) M Now, we can find the pH of the solution using the formula: \( \mathrm{pH} = - \log [\mathrm{H}^{+}] \) pH ≈ -log(3.33 × 10^(-3)) ≈ 2.48

Key concepts

Acid Dissociation Constant (Ka)

The acid dissociation constant, represented as \( K_a \), is a crucial parameter for understanding the strength of an acid in solution. It tells us how well an acid can donate protons (\( H^{+} \) ions) to the solution, indicating the degree of ionization of the acid. A larger \( K_a \) value means that the acid is stronger, as it dissociates more readily to release \( H^{+} \) ions. Conversely, a smaller \( K_a \) value corresponds to a weaker acid, which does not dissociate as much.

For instance, we have two acids: formic acid with \( K_{a1} = 1.77 \times 10^{-4} \) and propanoic acid with \( K_{a2} = 1.34 \times 10^{-5} \). These values suggest that formic acid is slightly stronger than propanoic acid, as it has a higher \( K_a \). However, both acids are weak compared to strong acids like hydrochloric acid, which have very large \( K_a \) values.

Understanding the \( K_a \) allows chemists to predict chemical behavior in equilibrium reactions.
It helps in calculating pH, a measure of acidity, through equilibrium expressions.

Equilibrium Expressions

Equilibrium expressions are mathematical representations that describe the concentrations of reactants and products at equilibrium for a chemical reaction.
They are derived from the law of mass action, which states that at equilibrium, the ratio of the concentration of products to reactants is constant for a given temperature.

For acids in aqueous solutions, the equilibrium expressions can be written using the acid dissociation constant \( K_a \). For example, for formic acid (HCOOH), the equilibrium expression is:
\( K_{a1} = \frac{[\mathrm{H}^{+}][\mathrm{HCOO}^{-}]}{[\mathrm{HCOOH}]} \)

This formula shows the relationship between the hydrogen ions, the conjugate base (\( \mathrm{HCOO}^{-} \)), and the undissociated acid (HCOOH). It provides the necessary framework to calculate the concentration of hydrogen ions at equilibrium, crucial for determining the pH.

Similarly, the equilibrium expression for propanoic acid (CH₃CH₂COOH) would be:
\( K_{a2} = \frac{[\mathrm{H}^{+}][\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{COO}^{-}]}{[\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{COOH}]} \)

These equations provide a systematic approach to solving the concentration contributions of \( \mathrm{H}^{+} \) ions and are essential while setting up ICE tables.

Initial-Change-Equilibrium (ICE) Table

The Initial-Change-Equilibrium (ICE) table is a valuable tool for tracking the concentration changes during a chemical reaction as it progresses to equilibrium. It provides a structured approach to understand how concentrations of reactants and products change.

Here's how you can use an ICE table for acid dissociation:

• Initial: Start by writing the initial concentrations of all species involved in the reaction. For example, for formic acid, you would note that initially, the concentration of \( \mathrm{HCOOH} \) is 0.0500 M and \( \mathrm{H}^{+} \, \text{and} \, \mathrm{HCOO}^{-} \) are zero.
• Change: Describe the changes in concentration as the reaction moves towards equilibrium. Generally, let \( x \) represent the change in \( \mathrm{H}^{+} \) ion concentration from formic acid, illustrating how the species concentrations are affected.
• Equilibrium: Determine the concentrations of each species when the reaction reaches equilibrium. Use these to set up the equilibrium expressions. For formic acid, those could be \( [\mathrm{H}^{+}] = x \) and \( [\mathrm{HCOOH}] = 0.0500 - x \).

By applying this table, we can convert the chemical problem into mathematical expressions that show how to solve for unknowns, such as \( x \), the concentration of \( \mathrm{H}^{+} \). In the end, this lets us calculate the pH of the solution, a key measure of its acidity.