Problem 190 - Chemistry An Atoms First Approach

Question

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

Step-by-Step Solution

Answer

The pH of the aqueous solution containing 0.0500 $M$ HCOOH and 0.150 $M$ CH3CH2COOH is approximately 2.44.

Step 1: Write the ionization reactions for both acids

The ionization reactions for each acid are as follows: For HCOOH: \(HCOOH \rightleftharpoons H^+ + HCOO^-\) For CH3CH2COOH: \(CH_{3}CH_{2}COOH \rightleftharpoons H^+ + CH_{3}CH_{2}COO^-\)

Step 2: Set up the initial and equilibrium concentrations for both reactions

Let x = [H+] contributed by HCOOH Let y = [H+] contributed by CH3CH2COOH | HCOOH | H+ | HCOO- | Initial| 0.0500| 0 | 0 | Change | - x | +x | + x | Final |0.0500-x| x | x | | CH3CH2COOH | H+ | CH3CH2COO- | Initial| 0.150 | 0 | 0 | Change | -y | +y | +y | Final | 0.150-y | y | y |

Step 3: Write the Ka expressions for both acids

For HCOOH: \(K_{a1} = 1.77 \times 10^{-4} = \frac{x^2}{0.0500 - x}\) For CH3CH2COOH: \(K_{a2} = 1.34 \times 10^{-5} = \frac{y^2}{0.150 - y}\)

Step 4: Use the quadratic formula to find the equilibrium [H+] concentration

Solve for x and y using the quadratic formulas, such as the quadratic equation in terms of x: \(x^2 + 1.77 \times 10^{-4}x - 8.85 \times 10^{-6} = 0\) Solve for x, resulting in: \(x \approx 3.30 \times 10^{-3}\) Similarly, solve for y, resulting in: \(y \approx 3.56 \times 10^{-4}\) Sum the contributions of [H+] from both acids: \(x + y = 3.3 \times 10^{-3} + 3.56 \times 10^{-4} \approx 3.66 \times 10^{-3}\)

Step 5: Calculate the pH

Now, we can find the pH using the final [H+] concentration: \(pH = - \log{([H^+])} = - \log{(3.66 \times 10^{-3})} \approx 2.44\) Therefore, the pH of the solution is approximately 2.44.

Key Concepts

Ionization Reaction

Chemical ionization reactions play a crucial role in understanding how acidic compounds release protons (H+) into a solution. An ionization reaction occurs when an acid donates its proton to the surrounding water, transforming itself into its conjugate base. An example can be observed in the ionization of formic acid (HCOOH) and propionic acid (CH3CH2COOH).

• In the case of formic acid, the reaction occurs as: \( \text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^- \)
• For propionic acid, the equation looks like: \( \text{CH}_3\text{CH}_2\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{CH}_2\text{COO}^- \)

These reactions lead to the production of hydrogen ions and are essential for understanding the pH of an acidic solution. By identifying each acid's ability to ionize, one can estimate its contribution to the overall acidity.

Equilibrium Concentration

Equilibrium concentration is crucial in determining the concentrations of substances when a chemical reaction has reached a state of balance. At equilibrium, the rate at which reactants transform into products equals the rate of the reverse process.

• Initially, acids such as HCOOH and CH3CH2COOH start with predefined concentrations and no ionized forms.
• As ionization proceeds, these acids reach equilibrium, where concentrations of H+, HCOO-, and CH3CH2COO- stabilize.

By deriving these equilibrium concentrations, one can measure acidity by determining how much acid dissociates at equilibrium. For example:

• The change in concentration is measured by variables \( x \) and \( y \) for HCOOH and CH3CH2COOH, respectively.
• These changes reflect protons released, affecting the solution's pH.

Understanding this concept is crucial for calculating the exact pH of the resulting solution.

Acid Dissociation Constant (Ka)

The acid dissociation constant, \( K_a \), measures an acid's strength by its ability to donate protons. A larger \( K_a \) value indicates a stronger acid, characterized by a greater tendency to lose its proton.
To illustrate, consider the chemical equilibrium of formic and propionic acids:

• For HCOOH, \( K_{a1} = 1.77 \times 10^{-4} \).
• For Ch3CH2COOH, \( K_{a2} = 1.34 \times 10^{-5} \).

These constants allow us to establish equations that quantify how much of each acid ionizes:

• \( K_{a1} = \frac{x^2}{0.0500 - x} \)
• \( K_{a2} = \frac{y^2}{0.150 - y} \)

These equations are pivotal for exploring each compound's ionization and effectively enabling calculations of equilibrium states. Understanding \( K_a \) facilitates predicting reaction behaviors and solving for proton concentrations, cementing the foundation for accurate pH calculations.

Quadratic Formula

The quadratic formula is a mathematical tool essential for solving complex equations that result from equilibrium expressions in chemistry. When an equilibrium expression yields a quadratic equation, the quadratic formula becomes useful:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, coefficients \( a \), \( b \), and \( c \) derive from rearranging the quadratic equation form: \( ax^2 + bx + c = 0 \).

• For formic acid's equilibrium, the equation is \( x^2 + 1.77 \times 10^{-4}x - 8.85 \times 10^{-6} = 0 \) yielding \( x \approx 3.30 \times 10^{-3} \).
• Similarly, solving for propionic acid, we compute \( y \approx 3.56 \times 10^{-4} \).

Applying this formula enables precise determination of how much each acid contributes to the H+ concentration. Its perfect applicability in equilibrium systems highlights its value, making it an indispensable method in chemistry.

H+ Concentration

The concentration of H+ ions directly influences a solution's pH, providing insight into its acidic or basic nature. To find the overall H+ concentration:
Consider individual contributions from each acid in a mixed solution:

• HCOOH contributes \( x = 3.30 \times 10^{-3} \) M.
• CH3CH2COOH contributes \( y = 3.56 \times 10^{-4} \) M.

Add these contributions to find the total H+ in the solution:
\( x + y = 3.30 \times 10^{-3} + 3.56 \times 10^{-4} = 3.66 \times 10^{-3} \) M.
Finally, link this concentration to pH by the equation:

• \( pH = - \log([H^+]) = - \log(3.66 \times 10^{-3}) \approx 2.44 \).

Understanding H+ concentration is key to evaluating the solution's acidity. It reflects how protons contributed by acids affect the pH, forming a core aspect of understanding solutions in chemistry.