Problem 59 - Chemistry The Molecular Science

Question

(a) Calculate the pH of a 0.050-M solution of benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} ; K_{\mathrm{a}}=1.2 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) (b) Calculate the percent of the acid that has ionized in this solution.

Step-by-Step Solution

Answer

(a) pH is 2.61; (b) 4.90% of the acid has ionized.

Step 1: Write the Ionization Equation

Benzoic acid ionizes in water according to the equation: \[ \text{C}_6\text{H}_5\text{COOH} \rightleftharpoons \text{C}_6\text{H}_5\text{COO}^- + \text{H}^+ \]

Step 2: Set Up the Expression for Ka

The acid dissociation constant \( K_a \) is given by \[ K_a = \frac{[\text{C}_6\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_6\text{H}_5\text{COOH}]} \]Given that \( K_a = 1.2 \times 10^{-4} \).

Step 3: Define Initial and Equilibrium Concentrations

Define the initial concentration of benzoic acid as \( [\text{C}_6\text{H}_5\text{COOH}] = 0.050 \) M, and assume \( x \) is the amount that ionizes. At equilibrium:- \([\text{C}_6\text{H}_5\text{COOH}] = 0.050 - x \)- \([\text{C}_6\text{H}_5\text{COO}^-] = x \)- \([\text{H}^+] = x \)

Step 4: Substitute Equilibrium Concentrations into Ka Expression

Substitute these values into the \( K_a \) expression:\[ 1.2 \times 10^{-4} = \frac{x \times x}{0.050 - x} \approx \frac{x^2}{0.050} \] Assuming \( x \) is small, simplify and solve for \( x \).

Step 5: Solve for x

Solving the simplified equation:\[ 1.2 \times 10^{-4} \times 0.050 = x^2 \]\[ 6 \times 10^{-6} = x^2 \]\[ x = \sqrt{6 \times 10^{-6}} \] \[ x \approx 0.00245 \] Thus, \( x = [\text{H}^+] \approx 0.00245 \) M.

Step 6: Calculate the pH

The pH is calculated using the formula:\[pH = -\log[\text{H}^+] = -\log(0.00245) \]\[ pH \approx 2.61 \]

Step 7: Calculate Percent Ionization

Percent ionization is calculated as:\[ \text{Percent Ionization} = \left(\frac{[\text{H}^+]}{0.050}\right) \times 100\% = \left(\frac{0.00245}{0.050}\right) \times 100\% \]\[ \text{Percent Ionization} \approx 4.90\% \]

Key Concepts

pH Calculation

Calculating the pH of a solution helps us understand how acidic or basic it is. pH is a measure of the hydrogen ion concentration \( [\text{H}^+] \) in a solution.
The scale ranges from 0 to 14:

• pH less than 7 is acidic
• pH equal to 7 is neutral
• pH greater than 7 is basic

To find the pH of benzoic acid, we first calculate \( x \), representing \( [\text{H}^+] \) at equilibrium. Using the formula: \[ pH = -\log[\text{H}^+] \] Substitute \( [\text{H}^+] = 0.00245 \) M: \[ pH = -\log(0.00245) approx 2.61 \] Now, we have a clear idea about the acidity of the 0.050 M benzoic acid solution.

Ionization of Acids

Ionization refers to how an acid dissociates into ions in water. With benzoic acid, this is key to understanding its behavior in solution.
The ionization equation is:\[ \text{C}_6\text{H}_5\text{COOH} \rightleftharpoons \text{C}_6\text{H}_5\text{COO}^- + \text{H}^+ \] This process affects the concentration of ions in the solution, impacting the pH.
To calculate the percent ionization, use: \[ \text{Percent Ionization} = \left(\frac{[\text{H}^+]}{0.050}\right) \times 100\% \] Substituting the known values: \[ \text{Percent Ionization} = \left(\frac{0.00245}{0.050}\right) \times 100\% approx 4.90\% \] This indicates the extent of ionization for benzoic acid, informing us about its strength as a weak acid.

Acid Dissociation Constant (Ka)

The acid dissociation constant, \( K_a \), is a vital part of acid-base chemistry.
It tells us how much an acid dissociates in a solution. Strong acids have large \( K_a \) values, while weak acids have smaller ones.
For benzoic acid: \[ K_a = \frac{[\text{C}_6\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_6\text{H}_5\text{COOH}]} \] Given \( K_a = 1.2 \times 10^{-4} \), which is typical for a weak acid.
We used this constant to find the hydrogen ion concentration, assuming minimal ionization \( (x) \). This simplification often helps in solving equilibrium problems efficiently.
The small \( K_a \) value reflects that benzoic acid doesn’t fully ionize, supporting our percent ionization findings.