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The problem asks for the **percent dissociation** of a weak acid, specifically **benzoic acid (C₆H₅COOH)**, in a 2.8 mM aqueous solution. To solve this, we need to apply the **concept of percent dissociation**, which measures how much of the acid ionizes in water.

Let me walk you through the necessary steps to calculate the answer.

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### **Formula for Percent Dissociation:**
\[
\% \, \text{dissociation} = \left( \frac{[\text{H}^+]}{[\text{C₆H₅COOH]}_0} \right) \times 100
\]

Where:
- \([H^+]\) is the concentration of hydrogen ions from dissociation.
- \([C₆H₅COOH]_0\) is the **initial concentration** of benzoic acid (in this case, \(2.8 \, \text{mM} = 2.8 \times 10^{-3} \, \text{M}\)).

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### **Step-by-Step Solution:**

1. **Look up the \(K_a\) value of benzoic acid:**
   The **acid dissociation constant** (\(K_a\)) of benzoic acid is typically around:
   \[ K_a = 6.3 \times 10^{-5} \]

2. **Set up the equilibrium expression:**

   For the dissociation of benzoic acid:
   \[ C₆H₅COOH \, \rightleftharpoons \, C₆H₅COO^- + H^+ \]

   Using the equilibrium expression:
   \[ K_a = \frac{[H^+]^2}{[C₆H₅COOH]_0 - [H^+]} \]

   Let \(x\) be the concentration of hydrogen ions dissociated. This gives:
   \[ K_a = \frac{x^2}{2.8 \times 10^{-3} - x} \]

3. **Make an approximation:**
   Since \(K_a\) is small, we assume \(x \ll 2.8 \times 10^{-3}\). Thus:
   \[ K_a \approx \frac{x^2}{2.8 \times 10^{-3}} \]

4. **Solve for \(x\):**
   \[ x^2 = K_a \times 2.8 \times 10^{-3} \]
   \[ x^2 = (6.3 \times 10^{-5})(2.8 \times 10^{-3}) \]
   \[ x^2 = 1.764 \times 10^{-7} \]
   \[ x = \sqrt{1.764 \times 10^{-7}} \]
   \[ x = 1.33 \times 10^{-4} \, \text{M} \]

5. **Calculate percent dissociation:**
   \[
   \% \, \text{dissociation} = \left( \frac{1.33 \times 10^{-4}}{2.8 \times 10^{-3}} \right) \times 100
   \]
   \[
   \% \, \text{dissociation} = (0.0475) \times 100 = 4.75\%
   \]

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### **Final Answer:**
The **percent dissociation** of benzoic acid in the 2.8 mM solution is approximately:
\[
\boxed{4.8\%}
\]
(rounded to 2 significant digits as required).

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Would you like further clarification on any step?  

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Here are 5 follow-up questions:
1. How do weak acids differ from strong acids in terms of dissociation?
2. What assumptions allow us to simplify equilibrium calculations for weak acids?
3. How does the concentration of the acid affect percent dissociation?
4. What are the units of the acid dissociation constant \(K_a\), and how does it relate to pKa?
5. How would adding a common ion (like sodium benzoate) affect the percent dissociation?

**Tip:** Approximations such as \(x \ll C_0\) are helpful for weak acids, but always double-check if the assumption holds after solving.

In a 2.8 mM aqueous solution of benzoic acid (C₆H₅COOH), what is the percentage of benzoic acid that is dissociated? Round your answer to 2 significant digits.

Learn how to calculate the percent dissociation of benzoic acid in a 2.8 mM solution using the equilibrium expression and acid dissociation constant (Ka). This detailed solution walks you through step-by-step with key concepts and formulas. Ideal for undergraduate chemistry students.

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